These notes synthesize material from multiple sources: Ben Webster’s PMATH 446/646 lectures at the University of Waterloo, David Eisenbud’s Commutative Algebra: with a View Toward Algebraic Geometry (Springer GTM 150, the primary textbook), Mel Hochster’s commutative algebra lecture notes (University of Michigan), Pete Clark’s commutative algebra notes (University of Georgia), and MIT OpenCourseWare 18.705. All enrichment attempts to preserve the spirit of Webster’s course.

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Chapter 1: Rings and Modules

Section 1.1: Motivation and Background

Commutative algebra sits at the confluence of two grand traditions in mathematics. On one side stands number theory, which studies the ring \(\mathbb{Z}\) and the more general rings of integers arising from algebraic number fields — rings like \(\mathbb{Z}[i]\), \(\mathbb{Z}[\sqrt{-5}]\), and \(\mathbb{Z}[\zeta_p]\) that arise when you try to factor rational integers using algebraic numbers. On the other side stands algebraic geometry, which encodes the solution sets of polynomial systems in algebraic structure.

Concretely, if \(f_1, \dots, f_k \in \mathbb{C}[x_1, \dots, x_n]\) and \(X = \{\mathbf{a} \in \mathbb{C}^n \mid f_i(\mathbf{a}) = 0 \text{ for all } i\}\), then the geometry of \(X\) is completely governed by the coordinate ring \(\mathbb{C}[X] = \mathbb{C}[x_1, \dots, x_n]/(f_1, \dots, f_k)\). Questions about dimension, smoothness, irreducibility, and intersections all translate into purely algebraic statements about this ring. This duality between geometry and algebra is the central theme of the course.

Why should a geometer care about algebra? Here is a concrete example. Consider the parabola \(X = V(y - x^2) \subset \mathbb{C}^2\) and the cusp \(Y = V(y^2 - x^3) \subset \mathbb{C}^2\). Both are smooth curves (the parabola obviously so; the cusp has a singularity at the origin). Their coordinate rings are \(\mathbb{C}[X] = \mathbb{C}[x, y]/(y - x^2) \cong \mathbb{C}[x]\) (a polynomial ring — a PID, hence in particular a UFD) and \(\mathbb{C}[Y] = \mathbb{C}[x,y]/(y^2 - x^3) \cong \mathbb{C}[t^2, t^3] \subset \mathbb{C}[t]\) (not integrally closed — \(t = y/x\) is integral over \(\mathbb{C}[Y]\) but does not lie in it). The singularity of \(Y\) at the origin is reflected in the failure of \(\mathbb{C}[Y]\) to be integrally closed, and the integral closure \(\mathbb{C}[t]\) is the coordinate ring of the normalization (i.e., the desingularization) of \(Y\). Algebraic properties of rings directly mirror geometric properties of varieties.

Throughout, every ring \(R\) is commutative with a multiplicative identity \(1 = 1_R\), and ring homomorphisms are required to send \(1\) to \(1\). Recall the hierarchy: fields \(\subset\) PIDs \(\subset\) UFDs \(\subset\) integral domains \(\subset\) rings. An ideal \(P \subsetneq R\) is prime if \(xy \in P \Rightarrow x \in P\) or \(y \in P\) (equivalently \(R/P\) is a domain), and maximal if there is no ideal strictly between \(P\) and \(R\) (equivalently \(R/\mathfrak{m}\) is a field). Every maximal ideal is prime; Zorn’s lemma guarantees that every proper ideal is contained in a maximal ideal.

The key examples to keep in mind throughout the course are: \(\mathbb{Z}\) (the archetypal Dedekind domain), \(k[x_1,\dots,x_n]\) (the coordinate ring of affine \(n\)-space), \(k[x_1,\dots,x_n]/(f_1,\dots,f_r)\) (the coordinate ring of a variety), and \(k[[x_1,\dots,x_n]]\) (the local ring of formal power series, capturing infinitesimal geometry near a point). As the course develops, we will see how each chapter of the theory illuminates properties of these examples.

Section 1.2: Modules

Before defining modules, let us explain why they matter. In linear algebra, we study vector spaces — that is, modules over a field. The power of linear algebra comes from the fact that over a field, every module (vector space) is free (has a basis). The study of modules over more general rings is a measure of how far a ring is from being a field. Modules appear everywhere: abelian groups are \(\mathbb{Z}\)-modules, ideals are \(R\)-submodules of \(R\), and the homology groups of a space are modules over various coefficient rings. In algebraic geometry, vector bundles on a variety correspond to locally free modules over the coordinate ring.

Let us pause on the example of \(\mathbb{Z}[i]\) as a \(\mathbb{Z}\)-module, since it is simultaneously simple and illuminating. The Gaussian integers \(\mathbb{Z}[i] = \{a + bi \mid a, b \in \mathbb{Z}\}\) are a free \(\mathbb{Z}\)-module of rank 2, with basis \(\{1, i\}\): every element is uniquely \(a \cdot 1 + b \cdot i\) for integers \(a, b\). But \(\mathbb{Z}[i]\) is also a ring in its own right, and the multiplication \((a + bi)(c + di) = (ac - bd) + (ad + bc)i\) gives it additional structure. As a \(\mathbb{Z}[i]\)-module, \(\mathbb{Z}[i] \cong \mathbb{Z}[i]^1\) is free of rank 1. The ideal \((1 + i)\mathbb{Z}[i]\) is a \(\mathbb{Z}\)-submodule of rank 2 (since \(\{(1+i), (1+i)i\} = \{1+i, -1+i\}\) is a \(\mathbb{Z}\)-basis) but a \(\mathbb{Z}[i]\)-submodule of rank 1 (generated by \(1+i\)). The quotient \(\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/2\mathbb{Z}\) — as a \(\mathbb{Z}[i]\)-module, it is killed by \(2 = (1+i)(1-i)\).

Another instructive example: the ring \(\mathbb{Z}[x]/(x^2 - 2)\) as a \(\mathbb{Z}\)-module. Every element can be written uniquely as \(a + b\sqrt{2}\) for \(a, b \in \mathbb{Z}\) (where we write \(\sqrt{2}\) for the image of \(x\)), so this is a free \(\mathbb{Z}\)-module of rank 2 with basis \(\{1, \sqrt{2}\}\). Multiplication is determined by \(\sqrt{2} \cdot \sqrt{2} = 2\). As a ring, \(\mathbb{Z}[x]/(x^2 - 2) \cong \mathbb{Z}[\sqrt{2}]\) — a real quadratic integer ring. The distinction between viewing it as a \(\mathbb{Z}\)-module (free of rank 2) versus a \(\mathbb{Z}[\sqrt{2}]\)-module (free of rank 1) illustrates how changing the base ring changes the module structure.

The standard isomorphism theorems hold for modules exactly as for groups and rings: the second isomorphism theorem says \((M + N)/N \cong M/(M \cap N)\), and the third says \((M/L)/(N/L) \cong M/N\) for \(L \subseteq N \subseteq M\).

Generating sets and free modules. A subset \(S \subseteq M\) generates \(M\) if every element of \(M\) is an \(R\)-linear combination of elements of \(S\). A module is finitely generated if it has a finite generating set. A set \(S\) is \(R\)-linearly independent if \(\sum_{s \in S} r_s s = 0\) implies each \(r_s = 0\). A free module is one possessing a basis (a linearly independent generating set). The free \(R\)-module on a set \(X\) is \(R^{(X)} = \bigoplus_{x \in X} R\).

Note how drastically different modules can be from vector spaces: the \(\mathbb{Z}\)-module \(\mathbb{Z}/6\mathbb{Z}\) is finitely generated (by \(1\)) but not free (since \(6 \cdot 1 = 0\) gives a relation). The \(\mathbb{Z}\)-module \(\mathbb{Q}\) is torsion-free (no nonzero element is annihilated by a nonzero integer) but also not free — it is not even finitely generated as a \(\mathbb{Z}\)-module, since any finitely many rationals live in a submodule generated by a single common denominator.

Annihilators. The annihilator of a module \(M\) is the ideal \(\mathrm{Ann}_R(M) = \{r \in R \mid rm = 0 \text{ for all } m \in M\}\). A module is faithful if \(\mathrm{Ann}(M) = 0\). Over \(R/\mathrm{Ann}(M)\), the module \(M\) becomes faithful. For example, \(\mathrm{Ann}_\mathbb{Z}(\mathbb{Z}/n) = (n)\), and \(\mathrm{Ann}_\mathbb{Z}(\mathbb{Q}) = (0)\).

Let us compute \(\mathrm{Ann}_\mathbb{Z}(\mathbb{Z}/6)\) explicitly. An integer \(n\) annihilates \(\mathbb{Z}/6\) if and only if \(n \cdot \bar{m} = \overline{nm} = 0\) in \(\mathbb{Z}/6\) for every \(\bar{m} \in \mathbb{Z}/6\). This means \(6 \mid nm\) for every integer \(m\). Taking \(m = 1\), we need \(6 \mid n\). Conversely, if \(6 \mid n\), then \(n \cdot \bar{m} = \overline{nm} = 0\) since \(6 \mid nm\). So \(\mathrm{Ann}_\mathbb{Z}(\mathbb{Z}/6) = 6\mathbb{Z} = (6)\). More generally, \(\mathrm{Ann}_\mathbb{Z}(\mathbb{Z}/n) = (n)\). The annihilator of the direct sum \(\mathbb{Z}/m \oplus \mathbb{Z}/n\) is \(\mathrm{Ann}(\mathbb{Z}/m) \cap \mathrm{Ann}(\mathbb{Z}/n) = (m) \cap (n) = (\mathrm{lcm}(m,n))\).

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Chapter 2: Direct Sums, Exact Sequences, and PIDs

Section 2.1: Direct Sums and Products

The direct sum and direct product are the fundamental ways to build new modules from old. Their distinction matters only for infinite families, but understanding both is essential for the theory of free modules.

For example, \(\bigoplus_{n \geq 1} \mathbb{Z}\) is the abelian group of sequences of integers that are eventually zero, while \(\prod_{n \geq 1} \mathbb{Z}\) is all sequences of integers. The direct sum is the “right” notion for free modules: the free module on a set \(X\) is \(\bigoplus_{x \in X} R\), not the product.

The rank of a finitely generated free module \(F \cong R^n\) is the integer \(n\). For commutative rings, this is well-defined: any two bases have the same cardinality (proved by tensoring with a residue field). This fails for noncommutative rings — there exist rings with \(R \cong R^2\) as \(R\)-modules.

Universal properties. The direct sum satisfies a universal property with respect to inclusions: given maps \(f_\alpha: M_\alpha \to N\), there is a unique map \(\bigoplus_\alpha M_\alpha \to N\) extending all \(f_\alpha\). The direct product satisfies the dual property with respect to projections: given maps \(g_\alpha: N \to M_\alpha\), there is a unique map \(N \to \prod_\alpha M_\alpha\). This is the categorical coproduct vs. product distinction.

Section 2.2: Exact Sequences

Exact sequences are the fundamental language for expressing relationships between modules. They encode injectivity, surjectivity, and quotients all at once, and they provide the scaffold for homological algebra.

Reading a short exact sequence \(0 \to M'' \to M \to M' \to 0\): the module \(M\) is an “extension” of \(M'\) by \(M''\). It contains \(M''\) as a submodule, and the quotient is \(M'\). The question of how many non-isomorphic such extensions exist is answered by \(\mathrm{Ext}^1_R(M', M'')\), which we will encounter later.

The four fundamental short exact sequences to know:

1. \(0 \to \mathbb{Z} \xrightarrow{\times n} \mathbb{Z} \to \mathbb{Z}/n \to 0\) — expresses \(\mathbb{Z}/n\) as a quotient of \(\mathbb{Z}\).
2. \(0 \to I \to R \to R/I \to 0\) — for any ideal \(I \subseteq R\).
3. \(0 \to \ker f \to M \xrightarrow{f} N \to \mathrm{im} f \to 0\) — the “factored” version of any map.
4. \(0 \to M \to M \oplus N \to N \to 0\) — the split sequence.

Why \(0 \to \mathbb{Z} \xrightarrow{\times 2} \mathbb{Z} \to \mathbb{Z}/2 \to 0\) does not split. Suppose for contradiction there is a section \(\sigma: \mathbb{Z}/2 \to \mathbb{Z}\) with the quotient map composed with \(\sigma\) being the identity. Then \(\sigma(\bar{1})\) must be an odd integer, and \(\sigma(\bar{0}) = \sigma(2 \cdot \bar{1}) = 2 \sigma(\bar{1})\) must equal \(0\) in \(\mathbb{Z}\), so \(\sigma(\bar 1)\) must satisfy \(2\sigma(\bar{1}) = 0\), hence \(\sigma(\bar{1}) = 0\). But then \(\sigma(\bar{1}) = 0\) is even, not odd — contradiction. More conceptually: \(\mathbb{Z}/2\) has a nonzero element of order 2, but \(\mathbb{Z}\) has no such element. A direct sum \(\mathbb{Z} \oplus \mathbb{Z}/2\) would have an element of order 2 in the \(\mathbb{Z}/2\) factor, but the sequence says \(M = \mathbb{Z}\) — and \(\mathbb{Z}\) is torsion-free. So no splitting is possible.

\[ \begin{array}{ccccccccc} 0 & \to & A & \xrightarrow{\alpha} & B & \xrightarrow{\beta} & C & \to & 0 \\ & & \downarrow f & & \downarrow g & & \downarrow h & & \\ 0 & \to & A' & \xrightarrow{\alpha'} & B' & \xrightarrow{\beta'} & C' & \to & 0 \end{array} \]

the Snake Lemma produces a long exact sequence \(0 \to \ker f \to \ker g \to \ker h \xrightarrow{\partial} \mathrm{coker}\, f \to \mathrm{coker}\, g \to \mathrm{coker}\, h \to 0\). The “connecting homomorphism” \(\partial\) is the key non-trivial piece. This is the prototype for the long exact sequences in homology theory.

\[ \begin{array}{ccccccccc} 0 & \to & \mathbb{Z} & \xrightarrow{\times 2} & \mathbb{Z} & \to & \mathbb{Z}/2 & \to & 0 \\ & & \downarrow \times 3 & & \downarrow \times 3 & & \downarrow \bar{3} & & \\ 0 & \to & \mathbb{Z} & \xrightarrow{\times 2} & \mathbb{Z} & \to & \mathbb{Z}/2 & \to & 0 \end{array} \]

where the vertical maps are multiplication by 3. Here \(f = g = \times 3: \mathbb{Z} \to \mathbb{Z}\) and \(h = \bar 3: \mathbb{Z}/2 \to \mathbb{Z}/2\). We compute: \(\ker f = 0\), \(\ker g = 0\), \(\ker h = 0\) (since \(3 \cdot \bar 1 = \bar 3 = \bar 1 \neq 0\) in \(\mathbb{Z}/2\)). The cokernels: \(\mathrm{coker}\, f = \mathbb{Z}/3\), \(\mathrm{coker}\, g = \mathbb{Z}/3\), \(\mathrm{coker}\, h = \mathbb{Z}/2 / (3\mathbb{Z}/2) = \mathbb{Z}/2 / \mathbb{Z}/2 = 0\). The Snake Lemma gives \(0 \to 0 \to 0 \to 0 \xrightarrow{\partial} \mathbb{Z}/3 \to \mathbb{Z}/3 \to 0 \to 0\), reflecting that \(\mathrm{coker}\, f \to \mathrm{coker}\, g\) is the induced map (which is the identity here).

For a less trivial example, take rows \(0 \to \mathbb{Z} \xrightarrow{\times 6} \mathbb{Z} \to \mathbb{Z}/6 \to 0\) and vertical maps \(\times 2\). Then \(\ker(\bar 2 : \mathbb{Z}/6 \to \mathbb{Z}/6) = \{0, 3\} \cong \mathbb{Z}/2\), \(\mathrm{coker}(\bar 2 : \mathbb{Z}/6 \to \mathbb{Z}/6) = \mathbb{Z}/\{0,2,4\} \cong \mathbb{Z}/2\). The connecting homomorphism \(\partial: \ker h \to \mathrm{coker}\, f\) sends \(\bar 3 \mapsto\) (lift \(\bar 3\) to \(\mathbb{Z}\), get \(3\), apply \(g\) to get \(6\), then project to \(\mathrm{coker}\, g = \mathbb{Z}/2\)) giving \(6 \bmod 2 = 0\). So \(\partial(\bar 3) = 0\) in this case.

Projective and injective modules. A module \(P\) is projective if every short exact sequence \(0 \to N \to M \to P \to 0\) splits. Equivalently, \(\mathrm{Hom}(P, -)=\) is exact. Free modules are projective. A module \(I\) is injective if every short exact sequence \(0 \to I \to M \to N \to 0\) splits. Equivalently, \(\mathrm{Hom}(-, I)\) is exact. The module \(\mathbb{Q}\) is injective over \(\mathbb{Z}\), and more generally every divisible abelian group is injective.

Section 2.3: Principal Ideal Domains

In a PID, the divisibility theory is fully tamed: every nonzero element factors uniquely into irreducibles (up to units and reordering), and greatest common divisors exist with Bezout coefficients. This makes the theory of modules over PIDs as clean as possible.

Why \((2, x)\) is not principal in \(\mathbb{Z}[x]\): a complete proof. Suppose \((2, x) = (f)\) for some \(f \in \mathbb{Z}[x]\). Since \(2 \in (f)\), we have \(f \mid 2\) in \(\mathbb{Z}[x]\), so \(f\) is a constant: \(f \in \{\pm 1, \pm 2\}\). If \(f = \pm 1\), then \((f) = \mathbb{Z}[x]\), so \(1 \in (2, x)\), meaning \(1 = 2g(x) + xh(x)\) for some \(g, h \in \mathbb{Z}[x]\). Substituting \(x = 0\): \(1 = 2g(0)\), so \(2 \mid 1\) in \(\mathbb{Z}\) — contradiction. If \(f = \pm 2\), then \((f) \subseteq 2\mathbb{Z}[x]\), but \(x \in (2, x)\) is not divisible by 2 — contradiction. So \((2, x)\) is not principal, confirming \(\mathbb{Z}[x]\) is not a PID.

The key distinguishing property of a PID that makes modules so tractable: given any submodule \(N\) of the free module \(R^n\), we can find a basis of \(R^n\) adapted to \(N\). This is the content of the Smith Normal Form theorem, which we turn to next.

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Chapter 3: Structure Theorem for Modules over a PID

The structure theorem for modules over a PID is one of the most powerful classification results in algebra. It says that the only finitely generated modules over a PID — up to isomorphism — are the “obvious” ones: free parts and cyclic torsion parts. Before we state it, let us compute an example by hand to see what the theorem is predicting.

\[ A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{pmatrix} \]

(whose columns are the generators of \(N\)). Performing integer row and column operations to put \(A\) in Smith Normal Form:

\[ \to \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix}. \]

The Smith Normal Form is \(\mathrm{diag}(1, 2, 3)\) (with \(1 \mid 2 \mid 3\) after reordering). Therefore \(M \cong \mathbb{Z}/1 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/3 \cong \mathbb{Z}/2 \oplus \mathbb{Z}/3 \cong \mathbb{Z}/6\). (We used that \(\mathbb{Z}/1 = 0\) and that \(\mathbb{Z}/2 \oplus \mathbb{Z}/3 \cong \mathbb{Z}/6\) by CRT since \(\gcd(2,3) = 1\).)

\[ \begin{pmatrix} 4 & 6 \\ 2 & 3 \end{pmatrix} \xrightarrow{R_1 \leftarrow R_1 - 2R_2} \begin{pmatrix} 0 & 0 \\ 2 & 3 \end{pmatrix} \xrightarrow{\text{swap}} \begin{pmatrix} 2 & 3 \\ 0 & 0 \end{pmatrix} \xrightarrow{C_2 \leftarrow C_2 - C_1} \begin{pmatrix} 2 & 1 \\ 0 & 0 \end{pmatrix} \xrightarrow{C_1 \leftrightarrow C_2} \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} \xrightarrow{C_2 \leftarrow C_2 - 2C_1} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. \]

The Smith Normal Form is \(\mathrm{diag}(1, 0)\). So the cokernel of this map \(\mathbb{Z}^2 \to \mathbb{Z}^2\) is \(\mathbb{Z}/1 \oplus \mathbb{Z}/0 \cong \mathbb{Z}\) — a free module of rank 1.

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Chapter 4: Tensor Products

Section 4.1: Construction and Universal Property

The tensor product is perhaps the most fundamental construction in commutative algebra and algebraic geometry. At its core, it is the universal way to “bilinearize” a map — to turn a bilinear map \(M \times N \to P\) into a linear map \(M \otimes N \to P\). But its importance goes far beyond this: tensor products are used to change the base ring (scalar extension), to compute fiber products of varieties, and to describe the interaction between different modules.

Before the formal definition, let us think about what \(\mathbb{Z}/m \otimes_\mathbb{Z} \mathbb{Z}/n\) should be. Both modules are cyclic, generated by \(1\). So \(\mathbb{Z}/m \otimes \mathbb{Z}/n\) should be generated by \(\bar{1} \otimes \bar{1}\). What is the annihilator of this generator? We need \(m(\bar{1} \otimes \bar{1}) = (m\bar{1}) \otimes \bar{1} = \bar{0} \otimes \bar{1} = 0\) and \(n(\bar{1} \otimes \bar{1}) = \bar{1} \otimes (n\bar{1}) = \bar{1} \otimes \bar{0} = 0\). So \(\bar{1} \otimes \bar{1}\) is annihilated by both \(m\) and \(n\), hence by \(\gcd(m,n)\). One can check the generator has order exactly \(\gcd(m,n)\), so \(\mathbb{Z}/m \otimes_\mathbb{Z} \mathbb{Z}/n \cong \mathbb{Z}/\gcd(m,n)\).

A warning: not every element of \(M \otimes N\) is a simple tensor. In \(\mathbb{R}^2 \otimes_\mathbb{R} \mathbb{R}^2\), the element \(e_1 \otimes e_1 + e_2 \otimes e_2\) is not a simple tensor (it is the matrix \(\begin{pmatrix}1&0\\0&1\end{pmatrix}\) under the identification \(\mathbb{R}^2 \otimes \mathbb{R}^2 \cong M_{2\times 2}(\mathbb{R})\), and only rank-1 matrices correspond to simple tensors).

The universal property is the “correct” definition of the tensor product — it characterizes \(M \otimes_R N\) up to unique isomorphism without reference to the presentation by generators and relations. This is the preferred point of view in category theory.

\[ \mathbb{Z}/m \xrightarrow{\times n} \mathbb{Z}/m \to \mathbb{Z}/m \otimes_\mathbb{Z} \mathbb{Z}/n \to 0. \]

The map \(\times n: \mathbb{Z}/m \to \mathbb{Z}/m\) sends \(\bar a \mapsto \overline{na}\). Its image is \(\{na \bmod m \mid a \in \mathbb{Z}\} = \gcd(n,m)\mathbb{Z}/m\mathbb{Z} \cong d\mathbb{Z}/m\mathbb{Z}\). Its kernel is \(\{a \mid na \equiv 0 \pmod m\} = (m/d)\mathbb{Z}/m\mathbb{Z}\). The cokernel (which equals \(\mathbb{Z}/m \otimes \mathbb{Z}/n\)) is \(\mathbb{Z}/m / (d\mathbb{Z}/m) \cong \mathbb{Z}/d\). This completes the proof.

\[q \otimes \bar k = \frac{r}{n} \otimes \bar k = r \cdot \frac{1}{n} \otimes \bar k = r \cdot \left(\frac{1}{n} \otimes \bar k\right).\]

But \(\frac{1}{n} \otimes \bar k = \frac{1}{n} \otimes n \cdot \frac{\bar k}{1} = \frac{n}{n} \otimes \frac{\bar k}{1}\) — let us be more careful. We have \(\frac{1}{n} \otimes \bar k = \frac{1}{n} \otimes (n \cdot \overline{k/n})\) if \(n \mid k\)… Actually the clean argument is: \(q \otimes \bar k = \frac{q}{n} \cdot n \otimes \bar k = \frac{q}{n} \otimes n\bar k = \frac{q}{n} \otimes \bar 0 = 0\), since \(n \bar k = 0\) in \(\mathbb{Z}/n\). Here we used that we can always divide \(q\) by \(n\) in \(\mathbb{Q}\). Every simple tensor is 0, so the module is 0.

Section 4.2: Functorial Properties

\[ \left(\bigoplus_\alpha M_\alpha\right) \otimes_R N \cong \bigoplus_\alpha (M_\alpha \otimes_R N). \]

The formula \((R/I) \otimes_R M \cong M/IM\) is used constantly. For example, for a local ring \((R, \mathfrak{m})\) and a finitely generated module \(M\), the module \(M/\mathfrak{m}M \cong (R/\mathfrak{m}) \otimes_R M\) is a vector space over the residue field — this is the fiber of \(M\) at the closed point of \(\mathrm{Spec}(R)\). Nakayama’s lemma says this vector space controls the minimal number of generators of \(M\).

\[ 0 \to \mathrm{Tor}_1(\mathbb{Z}/m, \mathbb{Z}/n) \to \mathbb{Z}/n \xrightarrow{\times m} \mathbb{Z}/n \to \mathbb{Z}/m \otimes \mathbb{Z}/n \to 0. \]

The kernel of \(\times m : \mathbb{Z}/n \to \mathbb{Z}/n\) is \(\{a \in \mathbb{Z}/n \mid ma \equiv 0 \pmod n\} = (n/\gcd(m,n))\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/\gcd(m,n)\). So \(\mathrm{Tor}_1^\mathbb{Z}(\mathbb{Z}/m, \mathbb{Z}/n) \cong \mathbb{Z}/\gcd(m,n)\).

This is the same as \(\mathbb{Z}/m \otimes \mathbb{Z}/n\) — no coincidence! By symmetry of \(\mathrm{Tor}\), \(\mathrm{Tor}_1(M, N) \cong \mathrm{Tor}_1(N, M)\).

Section 4.3: Tensor-Hom Adjunction

This adjunction encodes a fundamental asymmetry: \(\otimes\) is right exact and \(\mathrm{Hom}\) is left exact. Their higher derived functors are \(\mathrm{Tor}\) and \(\mathrm{Ext}\), respectively. The relationship between them (via e.g. Universal Coefficient Theorems) is a cornerstone of homological algebra.

Algebras and base change. An \(R\)-algebra is a ring \(S\) together with a ring homomorphism \(\alpha: R \to S\). Given an \(R\)-module \(M\), the scalar extension or base change to \(S\) is the \(S\)-module \(S \otimes_R M\). The tensor product \(A \otimes_R B\) of two \(R\)-algebras is itself an \(R\)-algebra via \((a_1 \otimes b_1)(a_2 \otimes b_2) = (a_1 a_2) \otimes (b_1 b_2)\).

Geometric interpretation. If \(X = \mathrm{Spec}(A)\) and \(Y = \mathrm{Spec}(B)\) are affine \(k\)-varieties, then \(A \otimes_k B\) is the coordinate ring of the product variety \(X \times_k Y = \mathrm{Spec}(A \otimes_k B)\). The fiber of \(X \to \mathrm{Spec}(k')\) over a field extension \(k'/k\) is \(X_{k'} = \mathrm{Spec}(A \otimes_k k')\). This explains why tensor products appear whenever you change the base field or take fiber products of varieties.

Example: base change and splitting behavior. The variety \(X = V(x^2 + y^2 - 1) \subset \mathbb{R}^2\) is a circle, which is irreducible over \(\mathbb{R}\) — its coordinate ring \(\mathbb{R}[x,y]/(x^2 + y^2 - 1)\) is a domain. But over \(\mathbb{C}\), we can factor \(x^2 + y^2 = (x+iy)(x-iy)\), so the base change to \(\mathbb{C}\) gives \(\mathbb{C}[x,y]/(x^2+y^2-1)\), which is still a domain (the circle is irreducible over \(\mathbb{C}\)). In contrast, the variety \(V(x^2 - 2)\) over \(\mathbb{Q}\) is irreducible, but over \(\mathbb{Q}(\sqrt{2})\) it splits into two points: \(\mathbb{Q}(\sqrt{2})[x]/(x^2 - 2) \cong \mathbb{Q}(\sqrt{2})[x]/((x - \sqrt{2})(x+\sqrt{2})) \cong \mathbb{Q}(\sqrt{2}) \times \mathbb{Q}(\sqrt{2})\).

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Chapter 5: Localization

Section 5.1: Construction

Localization is the algebraic counterpart of “zooming in” on part of a geometric space. If \(R = k[x,y]\) is the coordinate ring of the affine plane \(\mathbb{A}^2\), then localizing at a prime \(P = (x,y)\) (the origin) gives the local ring \(R_{(x,y)}\) whose elements are rational functions \(f/g\) with \(g(0,0) \neq 0\) — functions defined in a neighborhood of the origin. This is the germ of the structure sheaf at that point.

More generally, if we invert all elements outside a prime \(P\), we are focusing attention on the “behavior near \(P\).” If we invert a single element \(f\), we get functions defined wherever \(f \neq 0\) — the principal open set \(D(f) = \mathrm{Spec}(R) \setminus V(f)\).

Three fundamental examples to build intuition. Before the definition, let us work through three key examples.

(a) Inverting powers of 2. Let \(R = \mathbb{Z}\) and \(S = \{1, 2, 4, 8, \ldots\} = \{2^n \mid n \geq 0\}\). Then \(S^{-1}\mathbb{Z} = \mathbb{Z}[1/2] = \{a/2^n \mid a \in \mathbb{Z}, n \geq 0\}\). This ring contains all fractions whose denominator is a power of 2. For example, \(1/2, 3/4, 7/8 \in \mathbb{Z}[1/2]\), but \(1/3 \notin \mathbb{Z}[1/2]\). The prime ideals of \(\mathbb{Z}[1/2]\) correspond to primes of \(\mathbb{Z}\) not meeting \(S = \{2^n\}\), i.e., odd primes \((3), (5), (7), \ldots\) and the zero ideal. The prime \((2)\) has “disappeared” — it is no longer a prime in \(\mathbb{Z}[1/2]\) because 2 is now a unit!

(b) Localizing at a prime. Let \(R = \mathbb{Z}\) and \(P = (p)\) for a prime \(p\). Set \(S = \mathbb{Z} \setminus (p) = \{n \in \mathbb{Z} \mid p \nmid n\}\). Then \(S^{-1}\mathbb{Z} = \mathbb{Z}_{(p)} = \{a/b \in \mathbb{Q} \mid p \nmid b\}\). This ring consists of all rationals with denominator prime-to-\(p\). For \(p = 5\): \(\mathbb{Z}_{(5)}\) contains \(1/2, 3/7, 6/11\) but not \(1/5\) or \(3/25\). The ring \(\mathbb{Z}_{(5)}\) is a local ring with unique maximal ideal \((5)\mathbb{Z}_{(5)} = \{a/b \mid 5 \mid a, 5 \nmid b\}\) and residue field \(\mathbb{Z}_{(5)}/(5) \cong \mathbb{F}_5\). Geometrically, localizing at a prime “zooms in” on the closed point \((5) \in \mathrm{Spec}(\mathbb{Z})\).

(c) The coordinate ring example. For \(R = k[x]\) and \(f = x^2 - 1 = (x-1)(x+1)\), the localization \(R_f = k[x, (x^2-1)^{-1}]\) consists of rational functions \(p(x)/q(x)\) where the denominator \(q(x)\) is a polynomial in powers of \(x^2-1\). Geometrically, \(\mathrm{Spec}(R_f) = \mathbb{A}^1 \setminus \{1, -1\}\) — the affine line with the two points \(\pm 1\) removed.

Note the subtlety in the equivalence relation: in a domain, \((r_1, s_1) \sim (r_2, s_2) \iff r_1 s_2 = r_2 s_1\) (the factor \(t\) is not needed). But in a general ring with zero-divisors, we need the factor \(t\) to make transitivity work. For example, in \(\mathbb{Z}/6\mathbb{Z}\) with \(S = \{1, 3\}\), we have \(1/3 = 3/(3 \cdot 3) = 3/9 = 3/3\) since \(3(1 \cdot 3 - 3 \cdot 1) = 0\).

Let us work through a more concrete localization computation. Consider \(R = k[x,y]/(xy)\), the coordinate ring of the union of the two axes \(V(xy) = \{x\text{-axis}\} \cup \{y\text{-axis}\} \subset \mathbb{A}^2\). The prime ideals of \(R\) include \(\mathfrak{p}_1 = (x)/(xy)\) (corresponding to the \(y\)-axis) and \(\mathfrak{p}_2 = (y)/(xy)\) (corresponding to the \(x\)-axis), and the maximal ideals \(\mathfrak{m}_a = (x, y-a)\) for \(a \neq 0\) (points on the \(x\)-axis other than the origin), \(\mathfrak{m}_b = (x-b, y)\) for \(b \neq 0\) (points on the \(y\)-axis), and \(\mathfrak{m}_0 = (x,y)\) (the origin). Localizing at \(\mathfrak{p}_1 = (x)/(xy)\): the elements not in \(\mathfrak{p}_1\) include all polynomials with nonzero constant term mod \(xy\), and \(R_{\mathfrak{p}_1} \cong (k[y])_{(0)} = k(y)\)… actually let us think more carefully. In \(R = k[x,y]/(xy)\), the element \(y\) is not in \(\mathfrak{p}_1 = (x)/(xy)\), so \(y/1\) is a unit in \(R_{\mathfrak{p}_1}\). We have \(xy = 0\) in \(R\), so in \(R_{\mathfrak{p}_1}\), \(x \cdot y/1 = 0\) and \(y\) is a unit, forcing \(x = 0\). So \(R_{\mathfrak{p}_1} \cong k[y]_{(0)} = k(y)\). The localization has “forgotten” the \(x\)-axis component — it only sees the \(y\)-axis.

Section 5.2: Properties of Localization

Ideals in localizations. The ideals of \(S^{-1}R\) are exactly the ideals of the form \(S^{-1}I = \{r/s \mid r \in I, s \in S\}\) for ideals \(I \subseteq R\) with \(I \cap S = \emptyset\). The correspondence \(I \leftrightarrow S^{-1}I\) restricts to a bijection between prime ideals of \(S^{-1}R\) and prime ideals of \(R\) disjoint from \(S\). In particular, primes of \(R_P\) correspond exactly to primes of \(R\) contained in \(P\). This is the algebraic version of the intuition that the local ring at a point sees only the subvarieties passing through that point.

Localized modules. For an \(R\)-module \(M\), define \(S^{-1}M = S^{-1}R \otimes_R M\), equivalently equivalence classes of pairs \((m, s) \in M \times S\) under \((m_1, s_1) \sim (m_2, s_2) \iff t(s_2 m_1 - s_1 m_2) = 0\) for some \(t \in S\). Write \(M_P = R_P \otimes_R M\).

The proof of exactness is not hard: injectivity of \(S^{-1}M' \to S^{-1}M\) follows because if \(m'/s\) maps to \(0\), then \(m'/1 = 0\) in \(S^{-1}M\), meaning \(tm' = 0\) for some \(t \in S\), and since \(M' \hookrightarrow M\) is injective, \(tm' = 0\) in \(M'\) too, so \(m'/s = 0\) in \(S^{-1}M'\). Flatness follows from exactness.

This theorem formalizes the idea that ring-theoretic properties can be checked locally. It is the algebraic analogue of the fact that a sheaf is determined by its stalks. The local-to-global principle is used constantly: to show a map is injective, it suffices to show all its localizations are injective. To show a module is zero, it suffices to show it vanishes at every prime.

Geometric interpretation of localization, revisited. Let us summarize the geometric dictionary more explicitly. Suppose \(X = \mathrm{Spec}(k[x_1,\ldots,x_n]/I)\) is an affine variety with coordinate ring \(A\). Then:

• A point \(p \in X\) corresponds to a maximal ideal \(\mathfrak{m}_p \subset A\). The localization \(A_{\mathfrak{m}_p}\) is the local ring of \(X\) at \(p\), whose elements are “germs of regular functions at \(p\).” The maximal ideal \(\mathfrak{m}_p A_{\mathfrak{m}_p}\) is the “functions vanishing at \(p\).” The residue field \(A_{\mathfrak{m}_p}/\mathfrak{m}_p \cong k\) is the field of constants.
• An irreducible subvariety \(Y \subset X\) corresponds to a prime \(P \subset A\). The localization \(A_P\) is the local ring along \(Y\), whose elements are germs of functions defined on (an open subset of) \(Y\).
• The localization \(A_f = A[f^{-1}]\) corresponds to the principal open set \(D(f) = X \setminus V(f)\), the part of \(X\) where \(f \neq 0\).

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Chapter 6: Local Rings — Jacobson Radical and Nakayama’s Lemma

Section 6.1: The Jacobson Radical

A local ring \((R, \mathfrak{m})\) is a ring with a unique maximal ideal \(\mathfrak{m}\). Such rings arise naturally as localizations \(R_P\) of a ring \(R\) at a prime \(P\), as completions \(\hat{R}\) of a local ring, and as formal power series rings \(k[[x_1, \dots, x_n]]\). Working locally is a recurring strategy in commutative algebra: many questions are easier over a local ring, and by the local-to-global principle, local results glue together to give global results.

The Jacobson radical captures “how non-local” a ring is — it is the intersection of all maximal ideals, and it contains precisely those elements that are “small” in the sense that \(1 - ax\) is invertible for all \(a\).

Proof of the characterization. If \(x \in J(R)\) and \(1 - ax\) is not a unit, then \(1 - ax\) lies in some maximal ideal \(\mathfrak{m}\). But then \(x \in J(R) \subseteq \mathfrak{m}\), so \(ax \in \mathfrak{m}\), giving \(1 = (1 - ax) + ax \in \mathfrak{m}\) — contradiction. Conversely, if \(x \notin \mathfrak{m}\) for some maximal ideal \(\mathfrak{m}\), then \((x, \mathfrak{m}) = R\), so \(ax + m = 1\) for some \(a \in R\), \(m \in \mathfrak{m}\), giving \(1 - ax = m \in \mathfrak{m}\), a non-unit.

Computing the Jacobson radical. For \(R = \mathbb{Z}\), the maximal ideals are \((p)\) for each prime \(p\), so \(J(\mathbb{Z}) = \bigcap_p (p) = (0)\). For \(R = k[x]\), similarly \(J(k[x]) = (0)\). For the local ring \(R = k[[x]]\) (formal power series), the unique maximal ideal is \((x)\), so \(J(k[[x]]) = (x)\). For \(R = \mathbb{Z}/6\mathbb{Z}\), the maximal ideals are \((2)/(6)\) and \((3)/(6)\), so \(J(\mathbb{Z}/6) = (2) \cap (3) / (6) = (6)/(6) = 0\)… wait: the Jacobson radical of \(\mathbb{Z}/6\) is the intersection of its maximal ideals: \((2)\mathbb{Z}/6\mathbb{Z} \cap (3)\mathbb{Z}/6\mathbb{Z} = \{0, 2, 4\} \cap \{0, 3\} = \{0\} = 0\).

For \(R = \mathbb{Z}/12\mathbb{Z}\), the maximal ideals are \((2)/(12)\) and \((3)/(12)\), so \(J(\mathbb{Z}/12) = 2\mathbb{Z}/12 \cap 3\mathbb{Z}/12 = \{0,2,4,6,8,10\} \cap \{0,3,6,9\} = \{0, 6\} = 6\mathbb{Z}/12\mathbb{Z} \cong \mathbb{Z}/2\). Here \(J \neq 0\) but \(J\) is nilpotent: \(6^2 = 36 \equiv 0 \pmod{12}\).

Jacobson rings. A ring \(R\) is a Jacobson ring (or Hilbert ring) if every radical ideal is an intersection of maximal ideals, equivalently if \(\sqrt{I} = \bigcap_{\mathfrak{m} \supseteq I} \mathfrak{m}\) for every ideal \(I\). Examples include \(\mathbb{Z}\), any field, any finitely generated algebra over a field, and polynomial rings over Jacobson rings. The rings \(\mathbb{Z}_{(p)}\) and \(\mathbb{Q}[[x]]\) are not Jacobson (in \(\mathbb{Z}_{(p)}\), the prime \((0)\) is not a maximal ideal but is not an intersection of maximal ideals).

Section 6.2: Nakayama’s Lemma

Nakayama’s Lemma is one of the most powerful tools in commutative algebra, used in hundreds of arguments throughout the subject. At its heart, it says: if a finitely generated module is “killed at the level of the residue field” (i.e., \(IM = M\) for an ideal in the radical), then it is zero. The intuition is that working modulo the radical is like working over the residue field, and over a field, this would give \(M = 0\); Nakayama tells us that over a local ring, the situation is just as good.

Why finiteness is necessary. The module \(M = \bigoplus_{n \geq 0} k\) (infinite direct sum) over \(R = k[x]\) satisfies \(xM = M\) (every element of \(M \oplus 0 \oplus \cdots\) lies in \(xM\) since each summand is in \(xM\)) but \(M \neq 0\). Finiteness of generation is essential.

Corollary (2) is extremely useful: to find generators of a module over a local ring, it suffices to find generators modulo the maximal ideal — that is, to find generators “at the closed point.” For example, if \(R = k[[x,y]]\) and \(M = (x,y)R\), then \(M/\mathfrak{m}M \cong k^2\) (spanned by \(\bar x\) and \(\bar y\)), so \(M\) requires exactly 2 generators — namely \(x\) and \(y\).

Concrete Nakayama application: lifting generators. Let \(R = \mathbb{Z}_p[[t]]\) (the ring of formal power series over the \(p\)-adic integers), and let \(M\) be a finitely generated \(R\)-module. Suppose we know \(M/pM\) as an \(\mathbb{F}_p[[t]]\)-module. If \(M/pM\) is free of rank \(r\) over \(\mathbb{F}_p[[t]]\) with generators \(\bar m_1, \ldots, \bar m_r\), then by Nakayama’s Lemma (applied with \(I = (p) \subset J(R)\)), any lifts \(m_1, \ldots, m_r \in M\) of \(\bar m_1, \ldots, \bar m_r\) generate \(M\) over \(R\). This is the key step in proving that \(M\) itself is free of rank \(r\).

More concretely: suppose \((R, \mathfrak{m}) = (\mathbb{Z}_{(5)}, (5))\) and \(M\) is a finitely generated \(R\)-module such that \(M/5M \cong \mathbb{F}_5^3\) (a 3-dimensional vector space). Then \(M\) is generated by 3 elements over \(R\). To check it is free, we verify the Syzygy: find generators \(m_1, m_2, m_3\) and check that any relation \(a_1 m_1 + a_2 m_2 + a_3 m_3 = 0\) (with \(a_i \in R\)) implies all \(a_i = 0\). If \(M/5M\) is free and the higher torsion \(M/5^k M\) is free for all \(k\) (verifiable inductively), then \(M\) is free.

Application: locally free modules. A finitely generated module \(M\) is locally free of rank \(n\) if \(M_\mathfrak{m} \cong R_\mathfrak{m}^n\) for every maximal ideal \(\mathfrak{m}\). Such modules need not be free globally — the failure of local freeness to imply global freeness is measured by the Picard group. The key example: the ideal \((2, 1 + \sqrt{-5})\) in \(\mathbb{Z}[\sqrt{-5}]\) is locally free of rank 1 (since every localization at a prime is a PID, so every invertible module is free), but is not free as a \(\mathbb{Z}[\sqrt{-5}]\)-module — it represents a nontrivial element of the class group.

Application: lifting idempotents. Over a local ring \((R, \mathfrak{m})\), the only idempotents are \(0\) and \(1\). This is because \(R/\mathfrak{m}\) is a field (idempotents are \(0,1\)) and idempotents lift uniquely from \(R/\mathfrak{m}\) to \(R\) (by Hensel’s lemma, or directly). This implies local rings are indecomposable as rings.

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Chapter 7: Noetherian Rings and Modules

Section 7.1: Definitions and Equivalences

The Noetherian condition is the fundamental finiteness hypothesis in commutative algebra. Without it, pathological things happen: rings can have non-finitely-generated ideals, modules can have complicated ascending chain behavior, and key theorems like the primary decomposition fail. The Noetherian condition is mild enough to be satisfied by all rings arising in practice (polynomial rings, rings of integers, completions), yet strong enough to force powerful structural results.

To see why we need finiteness, consider the ring \(R = \mathbb{Q}[x_1, x_2, x_3, \dots]\) (a polynomial ring in countably many variables). The ascending chain of ideals \((x_1) \subsetneq (x_1, x_2) \subsetneq (x_1, x_2, x_3) \subsetneq \cdots\) never stabilizes — this ring is not Noetherian. The ideal \(I = (x_1, x_2, x_3, \dots)\) is not finitely generated. In contrast, any quotient of \(\mathbb{Q}[x_1, \dots, x_n]\) is Noetherian by the Hilbert Basis Theorem.

Proof of equivalence. (ACC \(\Rightarrow\) Maximum Condition): If a nonempty collection \(\mathcal{C}\) had no maximal element, we could build an infinite strictly ascending chain by starting with any \(I_1 \in \mathcal{C}\) and at each step picking \(I_{n+1} \in \mathcal{C}\) strictly containing \(I_n\) (since \(I_n\) is not maximal). (Maximum Condition \(\Rightarrow\) Finite Generation): Given an ideal \(I\), the collection of all finitely generated subideals of \(I\) has a maximal element \(J = (a_1, \dots, a_n)\). For any \(a \in I\), the ideal \((a_1, \dots, a_n, a)\) is a finitely generated subideal of \(I\) containing \(J\), so equals \(J\) by maximality, meaning \(a \in J\). Thus \(I = J\) is finitely generated. (Finite Generation \(\Rightarrow\) ACC): Given an ascending chain \(I_1 \subseteq I_2 \subseteq \cdots\), the union \(I = \bigcup_n I_n\) is an ideal, hence finitely generated by some \(a_1, \dots, a_k\). Each \(a_i\) lies in some \(I_{n_i}\), so all lie in \(I_N\) for \(N = \max n_i\), giving \(I = I_N = I_{N+1} = \cdots\).

Consequences. The Hilbert Basis Theorem immediately gives us that every ideal in \(k[x_1, \dots, x_n]\) is finitely generated — this is the algebraic content of the fact that every algebraic variety is defined by finitely many polynomial equations. More generally:

Section 7.2: Noetherian Modules and Primary Decomposition Setup

An \(R\)-module \(M\) is Noetherian if every ascending chain of submodules stabilizes, equivalently every submodule of \(M\) is finitely generated. Over a Noetherian ring, every finitely generated module is Noetherian. Short exact sequences preserve Noetherianness: if \(0 \to N \to M \to M/N \to 0\) is exact, then \(M\) is Noetherian if and only if both \(N\) and \(M/N\) are Noetherian.

The equality \(\sqrt{I} = \bigcap_{P \supseteq I} P\) (where the intersection is over prime ideals containing \(I\)) is a fundamental theorem: it says that the radical of an ideal is the intersection of all prime ideals containing it. Geometrically, \(\sqrt{I}\) is the ideal of all polynomial functions vanishing on \(V(I)\), and the primes over \(I\) correspond to the irreducible components of \(V(I)\).

Example: computing radicals explicitly. Consider \(I = (x^2 y, x y^2)\) in \(k[x,y]\). Then \(x^3 = x \cdot x^2 = x \cdot (x^2/y) \cdot y\)… let us be direct. The variety \(V(I)\) is the set where \(x^2 y = 0\) and \(x y^2 = 0\), which means either \(x = 0\) or \(y = 0\) (or both). So \(V(I) = V(x) \cup V(y)\), and the radical is \(\sqrt{I} = I(V(I)) = (x) \cap (y) = (xy)\). Let us verify: \((xy) \subseteq \sqrt{I}\) since \((xy)^2 = x^2 y^2 = y \cdot x^2 y \in I\). And \(\sqrt{I} \subseteq (x) \cap (y)\) since any prime containing \(I\) contains \(x^2 y\) and \(xy^2\), hence contains \(x\) or \(y\), and a prime over \(I\) must contain both components.

Artinian modules and the Fitting lemma. A module \(M\) is Artinian if every descending chain of submodules stabilizes. The Fitting’s Lemma states: if \(M\) is a Noetherian and Artinian module over a ring \(R\), and \(\phi: M \to M\) is an endomorphism, then \(M = \ker(\phi^n) \oplus \mathrm{im}(\phi^n)\) for sufficiently large \(n\). This generalizes the decomposition \(V = \ker(T - \lambda)^\infty \oplus \mathrm{im}(T - \lambda)^\infty\) from linear algebra (Jordan decomposition).

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Chapter 8: The Prime Spectrum and Zariski Topology

Section 8.1: Spec and MaxSpec

The prime spectrum is the bridge between algebra and geometry. To a ring \(R\), we associate the topological space \(\mathrm{Spec}(R)\) whose points are the prime ideals of \(R\). This might seem abstract, but let us verify that this correctly captures geometric intuition:

• For \(R = k[x_1, \dots, x_n]/I\) (the coordinate ring of a variety \(X \subset \mathbb{A}^n\)), the closed points of \(\mathrm{Spec}(R)\) (the maximal ideals) correspond to the points of \(X\) (by the Nullstellensatz). The non-closed points (the non-maximal primes) correspond to irreducible subvarieties: the prime \(P = (f_1, \dots, f_r)\) in \(R\) corresponds to the subvariety \(V(P) \subset X\).
• For \(R = \mathbb{Z}\), the prime spectrum has a closed point for each prime \(p\) (corresponding to the prime \((p)\)) and a generic point \((0)\) that is dense.

The generic point phenomenon. In classical algebraic geometry (using only maximal ideals), the “generic point” of a variety was not a point at all — it was a conceptual tool. In modern algebraic geometry (using \(\mathrm{Spec}\)), the generic point is an actual point of the space, corresponding to the prime \((0)\) in the coordinate ring (when the ring is a domain). The generic point specializes to every other point: the closure of \(\{(0)\}\) in the Zariski topology is the entire spectrum. This is the algebraic counterpart of the statement that a “general” point on a variety carries all the information.

Section 8.2: Zariski Topology

The axioms for closed sets are satisfied:

• \(V((0)) = \mathrm{Spec}(R)\), \(V((1)) = \emptyset\)
• \(V(I) \cup V(J) = V(IJ) = V(I \cap J)\)
• \(\bigcap_\alpha V(I_\alpha) = V\!\left(\sum_\alpha I_\alpha\right)\)

The first equation \(V(I) \cup V(J) = V(IJ)\) deserves a word: \(P \supseteq IJ \iff P \supseteq I\) or \(P \supseteq J\) (by primality), which is why the union of closed sets corresponds to the product of ideals. Geometrically, the union of two varieties \(X = V(I)\) and \(Y = V(J)\) is the variety \(V(I \cap J) = V(IJ)\) (since \(\sqrt{I \cap J} = \sqrt{IJ}\)).

Proof of quasi-compactness. Suppose \(\mathrm{Spec}(R) = \bigcup_\alpha D(f_\alpha)\). Then \(\bigcap_\alpha V(f_\alpha) = \emptyset\), so \(\sum_\alpha (f_\alpha) = R\). Hence \(1 = \sum_{i=1}^n r_i f_{\alpha_i}\) for some finite set, giving \(\mathrm{Spec}(R) = \bigcup_{i=1}^n D(f_{\alpha_i})\).

Proof of irreducibility criterion. \(V(I)\) is irreducible iff whenever \(V(I) \subseteq V(J_1) \cup V(J_2)\) we have \(V(I) \subseteq V(J_1)\) or \(V(I) \subseteq V(J_2)\). This is equivalent to: whenever \(\sqrt{I} \supseteq J_1 J_2\) we have \(\sqrt{I} \supseteq J_1\) or \(\sqrt{I} \supseteq J_2\) — precisely the condition that \(\sqrt{I}\) is prime.

Jacobson rings and MaxSpec. For a Jacobson ring \(R\), the subspace \(\mathrm{MaxSpec}(R) \subset \mathrm{Spec}(R)\) is dense. This means that for any radical ideal \(I\), \(\sqrt{I} = \bigcap_{\mathfrak{m} \supseteq I} \mathfrak{m}\) — maximal ideals see all of the algebraic structure. This is why, for algebraic geometry over algebraically closed fields, one can work entirely with closed points (the classical approach) and recover full information.

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Chapter 9: Integral Extensions and the Nullstellensatz

Section 9.1: Integral Elements

Integrality is the generalization of “algebraic” from fields to rings. Over a field, every element of an algebraic extension satisfies a polynomial equation. Over a ring, an integral element satisfies a monic polynomial equation — the monic condition is crucial, as it ensures that the leading coefficient is 1 and the relation is “tight.”

Integrality governs the relationship between a ring and its overrings in the same way that algebraic closure governs fields. It appears in:

• Number theory: the ring of integers \(\mathcal{O}_K\) is the integral closure of \(\mathbb{Z}\) in a number field \(K\).
• Algebraic geometry: the normalization of a variety is obtained by taking the integral closure of its coordinate ring.
• Going-up/Going-down: the prime ideal structure of an integral extension is closely controlled by the base ring.

Examples. The element \(\sqrt{2} \in \mathbb{R}\) is integral over \(\mathbb{Z}\) (it satisfies \(t^2 - 2 = 0\)), but \(1/2 \in \mathbb{Q}\) is not integral over \(\mathbb{Z}\) (if \((1/2)^n + r_1(1/2)^{n-1} + \cdots + r_n = 0\) with \(r_i \in \mathbb{Z}\), multiplying by \(2^n\) gives \(1 + 2r_1 + 4r_2 + \cdots + 2^n r_n = 0\), so \(1 \equiv 0 \pmod 2\) — contradiction). This confirms that \(\mathbb{Z}\) is integrally closed in \(\mathbb{Q}\).

The element \(i = \sqrt{-1}\) is integral over \(\mathbb{Z}\) (satisfies \(t^2 + 1 = 0\)), so \(\mathbb{Z}[i]\) is contained in the integral closure of \(\mathbb{Z}\) in \(\mathbb{Q}(i)\). In fact \(\mathbb{Z}[i]\) is the full ring of integers \(\mathcal{O}_{\mathbb{Q}(i)}\).

The golden ratio \(\phi = (1+\sqrt{5})/2\). Does \(\phi\) lie in the integral closure of \(\mathbb{Z}\) in \(\mathbb{Q}(\sqrt{5})\)? We compute: \(\phi^2 = (1 + 2\sqrt{5} + 5)/4 = (6 + 2\sqrt{5})/4 = (3+\sqrt{5})/2\). Then \(\phi^2 - \phi - 1 = (3+\sqrt{5})/2 - (1+\sqrt{5})/2 - 1 = (3+\sqrt{5} - 1 - \sqrt{5})/2 - 1 = 1 - 1 = 0\). So \(\phi\) satisfies the monic relation \(t^2 - t - 1 = 0\) over \(\mathbb{Z}\), hence \(\phi\) is integral. This confirms that \(\mathcal{O}_{\mathbb{Q}(\sqrt{5})} = \mathbb{Z}[(1+\sqrt{5})/2]\).

This characterization is extremely useful for proving closure properties.

The integral closure of \(\mathbb{Z}\) in \(\mathbb{Q}(\sqrt{-5})\). The ring \(\mathbb{Z}[\sqrt{-5}]\) is the ring of integers of \(\mathbb{Q}(\sqrt{-5})\) (since \(-5 \equiv 3 \pmod 4\), we have \(\mathcal{O}_K = \mathbb{Z}[\sqrt{-5}]\)). This ring is integrally closed (it is a Dedekind domain) but not a UFD: \(6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})\) are two distinct factorizations into irreducibles. The elements \(2, 3, 1\pm\sqrt{-5}\) are all irreducible in \(\mathbb{Z}[\sqrt{-5}]\) but none are prime (e.g., \(2\) divides \((1+\sqrt{-5})(1-\sqrt{-5}) = 6\) but \(2 \nmid 1 \pm \sqrt{-5}\) in the ring). The failure is measured by the class group \(\mathrm{Cl}(\mathbb{Z}[\sqrt{-5}]) \cong \mathbb{Z}/2\mathbb{Z}\).

Section 9.2: Nullstellensatz

The Nullstellensatz (German: “theorem of zeros”) is the fundamental theorem relating algebra and geometry over algebraically closed fields. It has two forms:

• The Weak Nullstellensatz says that a system of polynomial equations over an algebraically closed field has a solution unless the equations are “trivially inconsistent” (i.e., \(1\) is a combination of them).
• The Strong Nullstellensatz identifies the ideal of all polynomials vanishing on a variety \(V(I)\) as the radical \(\sqrt{I}\).

The strong form implies: the map \(I \mapsto V(I)\) gives a bijection between radical ideals of \(k[x_1, \dots, x_n]\) and algebraic subsets of \(k^n\) (for algebraically closed \(k\)). This is the algebraic geometry dictionary in its most basic form.

An explicit Nullstellensatz computation. Consider \(I = (x^2 - y, y^2 - x) \subset \mathbb{C}[x,y]\). What is \(V(I)\)? We need \(y = x^2\) and \(y^2 = x\). Substituting: \(x^4 = x\), so \(x^4 - x = x(x-1)(x^2+x+1) = 0\). The roots are \(x = 0\) and \(x = 1\) (the other roots involve \(\omega = e^{2\pi i/3}\)). For \(x = 0\): \(y = 0\). For \(x = 1\): \(y = 1\). For \(x = \omega\): \(y = \omega^2\). For \(x = \omega^2\): \(y = \omega^4 = \omega\). So \(V(I) = \{(0,0), (1,1), (\omega, \omega^2), (\omega^2, \omega)\}\), a set of four points.

The radical of \(I\) is \(\sqrt{I} = \mathcal{I}(V(I)) = (x - 0)(x - 1)(x - \omega)(x - \omega^2) \cap \text{(something involving } y\text{)} = (x^4 - x, y - x^2)\)… actually \(\sqrt{I} = (x^4 - x, y - x^2)\). We have \(x^4 - x = x(x-1)(x^2+x+1)\), and the ideal \((x^4 - x, y - x^2)\) is radical since its variety is 4 reduced points. But note that \(I\) itself may not equal \(\sqrt{I}\) — we may need to take the radical. Let’s check: \(y^2 - x = (y-x^2)(y+x^2) + x^4 - x + 2x^2(y-x^2) - 2x^2 y + 2x^4 + ...\) this gets complicated; the point is that \(\sqrt{I}\) corresponds to the 4 points and is a radical ideal.

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Chapter 10: Krull Dimension and Noether Normalization

Section 10.1: Krull Dimension

Dimension is one of the most fundamental invariants of a topological space or algebraic variety. For a variety \(X\), we expect the dimension to measure the “degrees of freedom” — a curve has dimension 1, a surface dimension 2, etc. Krull dimension translates this into purely algebraic language by counting the length of the longest chain of irreducible subvarieties.

Recall the geometric picture: prime ideals of \(k[x_1, \dots, x_n]/I\) correspond to irreducible subvarieties of \(V(I) \subset \mathbb{A}^n\). A chain of primes \(P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_d\) corresponds to a chain of irreducible subvarieties \(V(P_0) \supsetneq V(P_1) \supsetneq \cdots \supsetneq V(P_d)\). The length of the longest chain is the dimension.

A non-trivial example: dimension of \(k[x,y,z]/(xy, xz)\). Consider \(R = k[x,y,z]/(xy, xz)\). What is \(\dim R\)? The variety \(V(xy, xz) = \{x = 0\} \cup \{y = 0, z = 0\}\) decomposes as the plane \(\{x = 0\} \cong \mathbb{A}^2\) (dimension 2) and the line \(\{y = z = 0\} \cong \mathbb{A}^1\) (dimension 1). The dimension of \(R\) equals the maximum of the dimensions of the irreducible components, which is 2.

\[ (0) \subsetneq \overline{(y,z)} \subsetneq \overline{(x, y, z)}, \]

where bars denote images in \(R\). Here \(\overline{(y,z)} = (y,z)/(xy,xz)\) is a prime: in \(R/(y,z) \cong k[x]\) (setting \(y = z = 0\) in \(k[x,y,z]/(xy,xz)\) gives \(k[x]\)), which is a domain, so \((y,z)\) maps to a prime. And \(\overline{(x,y,z)}\) is the maximal ideal corresponding to the origin. The chain has length 2, confirming \(\dim R \geq 2\). To see \(\dim R = 2\), one can use Noether normalization.

Now the irreducible component \(\{x = 0\}\) corresponds to the prime \(\overline{(x)}\) in \(R\), and \(R/\overline{(x)} \cong k[y,z]\), which has dimension 2. This gives a maximal chain \((0) \subsetneq \overline{(x)} \subsetneq \overline{(x,y)} \subsetneq \overline{(x,y,z)}\) of length 3 — but wait, is \(\overline{(x)}\) a prime? In \(R/\overline{(x)} = k[x,y,z]/(xy,xz,x) \cong k[y,z]\), yes, it is. So we have a chain of length 3 starting from \((0)\)! But that would give \(\dim R \geq 3\), contradicting our earlier claim. Let me recheck: \((0) \subsetneq \overline{(x)} \subsetneq \overline{(x,y)} \subsetneq \overline{(x,y,z)}\). Is \(\overline{(x)}\) actually prime? In \(k[x,y,z]/(xy,xz)\), if we mod out by \((x)\), we get \(k[y,z]\), a domain. Yes, \(\overline{(x)}\) is prime. The chain \((0) \subsetneq \overline{(x)} \subsetneq \overline{(x,y)} \subsetneq \overline{(x,y,z)}\) has length 3, showing \(\dim R \geq 3\). But since \(R\) is a quotient of \(k[x,y,z]\) (dimension 3), we have \(\dim R \leq 3\). In fact \(\dim R = 2\) … hold on, let me reconsider. The Krull dimension of a polynomial ring modulo an ideal can be lower than the Krull dimension of the polynomial ring. Here, \(k[x,y,z]/(xy,xz)\) maps the chain \((0) \subset (x) \subset (x,y) \subset (x,y,z)\) from the polynomial ring to a chain of the same length in the quotient (since none of these ideals contain \((xy,xz)\) except at the full ideal). So indeed \(\dim R = 2\) is incorrect — the dimension is actually 2 in terms of the irreducible components, but the chain above has length 3. Actually dimension equals 2: the component \(\{x=0\}\) is 2-dimensional. Let me re-examine.

The variety \(V(xy,xz) \subset \mathbb{A}^3\) has two irreducible components: \(W_1 = V(x)\) (the plane \(x=0\), 2-dimensional) and \(W_2 = V(y,z)\) (the line \(y=z=0\), 1-dimensional). The ring \(R\) is the coordinate ring of this reducible variety. The chains of primes in \(R\) correspond to chains of irreducible subvarieties of \(V(xy,xz)\). The longest chain in \(W_1\) is: a generic point of \(W_1\) (dim 2) $\supset$ a curve in \(W_1\) (dim 1) $\supset$ a point (dim 0), giving a chain of length 2 in \(R\). But the algebraic chain \((0) \subsetneq (x) \subsetneq (x,y) \subsetneq (x,y,z)\) in \(R\) has length 3 — does this mean \(\dim R = 3\)? Yes! Krull dimension counts the length of the longest prime chain, which is 3 here. This is a case where the Krull dimension (3) exceeds the geometric dimension of the irreducible components (max = 2). Actually, \((x)\) in \(R\) is prime of height 1, and \(R/(x) \cong k[y,z]\) has dimension 2, so \(\mathrm{ht}((x)) + \dim(R/(x)) = 1 + 2 = 3 = \dim R\). So \(\dim R = 3\). The original claim of “dim = 2” was incorrect; the Krull dimension of \(k[x,y,z]/(xy,xz)\) is indeed 3 (same as \(k[x,y,z]\), since the quotient does not drop dimension in this case). This is because the irreducible component \(W_1 = V(x)\) is itself a 2-dimensional variety with a full chain of primes.

Section 10.2: Noether Normalization

Noether normalization is the key structure theorem for finitely generated algebras over a field. It says that any such algebra is a finite module over a polynomial subalgebra — that is, every algebraic variety is a “finite cover” of an affine space. This makes the geometry of a variety accessible via the simpler geometry of affine space, and it is the basis for computing Krull dimensions and for the proof of the Nullstellensatz.

Geometric content. Noether normalization says that every \(d\)-dimensional variety \(X\) admits a finite surjective morphism \(X \to \mathbb{A}^d\). The map is finite (finite fibers) but not generally injective — it is a “finite cover” of affine space, ramified over some hypersurfaces. For curves (\(d=1\)), this gives a finite map \(X \to \mathbb{A}^1\), realizing every curve as a branched cover of the line.

An explicit Noether normalization. Consider the twisted cubic \(A = k[t, t^2, t^3] \subset k[t]\). This is a finitely generated \(k\)-algebra, isomorphic to \(k[x,y,z]/(xz-y^2, x^2 w - z, \ldots)\). As a \(k[t]\)-module it is just \(k[t]\) itself; but as a \(k\)-algebra generated by \(t, t^2, t^3\), the Noether normalization says we should find algebraically independent \(z_1 \in A\) such that \(A\) is finite over \(k[z_1]\). Take \(z_1 = t\). Then \(A = k[t]\) is a free \(k[t]\)-module of rank 1, and trivially finite. We get \(d = 1 = \dim A\).

A more interesting example: \(A = k[x,y]/(xy - 1) = k[x, x^{-1}]\). This is the coordinate ring of the hyperbola. The Noether normalization: take \(z_1 = x + y = x + x^{-1}\). Then \(x^2 - z_1 x + 1 = 0\), so \(x\) is integral over \(k[z_1]\). Since \(A = k[x,y]/(xy-1) = k[x, x^{-1}]\) and \(x\) is integral over \(k[z_1]\), the ring \(A\) is finite over \(k[z_1] \cong k[t]\). So \(d = 1\) and the map to \(\mathbb{A}^1\) is given by \((a, a^{-1}) \mapsto a + a^{-1}\), a 2-to-1 map branched over \(\pm 2\).

The formula \(\mathrm{ht}(P) + \dim(k[x_1, \dots, x_n]/P) = n\) is the algebraic version of the fact that a codimension-\(c\) subvariety of \(\mathbb{A}^n\) has dimension \(n - c\).

Section 10.3: Going Up, Lying Over, Incomparability

The theorems in this section describe how prime ideals behave under integral extensions. Geometrically, an integral extension \(R \subseteq S\) corresponds to a finite morphism \(\mathrm{Spec}(S) \to \mathrm{Spec}(R)\). The three theorems describe properties of this finite morphism:

• Lying Over: the map is surjective on closed points (every point of the base has a preimage).
• Going Up: the map is “closed” at the level of specialization (if a point \(P\) specializes to \(P'\), then any lift of \(P\) can be extended to a lift of \(P'\)).
• Incomparability: the fibers are discrete (no two points in the same fiber are comparable under specialization).

Proof of Lying Over. Localizing at \(P\), we may assume \((R, P)\) is local. It suffices to find a prime of \(S\) over \(P\), i.e., a prime \(Q\) with \(PS \cap R = P\) (i.e., \(Q\) doesn’t contract to something smaller). The key point: \(PS \neq S\) (if \(1 \in PS\), then \(1 = \sum_i p_i s_i\), and taking norms gives a contradiction). So \(PS\) is a proper ideal, hence contained in a maximal ideal \(Q\) of \(S\). Since \(S/Q\) is a field integral over \(R/(Q\cap R)\), the subring \(R/(Q \cap R)\) is also a field (by the key lemma from the Nullstellensatz section), so \(Q \cap R\) is maximal in \(R\), hence equals \(P\).

Going Down. For flat extensions (instead of integral ones), an analogous property holds: if \(R \to S\) is flat and \(P \supseteq P'\) are primes in \(R\) with \(Q'\) lying over \(P'\), there exists \(Q \subseteq Q'\) lying over \(P\). Localization maps are flat, which is why they have the Going Down property. Integral extensions need not have Going Down unless \(R\) is integrally closed (this is Cohen-Seidenberg’s Going Down theorem).

Cohen-Seidenberg Going Down. Let \(R \subseteq S\) be an integral extension with \(R\) integrally closed (i.e., normal). If \(P \supseteq P'\) are primes in \(R\) and \(Q\) is a prime of \(S\) over \(P\), then there exists \(Q' \subseteq Q\) with \(Q' \cap R = P'\). The integrality-closed hypothesis on \(R\) is essential: it fails for \(\mathbb{Z} \subset \mathbb{Z}[\sqrt{-3}]\) at the prime \((2)\) above \((2)\) in \(\mathbb{Z}\) if we chose a bad extension.

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Chapter 11: Gelfand–Kirillov Dimension

The Gelfand–Kirillov (GK) dimension is a refined measure of the “size” of an algebra, counting how fast the space of products of generators grows. For commutative algebras, it coincides with the Krull dimension, but GK dimension extends to noncommutative algebras where Krull dimension is harder to define.

Hilbert function. For a Noetherian local ring \((R, \mathfrak{m})\) with residue field \(F = R/\mathfrak{m}\), the Hilbert function \(H(n) = \dim_F(\mathfrak{m}^n / \mathfrak{m}^{n+1})\) is eventually a polynomial in \(n\) of degree \(\mathrm{Kdim}(R) - 1\). The associated graded ring \(\mathrm{gr}_\mathfrak{m}(R) = \bigoplus_{n \geq 0} \mathfrak{m}^n/\mathfrak{m}^{n+1}\) is a finitely generated graded \(F\)-algebra with \(\mathrm{Kdim}(\mathrm{gr}_\mathfrak{m}(R)) = \mathrm{Kdim}(R)\).

Example. For \(R = k[x,y]_{(x,y)}\) (polynomial ring localized at the origin), we have \(\mathfrak{m}^n/\mathfrak{m}^{n+1} \cong k^{n+1}\) (spanned by monomials of degree \(n\)), so \(H(n) = n + 1\), a polynomial of degree 1 = \(\dim R - 1 = 2 - 1\). For the node \(R = k[x,y]/(xy)_{(x,y)}\), we have \(\dim_k \mathfrak{m}^n/\mathfrak{m}^{n+1} = 2\) for all \(n \geq 1\), so \(H(n) = 2\) (constant = degree 0 polynomial, matching \(\dim R = 1\)).

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Chapter 12: Artinian Rings

The Artinian condition is, in a sense, the opposite of the Noetherian condition. A Noetherian ring has no infinite ascending chains; an Artinian ring has no infinite descending chains. The remarkable fact is that for commutative rings, Artinian is strictly stronger than Noetherian: every Artinian ring is Noetherian (but not vice versa), and Artinian commutative rings are exactly the Noetherian rings of dimension zero.

Artinian rings arise naturally as quotients of local rings by powers of the maximal ideal: \(R/\mathfrak{m}^n\) is Artinian for any local Noetherian ring \((R, \mathfrak{m})\). They are the “infinitesimal” objects in algebraic geometry — the scheme \(\mathrm{Spec}(k[x]/(x^n))\) is a “fat point” of multiplicity \(n\), capturing \(n-1\) orders of infinitesimal information.

Example: classifying local Artinian rings. A local Artinian ring \((R, \mathfrak{m})\) satisfies \(\mathfrak{m}^n = 0\) for some \(n\). The simplest examples over a field \(k\):

• \(k[x]/(x^n)\): the “fat point” of order \(n\). The maximal ideal is \((x)\) and \(\mathfrak{m}^i = (x^i)\).
• \(k[x,y]/(x^2, xy, y^2)\): a local ring with \(\mathfrak{m}^2 = 0\) and \(\mathfrak{m}/\mathfrak{m}^2 \cong k^2\).
• \(k[x,y]/(x^2, y^2)\): a local ring with \(\mathfrak{m}^3 = 0\) (since \(\mathfrak{m}^3\) involves degree-3 monomials in \(x,y\), but \(x^2 = y^2 = 0\) kills everything). Actually \(\mathfrak{m} = (x,y)\), \(\mathfrak{m}^2 = (x^2, xy, y^2) = (xy)\), \(\mathfrak{m}^3 = 0\). This ring has \(k\)-dimension 4: basis \(\{1, x, y, xy\}\).

Why Artinian implies Noetherian (but not conversely). The key argument: in an Artinian ring, every prime is maximal (since an Artinian domain is a field). So the Krull dimension is 0. Then the structure theorem decomposes \(R\) into a finite product of local Artinian rings. Each local Artinian ring is Noetherian (it has finite length as a module over itself). A finite product of Noetherian rings is Noetherian.

Artinian modules. An \(R\)-module \(M\) is Artinian if every descending chain of submodules stabilizes. Over a Noetherian ring, a finitely generated module is Artinian if and only if it has finite length (a composition series). The length \(\ell(M)\) of a module of finite length is the number of steps in any composition series; it is additive in short exact sequences: \(\ell(M) = \ell(M') + \ell(M'')\) for \(0 \to M' \to M \to M'' \to 0\).

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Chapter 13: Valuation Rings and Discrete Valuation Rings

Valuation rings arise naturally when we try to measure “how singular” a function is at a point. A discrete valuation on a field \(K\) assigns to each nonzero element an integer measuring its “order of vanishing” (if positive) or “order of pole” (if negative). The valuation ring consists of elements with non-negative valuation — functions that are regular (not poles) at the point.

The central example: for a smooth algebraic curve \(C\) and a point \(p \in C\), the local ring \(\mathcal{O}_{C,p}\) is a discrete valuation ring. The valuation is the order of vanishing at \(p\). Rational functions on \(C\) can be written as \(f = u \cdot t^n\) where \(t\) is a local coordinate (“uniformizer”) at \(p\), \(u\) is a unit in \(\mathcal{O}_{C,p}\), and \(n = \nu_p(f)\) is the order. This is why the theory of DVRs is central to the theory of curves.

The \(p\)-adic valuation in detail. For the prime \(p = 3\), we have:

• \(\nu_3(9) = 2\) (since \(9 = 3^2\))
• \(\nu_3(1/9) = -2\)
• \(\nu_3(18) = \nu_3(2 \cdot 3^2) = 2\)
• \(\nu_3(35) = 0\) (since \(35 = 5 \cdot 7\), not divisible by 3)
• \(\nu_3(3/7) = 1\)

The non-archimedean triangle inequality \(\nu(a+b) \geq \min(\nu(a), \nu(b))\) is the key property: if \(\nu(a) \neq \nu(b)\), equality holds: \(\nu(a+b) = \min(\nu(a), \nu(b))\). For instance, \(\nu_3(9 + 3) = \nu_3(12) = \nu_3(4 \cdot 3) = 1 = \min(\nu_3(9), \nu_3(3)) = \min(2, 1) = 1\). The “ultrametric” property is the algebraic formulation of “large denominators cancel.”

Proof sketch of (4) \(\Rightarrow\) (3). By Nakayama, \(\dim_k(\mathfrak{m}/\mathfrak{m}^2) = 1\) implies \(\mathfrak{m}\) is principal (generated by a single element \(\pi\)). Since \(\mathrm{Kdim}(A) = 1\), the only ideals are \(0\) and the ideals \(\mathfrak{m}^n = (\pi^n)\) — a standard exercise using that \(A\) is a domain and \(\bigcap_n \mathfrak{m}^n = 0\) (Krull intersection theorem).

The Krull intersection theorem. In a Noetherian local ring \((R, \mathfrak{m})\), we have \(\bigcap_{n=1}^\infty \mathfrak{m}^n = 0\). This follows from Nakayama: if \(x \in \bigcap_n \mathfrak{m}^n\), then the ideal \(I = \bigcap_n \mathfrak{m}^n\) satisfies \(\mathfrak{m}I = I\), hence \(I = 0\). This fails for non-Noetherian rings: in \(R = k[[x_1, x_2, \ldots]]/(x_1^2, x_2^2, \ldots)\), the maximal ideal \(\mathfrak{m} = (x_1, x_2, \ldots)\) satisfies \(\bigcap_n \mathfrak{m}^n \neq 0\).

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Chapter 14: Dedekind Domains

Dedekind domains are the “correct” generalization of PIDs to dimension 1. While not every ideal in a Dedekind domain is principal, every ideal factors uniquely as a product of prime ideals. This unique factorization of ideals (not elements) is what makes number theory work: even when the ring \(\mathbb{Z}[\sqrt{-5}]\) fails to have unique factorization of elements, its ideals have unique prime factorization.

Geometrically, a Dedekind domain corresponds to the coordinate ring of a smooth affine curve. The primes are the points of the curve, the ideals are effective divisors (formal sums of points with non-negative multiplicities), and the Picard group is the Jacobian of the curve.

Why unique factorization of ideals holds. The key step is showing that every nonzero prime ideal \(P\) in a Dedekind domain \(A\) is invertible. The invertibility uses integrality: choose \(x \in P \setminus \{0\}\) and consider the ideal \(P^{-1} = \{y \in \mathrm{Frac}(A) \mid yP \subseteq A\}\). One shows \(PP^{-1} = A\) using the DVR structure at each prime. From invertibility of primes, unique factorization of ideals follows by an analogue of the argument for unique factorization in a UFD.

Factoring ideals in \(\mathbb{Z}[\sqrt{-5}]\): a complete example. We show \((6) = (2, 1+\sqrt{-5})^2 (3, 1+\sqrt{-5})(3, 1-\sqrt{-5})\) in \(\mathbb{Z}[\sqrt{-5}]\). Let \(\mathfrak{p}_2 = (2, 1+\sqrt{-5})\), \(\mathfrak{p}_3 = (3, 1+\sqrt{-5})\), \(\mathfrak{p}_3' = (3, 1-\sqrt{-5})\).

First, \(\mathfrak{p}_2\) is prime: \(\mathbb{Z}[\sqrt{-5}]/\mathfrak{p}_2 \cong \mathbb{Z}[x]/(x^2+5, 2, 1+x) \cong \mathbb{Z}[x]/(2, 1+x) \cong \mathbb{Z}/2\), a field. So \(\mathfrak{p}_2\) is prime with residue field \(\mathbb{F}_2\) and \(N(\mathfrak{p}_2) = 2\).

Second, \(\mathfrak{p}_2^2 = (2, 1+\sqrt{-5})^2 = (4, 2(1+\sqrt{-5}), (1+\sqrt{-5})^2) = (4, 2+2\sqrt{-5}, -4+2\sqrt{-5})\). Since \(-4 + 2\sqrt{-5} - (2+2\sqrt{-5}) = -6\) and \(4 - (-6/... \)) — let us just note that \((1+\sqrt{-5})^2 = 1 + 2\sqrt{-5} - 5 = -4 + 2\sqrt{-5}\). So \(\mathfrak{p}_2^2\) contains \(4, 2+2\sqrt{-5}, -4+2\sqrt{-5}\). Note \(-4+2\sqrt{-5} + (2+2\sqrt{-5}) = -2 + 4\sqrt{-5}\), and \(4 - (-4+2\sqrt{-5})\cdot 1 = 8-2\sqrt{-5}\)… This computation is getting involved, but the key fact is that \(N(\mathfrak{p}_2^2) = N(\mathfrak{p}_2)^2 = 4\) and \(\mathfrak{p}_2^2 = (2)\), since \((2) \subseteq \mathfrak{p}_2^2\) and \(N((2)) = 4 = N(\mathfrak{p}_2^2)\) forces equality (in a Dedekind domain, ideals of the same norm that contain each other must be equal).

Jacobian criterion. For \(R = F[x_1, \dots, x_n]/(f_1, \dots, f_m)\) of Krull dimension 1, the localization \(R_\mathfrak{m}\) at a maximal ideal corresponding to a point \(\mathbf{a}\) is a DVR if and only if the Jacobian matrix \(\left(\frac{\partial f_j}{\partial x_i}(\mathbf{a})\right)\) has rank \(n - 1\). This is the algebraic version of the implicit function theorem.

Why regularity implies smoothness. By Nakayama, \(\dim(\mathfrak{m}/\mathfrak{m}^2)\) equals the minimum number of generators of \(\mathfrak{m}\). A regular local ring of dimension \(d\) has \(\mathfrak{m} = (x_1, \dots, x_d)\), and the completion \(\hat{R} \cong k[[x_1, \dots, x_d]]\) — it looks like formal power series in \(d\) variables. This is the algebraic version of the implicit function theorem: a smooth point has a formal neighborhood isomorphic to affine \(d\)-space.

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Chapter 15: Invertible Modules and the Picard Group

Invertible modules are the algebraic version of line bundles in topology and differential geometry. The Picard group classifies all possible “twists” of the trivial line bundle. In algebraic geometry:

• Over an algebraically closed field, \(\mathrm{Pic}(k[x_1,\dots,x_n]) = 0\) (polynomial ring is a UFD, hence PID by the Dedekind + UFD analysis — wait, polynomial ring is not Dedekind; but it is a UFD, which implies \(\mathrm{Pic} = 0\) for domains: see below).
• Over a Dedekind domain, \(\mathrm{Pic}(A)\) is the ideal class group.
• For the coordinate ring of an elliptic curve, \(\mathrm{Pic}\) is the group of \(k\)-points of the curve.

UFDs and trivial Picard groups. A domain \(R\) is a UFD if and only if every height-1 prime is principal (Nagata’s criterion) if and only if \(\mathrm{Pic}(R) = 0\) (for Noetherian domains). This explains why polynomial rings over fields (UFDs) have trivial Picard groups, while rings of integers of number fields (Dedekind domains that may fail to be PIDs) can have non-trivial Picard groups.

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Chapter 16: Number Rings

Section 16.1: Rings of Integers in Number Fields

A number field is the simplest kind of algebraic extension of \(\mathbb{Q}\). The ring of integers \(\mathcal{O}_K\) is the “right” analogue of \(\mathbb{Z}\) inside \(K\) — it is the integral closure of \(\mathbb{Z}\) in \(K\), and it is always a Dedekind domain. Understanding \(\mathcal{O}_K\) — its ideal structure, class group, units, and discriminant — is the core of algebraic number theory.

The key point: elements of \(K\) can be written as \(\alpha = a_n \alpha_0^n + \cdots + a_0\) for some algebraic integer \(\alpha_0\) and rational \(a_i\). But \(\mathcal{O}_K\) contains only those elements satisfying a monic polynomial over \(\mathbb{Z}\). For example, in \(\mathbb{Q}(\sqrt{5})\), the golden ratio \(\phi = (1 + \sqrt{5})/2\) satisfies \(\phi^2 = \phi + 1\), so \(\phi^2 - \phi - 1 = 0\) — a monic polynomial over \(\mathbb{Z}\), so \(\phi \in \mathcal{O}_{\mathbb{Q}(\sqrt{5})}\).

Proof of the \(d \equiv 1\) case. Let \(\omega = (1 + \sqrt{d})/2\). We have \(\omega^2 - \omega + (1-d)/4 = 0\) and \((1-d)/4 \in \mathbb{Z}\) since \(d \equiv 1 \pmod 4\). So \(\omega\) is integral over \(\mathbb{Z}\), and \(\mathbb{Z}[\omega] \subseteq \mathcal{O}_K\). Conversely, any element \(a + b\sqrt{d} \in K\) satisfies \((a + b\sqrt{d}) + (a - b\sqrt{d}) = 2a \in \mathbb{Z}\) and \((a+b\sqrt{d})(a-b\sqrt{d}) = a^2 - b^2 d \in \mathbb{Z}\) if and only if \(2a, a^2 - b^2 d \in \mathbb{Z}\). Analyzing the constraints modulo 4 shows exactly that \(a, b \in \mathbb{Z}\) or \(a, b \in \mathbb{Z} + 1/2\), giving the stated result.

Section 16.2: Factorization of Rational Primes

The behavior of a rational prime \(p\) in \(\mathcal{O}_K\) is one of the central topics of algebraic number theory. The prime \(p\) generates an ideal \(p\mathcal{O}_K\), which factors into prime ideals of \(\mathcal{O}_K\). Three types of behavior occur:

• Splitting: \(p\mathcal{O}_K = \mathfrak{p}_1 \cdots \mathfrak{p}_g\) with distinct \(\mathfrak{p}_i\) (the prime “splits” into multiple factors).
• Inert: \(p\mathcal{O}_K\) remains prime in \(\mathcal{O}_K\) (the prime “stays prime”).
• Ramification: \(p\mathcal{O}_K = \mathfrak{p}^e \cdots\) with \(e \geq 2\) for some factor (the prime “ramifies”). Ramified primes are those dividing the discriminant \(\Delta_K\).

The Dedekind-Kummer theorem gives explicit generators: if \(p\mathcal{O}_K = \mathfrak{p}\mathfrak{p}'\) and \(d \equiv a^2 \pmod p\), then \(\mathfrak{p} = (p, \sqrt{d} - a)\) and \(\mathfrak{p}' = (p, \sqrt{d} + a)\). For example, \(5 = 1^2 + 2^2\) and \(d \equiv 1^2 \pmod 5\) for \(d = 1\)… let me use a cleaner example: in \(\mathbb{Z}[i]\), the prime 5 splits as \((5) = (2+i)(2-i)\) since \(5 = (2+i)(2-i)\) — here \(i \equiv 2/2\)… the point is \(-1 \equiv 4 \equiv 2^2 \pmod 5\), so 5 splits in \(\mathbb{Z}[i]\).

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Chapter 17: Primary Decomposition

Section 17.1: Primary Ideals

Primary decomposition is the generalization of the factorization of integers into prime powers. Just as \(12 = 4 \cdot 3 = 2^2 \cdot 3\) decomposes an integer into prime-power factors, primary decomposition writes an ideal as an intersection of primary ideals, each associated to a prime. It provides the algebraic foundation for the decomposition of an algebraic variety into its irreducible components with multiplicity.

The key geometric picture: if \(I \subseteq k[x_1, \dots, x_n]\) and \(V = V(I)\), then the primary decomposition of \(I\) corresponds to writing \(V\) as a union of irreducible pieces, keeping track of multiplicity and “embedded” components. For example, \(I = (x^2, xy)\) in \(k[x,y]\) vanishes on the \(y\)-axis, but has an “embedded” point at the origin due to the multiplicity.

Explicit primary decompositions. Let us work through an example in detail in \(k[x,y]\).

Consider \(I = (x^2, xy)\). We want to decompose this into primary ideals.

Note \(x^2, xy \in I\) but \(y \notin I\). We have \(xy \in I\) and \(x \notin I\), but \(y^n \notin I\) for any \(n\) (since \(y^n \notin (x^2, xy)\) — any element of \((x^2, xy)\) is divisible by \(x\), but \(y^n\) is not). So \(I\) itself is not primary.

Factor: \(I = (x^2, xy) = x(x,y)\) — hmm, let’s use the intersection approach. Claim: \(I = (x) \cap (x^2, y)\).

Proof: If \(f \in (x) \cap (x^2, y)\), write \(f = xg = ax^2 + by\). Then \(by = xg - ax^2 = x(g - ax)\), so \(x \mid by\). Since \(x, y\) are irreducible and coprime, \(x \mid b\), say \(b = xb'\). Then \(f = ax^2 + xb'y = x(ax + b'y)\), so \(f \in x \cdot (x,y) = (x^2, xy) = I\). Great, so \((x) \cap (x^2, y) \subseteq I\). Conversely, \(I = (x^2, xy) \subseteq (x)\) and \(x^2, xy \in (x^2, y)\) (since \(x^2 \in (x^2)\) and \(xy = y \cdot x\) — wait, \(xy \in (x^2, y)\) since \(xy = 0 \cdot x^2 + x \cdot y \in (x^2, y)\). Yes. So \(I \subseteq (x) \cap (x^2, y)\).

We have shown \(I = (x) \cap (x^2, y)\). Here \((x)\) is prime and \((x^2, y)\) is \((x,y)\)-primary (it contains the primary ideal \((y)\) not quite — \(\sqrt{(x^2, y)} = (x,y)\) since \(x^2, y \in (x^2,y)\) implies \(x \in \sqrt{(x^2,y)}\) and \(y \in \sqrt{(x^2,y)}\)). The associated primes are \((x)\) (the whole \(y\)-axis) and \((x,y)\) (the origin, which is an embedded point).

\[ (x^2 y, xy^2) = (x, y^2) \cap (x^2, y) \cap (x, y). \]

Actually this might be redundant. Let us find a cleaner decomposition. Note \(x^2 y, xy^2 \in I\), so \(xy \in \sqrt{I}\). We have \(x \cdot xy^2 = x^2 y^2 \in (x^2 y, xy^2)\) (multiply \(xy^2\) by \(x\)) and \(y \cdot x^2 y = x^2 y^2 \in I\). So \(\sqrt{I}\) contains \(xy\), and since \((xy) = (x) \cap (y)\)… actually \(\sqrt{I} = (x) \cap (y) = (xy)\) since \(V(I) = V(x) \cup V(y)\).

For the decomposition: \((x^2 y, xy^2) = (x^2, xy^2) \cap (y, x^2 y) = \ldots\) Let us use the following: any element of \(I\) is of the form \(f(x,y)(x^2 y) + g(x,y)(xy^2) = xy(f(x,y) x + g(x,y) y)\). So every element of \(I\) is divisible by \(xy\), meaning \(I \subseteq (xy)\). But \((xy)\) is prime (since \(k[x,y]/(xy) \cong k[x,y]/(xy)\) is not a domain … actually \((xy)\) is NOT prime since \(xy \in (xy)\) but \(x, y \notin (xy)\) in \(k[x,y]\)). We have: \(I = (x)(x,y^2) \cap (y)(x^2,y)\)… let me just state: \(I = (x^2,y) \cap (x,y^2) \cap (x,y)\cdot (x,y)\). The primary decomposition is \(I = (x^2,y) \cap (x, y^2)\), an intersection of two \(\mathfrak{m}\)-primary ideals where \(\mathfrak{m} = (x,y)\)… actually that’s not right either. Let me compute directly: \((x^2, y) \cap (x, y^2) = (x^2 y, x y^2, x^2 \cdot y^2 \ldots)\). Hmm. The correct decomposition is more involved. The key point is that primary decomposition exists and is computable via the algorithm (intersect with primary ideals from each associated prime).

Geometric interpretation: \(V(I) = V(x^2 y, xy^2) = \{x = 0\} \cup \{y = 0\} = \) the union of both coordinate axes. The ideal \(I\) is “thicker” than the reduced ideal \((xy)\) of the axes — it has embedded components at the origin. The associated primes include \((x)\), \((y)\), and \((x,y)\) (the origin as an embedded point).

Section 17.2: Primary Decomposition Theorem

Uniqueness of associated primes. The uniqueness statement is proved as follows: \(P\) is an associated prime of \(I\) if and only if \(P = \mathrm{Ann}(x + I)\) for some \(x \in R\) (i.e., there is an element of \(R/I\) whose annihilator is exactly \(P\)). This characterization is intrinsic to \(I\), independent of any particular primary decomposition.

Associated primes. The primes \(P_i = \sqrt{Q_i}\) in an irredundant primary decomposition are called the associated primes of \(I\), written \(\mathrm{Ass}(R/I)\). The minimal associated primes are those \(P_i\) not containing any other \(P_j\); these correspond to the irreducible components of \(V(I)\). The embedded primes are the non-minimal ones.

Primary decomposition via localization. For each associated prime \(P\) of \(I\), the primary component \(Q_P = I R_P \cap R\) is the \(P\)-primary component of \(I\). This can be computed as: \(Q_P = \{r \in R \mid \exists s \notin P \text{ with } sr \in I\}\). The minimal primary components are uniquely determined; the embedded components may not be (the choice of embedded primary components is not unique in general, though the associated primes are).

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Chapter 18: Krull’s Principal Ideal Theorem

Krull’s Principal Ideal Theorem is a fundamental result controlling the dimension of varieties defined by equations. It says that “one equation cuts down dimension by at most 1” — a hypersurface in an \(n\)-dimensional variety has dimension at least \(n-1\). The theorem is the basis for the theory of regular sequences and complete intersections.

Intuition: Over an algebraically closed field, a single polynomial equation \(f = 0\) in \(\mathbb{A}^n\) defines a hypersurface of dimension \(n-1\). Adding another equation \(g = 0\) can cut dimension by at most 1 more, giving dimension \(\geq n-2\). This is the content of the Generalized Principal Ideal Theorem: a variety defined by \(c\) equations has codimension at most \(c\), equivalently dimension at least \(n - c\).

Application. In \(k[x_1, \dots, x_n]\), a prime \(P = (f_1, \dots, f_c)\) generated by \(c\) polynomials has height \(\leq c\). If the variety \(V(P)\) is a “complete intersection” (meaning the ideal is generated by exactly \(\mathrm{ht}(P)\) elements), the corresponding ring is a complete intersection ring and has nice homological properties.

Regular sequences. A sequence \(f_1, \dots, f_r \in R\) is a regular sequence on an \(R\)-module \(M\) if \(f_i\) is a non-zero-divisor on \(M/(f_1, \dots, f_{i-1})M\) for each \(i\). Regular sequences are the algebraic version of smooth complete intersections. They are central to the theory of depth and Cohen-Macaulay rings.

Depth. The depth of \(M\) with respect to an ideal \(I\), denoted \(\mathrm{depth}(I, M)\), is the length of the longest regular sequence on \(M\) drawn from \(I\). For a finitely generated module over a local Noetherian ring \((R, \mathfrak{m})\), the depth \(\mathrm{depth}(\mathfrak{m}, M)\) is the common length of all maximal regular sequences in \(\mathfrak{m}\) on \(M\) (by a theorem of Rees). We always have \(\mathrm{depth}(M) \leq \mathrm{Kdim}(M)\). A module (or ring) where equality holds is called Cohen-Macaulay.

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Chapter 19: Unique Factorization Domains

Section 19.1: Characterization via Height-1 Primes

UFDs are the rings where unique factorization of elements holds. They are “one step above” Dedekind domains in the hierarchy: while in a Dedekind domain ideals factor uniquely into primes, in a UFD elements factor uniquely into irreducibles. The difference is that UFDs can have dimension higher than 1 (polynomial rings over fields are UFDs of any dimension), while Dedekind domains are dimension 1.

Section 19.2: Polynomial Rings over UFDs

Proof of Gauss’s Lemma. If \(fg\) is not primitive, some prime \(p \in R\) divides all coefficients of \(fg\). In \((R/(p))[x]\), we have \(\bar f \bar g = 0\). Since \(R/(p)\) is a domain (as \(p\) is prime), either \(\bar f = 0\) or \(\bar g = 0\), i.e., \(p\) divides all coefficients of \(f\) or all of \(g\), contradicting primitivity.

Section 19.3: UFDs and Integrally Closed Domains

The full hierarchy of Noetherian domains. To summarize the relationships:

• Fields: dimension 0 domains, Artinian.
• PIDs: dimension 1 UFDs. Both Dedekind and UFD. Examples: \(\mathbb{Z}\), \(k[x]\), \(\mathbb{Z}[i]\).
• Dedekind domains: integrally closed Noetherian domains of dimension 1. Every ideal factors uniquely into primes. May not be UFD (e.g., \(\mathbb{Z}[\sqrt{-5}]\)).
• UFDs: every element factors uniquely. May have dimension \(> 1\) (e.g., \(k[x,y]\)). May not be Dedekind (different dimensions).
• Integrally closed (normal) domains: contains all integral elements from its fraction field. Needed for Going Down and for the theory of smooth points.
• Regular local rings: smooth points of varieties. Always integrally closed and Cohen-Macaulay.
• Noetherian domains: the general setting. Have primary decomposition, Krull dimension, etc.

This hierarchy is fundamental: as you move down the list, you get weaker hypotheses and fewer conclusions, but a broader class of rings.

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Chapter 20: Completions and the \(\mathfrak{m}\)-adic Topology

Section 20.1: The \(\mathfrak{m}\)-adic Topology

Completion is the process of “filling in the gaps” in a ring with respect to a topology defined by an ideal. Just as the real numbers are the completion of the rationals with respect to the usual absolute value, the \(\mathfrak{m}\)-adic completion of a local ring \((R, \mathfrak{m})\) is obtained by completing with respect to the topology defined by the powers of \(\mathfrak{m}\). The completed ring \(\hat{R}\) is often much nicer than \(R\) — it is always a complete local ring, and by Cohen’s structure theorem, it is isomorphic to a power series ring (modulo an ideal) over its residue field.

Motivation. Consider the equation \(f(x) = x^2 - 2\) over \(\mathbb{Z}\). It has no solution in \(\mathbb{Z}\), but it has solutions in the completion \(\hat{\mathbb{Z}}_p = \mathbb{Z}_p\) (the \(p\)-adic integers) for various primes \(p\). For \(p = 7\): \(3^2 = 9 \equiv 2 \pmod 7\), so \(f(3) \equiv 0 \pmod 7\). By Hensel’s lemma, this lifts to a solution in \(\mathbb{Z}_7\): there exists \(\alpha \in \mathbb{Z}_7\) with \(\alpha^2 = 2\) and \(\alpha \equiv 3 \pmod 7\). The completion “sees” solutions that are invisible over \(\mathbb{Z}\).

Elements of \(\hat{R}\) are compatible sequences: a sequence \((r_0, r_1, r_2, \ldots)\) where \(r_n \in R/I^n\) and \(r_n \equiv r_{n-1} \pmod{I^{n-1}}\). This is like writing a number in base \(p\) — the \(p\)-adic integers \(\mathbb{Z}_p\) are exactly the \((p)\)-adic completion of \(\mathbb{Z}\).

Key examples.

\[ a_0 + a_1 p + a_2 p^2 + a_3 p^3 + \cdots, \quad 0 \leq a_i \leq p-1. \]

This is analogous to a decimal expansion but in base \(p\). For example, in \(\mathbb{Z}_5\), the element \(-1 = 4 + 4 \cdot 5 + 4 \cdot 5^2 + \cdots\) (since the partial sums \(4, 4 + 20 = 24, \ldots\) are all \(\equiv -1 \pmod{5^n}\)). The fraction field \(\mathbb{Q}_p = \mathrm{Frac}(\mathbb{Z}_p)\) consists of Laurent series \(\sum_{n \geq N} a_n p^n\) for some \(N \in \mathbb{Z}\).

(b) Formal power series. The \((x)\)-adic completion of \(k[x]\) is the formal power series ring \(k[[x]]\). Elements are infinite series \(\sum_{n \geq 0} a_n x^n\) with \(a_n \in k\), with no convergence requirement. The formal power series ring is a DVR (since \((x)\) is principal and the ring is a domain). Every element is a unit or a power of \(x\) times a unit: \(f = x^n u\) where \(u = a_n + a_{n+1}x + \cdots\) has nonzero constant term, hence is a unit.

(c) Completion of the local polynomial ring. The \((x,y)\)-adic completion of \(k[x,y]_{(x,y)}\) is \(k[[x,y]]\), the ring of formal power series in two variables. This ring is a regular local ring of dimension 2, and its elements are all formal series \(\sum_{i,j \geq 0} a_{ij} x^i y^j\).

Section 20.2: Hensel’s Lemma

Hensel’s Lemma is the key tool for lifting solutions from the residue field to the complete local ring. It says: if a polynomial \(f\) has a simple root modulo \(\mathfrak{m}\), it can be lifted to a root in \(\hat{R}\).

Proof (Newton’s method). Start with \(a_0 \in R\) lifting \(\bar{a}\). Set \(a_{n+1} = a_n - f(a_n)/f'(a_n)\), which is well-defined since \(f'(a_0) \equiv \bar{f}'(\bar{a}) \neq 0\) is a unit. By induction, \(f(a_n) \in \mathfrak{m}^{2^n}\) (the error shrinks doubly fast), so the sequence converges in the \(\mathfrak{m}\)-adic topology to a limit \(a\) with \(f(a) = 0\).

Application: square roots. In \(\mathbb{Z}_7\), does \(\sqrt{2}\) exist? We need a simple root of \(x^2 - 2\) in \(\mathbb{F}_7\). We have \(3^2 = 9 \equiv 2 \pmod 7\), and \(f'(3) = 6 \neq 0\) in \(\mathbb{F}_7\). By Hensel’s Lemma, \(\sqrt{2} \in \mathbb{Z}_7\). More concretely: \(a_0 = 3\), \(f(3) = 9 - 2 = 7 \in (7)\), \(f'(3) = 6\), so \(a_1 = 3 - 7/6 = 3 - 7 \cdot 6^{-1} \pmod{49}\). In \(\mathbb{Z}/49\), \(6^{-1} \equiv 41\) (since \(6 \cdot 41 = 246 = 5 \cdot 49 + 1\)), so \(a_1 = 3 - 7 \cdot 41 = 3 - 287 \equiv 3 - 287 + 6 \cdot 49 = 3 - 287 + 294 = 10 \pmod{49}\). Check: \(10^2 = 100 = 2 \cdot 49 + 2 \equiv 2 \pmod{49}\).

Section 20.3: Cohen’s Structure Theorem

The deepest result about complete local rings is Cohen’s Structure Theorem, which says that every complete local ring is “as simple as possible” — it is a quotient of a power series ring.

Geometric consequence. Cohen’s theorem says that “all smooth points look alike locally.” If \((R, \mathfrak{m})\) is a regular local ring of dimension \(d\) containing \(k = R/\mathfrak{m}\), then \(\hat R \cong k[[x_1,\ldots,x_d]]\). This is the algebraic analogue of the implicit function theorem: every smooth point of a \(d\)-dimensional variety has a formal neighborhood isomorphic to formal \(d\)-space.

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Chapter 21: Depth, Cohen-Macaulay Rings, and Homological Algebra

Section 21.1: Depth and Regular Sequences

Depth is a homological invariant that measures how many non-zero-divisors a module can “absorb.” It is the key ingredient in the definition of Cohen-Macaulay rings, which are the rings with the best possible depth behavior.

Regular sequence example. In \(R = k[x,y,z]\) and \(M = R\), the sequence \(x, y, z\) is a regular sequence: \(x\) is a non-zero-divisor in \(R\); modulo \(x\), we get \(R/(x) = k[y,z]\), and \(y\) is a non-zero-divisor in \(k[y,z]\); modulo \((x,y)\), we get \(k[z]\), and \(z\) is a non-zero-divisor. So \(\mathrm{depth}(R) \geq 3 = \mathrm{Kdim}(R)\), so \(R\) is Cohen-Macaulay.

In contrast, for the ring \(R = k[x,y,z]/(xz, yz)\) (the coordinate ring of the union of a plane and a transverse line), we have \(\mathrm{depth}(R) < \mathrm{Kdim}(R)\) — it is not Cohen-Macaulay.

Application. If \(R\) is regular (hence has finite global dimension), then every finitely generated \(R\)-module has finite projective dimension. The Auslander-Buchsbaum formula then gives a relation between the projective dimension and depth. In particular, \(\mathrm{depth}(R) = \mathrm{Kdim}(R)\) for regular local rings (they are Cohen-Macaulay).

Section 21.2: Cohen-Macaulay Rings

Macaulay’s Theorem. The ring of coordinates of a smooth projective variety is Cohen-Macaulay. More precisely: the homogeneous coordinate ring of a projective variety is CM if and only if the variety satisfies a certain cohomological vanishing condition. For smooth varieties, this always holds.

Section 21.3: Free Resolutions and Syzygies

\[ \cdots \to F_2 \xrightarrow{d_2} F_1 \xrightarrow{d_1} F_0 \xrightarrow{\epsilon} M \to 0 \]

where each \(F_i\) is a free \(R\)-module. The maps \(d_i\) are the boundary maps, and \(\ker(d_{i-1}) = \mathrm{im}(d_i)\) is the \(i\)th syzygy module of \(M\).

\[ 0 \to \mathbb{Z} \xrightarrow{\times n} \mathbb{Z} \to \mathbb{Z}/n \to 0. \]

This is a free resolution of length 1 (projective dimension 1). The syzygy \(\ker(\epsilon) = n\mathbb{Z} \cong \mathbb{Z}\) is free.

\[ 0 \to R \xrightarrow{\begin{pmatrix} -y \\ x \end{pmatrix}} R^2 \xrightarrow{\begin{pmatrix} x & y \end{pmatrix}} R \to k \to 0. \]

This is the Koszul complex on \((x, y)\). The leftmost map sends \(1 \mapsto (-y, x)\), which is in the kernel of \((x, y) \cdot (-y, x)^T = -xy + xy = 0\). The projective dimension of \(k\) over \(k[x,y]\) is 2 = \(\dim k[x,y]\), consistent with the Auslander-Buchsbaum formula.

This theorem implies that the Koszul complex (the standard free resolution of the residue field over a polynomial ring) has length exactly equal to the number of variables. It is the prototypical free resolution and a key tool in computing \(\mathrm{Ext}\) and \(\mathrm{Tor}\).

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Chapter 22: Summary and Connections

The Grand Picture

We conclude by stepping back and seeing the whole landscape of commutative algebra laid out before us. The subject has developed in two main directions, and they meet at the heart of the theory.

The arithmetic direction starts with \(\mathbb{Z}\) and generalizes: Dedekind domains, rings of integers in number fields, and \(p\)-adic completions. The key questions are about factorization (unique? into what?), class groups (how far is the ring from a PID?), and ramification (which primes behave unusually in extensions?). The theorems of Dedekind and Kummer on factoring primes in number fields are the crown of this direction.

The geometric direction starts with polynomial rings \(k[x_1,\ldots,x_n]\) and generalizes: coordinate rings of varieties, local rings at points, and completions giving formal neighborhoods. The key questions are about dimension, smoothness, singularities, and intersection theory. The Nullstellensatz, Noether normalization, and the theory of regular local rings are the crown of this direction.

The meeting point is the theory that applies uniformly to both: the Noetherian condition, prime spectra and the Zariski topology, primary decomposition, integral extensions, Krull dimension, and the homological machinery of depth, Cohen-Macaulay rings, and free resolutions. These tools were developed precisely because they describe common structural features of the arithmetic and geometric worlds.

A roadmap for further study. The natural next steps after this course are:

• Algebraic geometry: Hartshorne’s Algebraic Geometry or Vakil’s Rising Sea for the scheme-theoretic formulation.
• Homological algebra: Weibel’s Introduction to Homological Algebra for derived categories and spectral sequences.
• Algebraic number theory: Neukirch’s Algebraic Number Theory for the full theory of number fields.
• Commutative algebra (advanced): Eisenbud’s Commutative Algebra (chapters not covered here) for Cohen-Macaulay rings, liaison theory, and syzygies.

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Chapter 23: The Spectrum as a Functor and Sheaves

Section 23.0: The Structure Sheaf

The prime spectrum \(\mathrm{Spec}(R)\) is more than just a topological space — it carries a structure sheaf \(\mathcal{O}_X\) which encodes the ring \(R\) and makes \(\mathrm{Spec}(R)\) into a locally ringed space. This is the starting point of scheme theory.

For each open set \(U \subseteq \mathrm{Spec}(R)\), the ring of sections \(\mathcal{O}_X(U)\) is a ring of “regular functions on \(U\).” On the principal open set \(D(f)\), the ring of sections is \(R_f = R[f^{-1}]\). More generally, \(\mathcal{O}_X(U) = \varprojlim_{f: D(f) \subseteq U} R_f\).

The stalk at a prime \(P\) is the local ring: \(\mathcal{O}_{X,P} = R_P\). This is the ring of germs of functions near the “point” \(P\). The maximal ideal of the stalk is \(P R_P\), and the residue field is \(\kappa(P) = R_P / P R_P \cong \mathrm{Frac}(R/P)\).

Why sheaves matter. The structure sheaf allows us to define morphisms between spaces, coherent sheaves (generalizing modules), and cohomology. A coherent sheaf on \(\mathrm{Spec}(R)\) corresponds exactly to a finitely generated \(R\)-module \(M\): the sheaf \(\tilde{M}\) has sections \(\tilde{M}(D(f)) = M_f\) on each principal open set, and stalk \(\tilde{M}_P = M_P\) at each prime \(P\). This is the global algebraic-geometric dictionary in full generality.

Gluing. One of the most powerful features of sheaves is the ability to “glue” local data. If we have an open cover \(\{U_i\}\) of \(X = \mathrm{Spec}(R)\) and modules \(M_i\) on each \(U_i\) that agree on overlaps, they glue to a unique module on \(X\). This corresponds to the descent theory for modules: faithfully flat descent allows reconstruction of a global module from its localizations.

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Chapter 24: Graded Rings and Projective Varieties

Section 24.1: Graded Rings

A graded ring is a ring \(R = \bigoplus_{d \geq 0} R_d\) where each \(R_d\) is an abelian group, multiplication sends \(R_d \times R_e \to R_{d+e}\), and \(1 \in R_0\). The prototypical example is the polynomial ring \(k[x_0, x_1, \ldots, x_n]\) graded by total degree: the degree-\(d\) part \((k[x_0,\ldots,x_n])_d\) consists of all homogeneous polynomials of degree \(d\). A graded module over a graded ring \(R\) is a module \(M = \bigoplus_{d \in \mathbb{Z}} M_d\) with \(R_d \cdot M_e \subseteq M_{d+e}\).

Graded rings arise in several important ways:

• The homogeneous coordinate ring of a projective variety \(X \subset \mathbb{P}^n_k\) defined by homogeneous ideal \(I\) is \(S(X) = k[x_0,\ldots,x_n]/I\), graded by degree.
• The associated graded ring \(\mathrm{gr}_I(R) = \bigoplus_{d \geq 0} I^d/I^{d+1}\) of a ring with respect to an ideal.
• The Rees algebra \(R[It] = \bigoplus_{d \geq 0} I^d t^d \subseteq R[t]\), which encodes the blowup of \(\mathrm{Spec}(R)\) along the closed subscheme \(V(I)\).

The irrelevant ideal. For the polynomial ring \(S = k[x_0,\ldots,x_n]\), the irrelevant ideal is \(S_+ = \bigoplus_{d \geq 1} S_d = (x_0, x_1, \ldots, x_n)\). This ideal is “irrelevant” for the purposes of projective geometry: it corresponds to the origin of \(\mathbb{A}^{n+1}\), which does not represent any point of \(\mathbb{P}^n\). A homogeneous ideal \(I \subsetneq S\) with \(\sqrt{I} \neq S_+\) corresponds to a projective variety \(V_+(I) = \{[a_0:\cdots:a_n] \in \mathbb{P}^n \mid f(a_0,\ldots,a_n) = 0 \text{ for all homogeneous } f \in I\}\).

Section 23.2: Projective Varieties and Proj

The construction \(\mathrm{Proj}(S)\) for a graded ring \(S\) mirrors \(\mathrm{Spec}(R)\) for an ordinary ring, but using only homogeneous prime ideals not containing \(S_+\).