# Commutative Algebra

Last updated: Tue, 22 Jan 2019

Convention: $A$ is used to denote a ring, $A[T]$ is a polynomial ring. $(f) \subset A[T]$ is a polynomial. $M$ is an A-module. $I$ and $J$ are ideals, $V$ is a variety, $\mathcal{V}$ is a Zariski topology. $L$, $M$, $N$ are modules. $\varphi$ is a homomorphism.

Direct sum $A = A_{1} \oplus A_{2}$ is defined if $A_{1} = Ax$ and $A_{2} = Ax'$; $x$ and $x'$ are complementary orthogonal idempotents, and $x' = 1 - x$, $xx' = 0$.

The bridge between algebra and geometry is established as follows: the maximal ideal $A = k[x_1, ..., x_N]$ corresponds to points in $X$.

Principal Ideal Domain $\subset$ Unique Factorization Domain $\subset$ Commutative Rings.

$\text{Spec } A = \{P \mid P \subset A \text{ is a prime ideal}\}$ $\text{m-Spec } A = \{P \mid P \subset A \text{ is a maximal ideal}\}$

These are obvious:

1. $m \text{ is maximal} \Leftrightarrow A/m \text{ is a field}$
2. $p \text{ is prime} \Leftrightarrow A/P \text{ is an integral domain}$
3. $\text{Maximal ideal} \Rightarrow \text{Prime ideal}$

Definition of radical: $\text{rad } I = \{f \in A \mid f^{n} = 0 \text{ for some } n\}$.

$\text {rad } I = \bigcap\limits_{\substack{P \subset \text{Spec } A \\ I \subset A}} P$, and this could be interpreted as a weak Nullstellansatz.

$A$ is an integral domain $\leftrightarrow$ 0 is a prime ideal in $A$ $\leftrightarrow$ $A$ has no divisors of 0.

$x \subset A_1$ and $1 - x \subset A_2$ are complementary orthogonal idempotents if $x$ is nilpotent; that is, $A$ is the direct sum $A = A_1 \oplus A_2$. Also, $x$ is nilpotent implies that $1 - x$ is invertible:

$\frac{1}{1 - x} = \sum_{i = 0}^{\infty} x^i$

Multiplication of ideals is defined as: $IJ = \sum f_i g_i\; \colon f_i \subset I, g_i \subset J$

$I$ and $J$ are strongly coprime if $I + J = A$, implying: $\begin{cases} I + J = I \cap J \\ A/IJ \cong A/I \times A/J \end{cases}$

## Modules

Modules let us do linear algebra on general rings, even non-commutative ones. For instances of non-commutative rings, there's no need to look further than matrices over $\mathbb{R}$, the field of real numbers. Another instance of non-commutative algebra can be found in Quaternions, that have different properties when cycling clockwise and counter-clockwise.

A module $M$ over ring $A$ is formally defined as the multiplication map $A \times M \rightarrow AM$, or $(f, m) \mapsto fm$ satisfying the following axioms:

1. $(f + g) m = fm + gm$
2. $f (m + n) = fm + fn$
3. $(fg) m = f (gm)$
4. $1_A m = m$

Theory of modules $\leftrightarrow$ Theory of linear algebra over $A$. A-modules over a field $k$ $\leftrightarrow$ $k$-vector space, exactly.

Submodule $I \subset A$ $\leftrightarrow$ ideal $I$ in $A$.

Free modules of $A[X, Y]$ are generated by:

$A^{\text{card} \lambda} = \left\{\sum_{\lambda \subset \Lambda} a_{\lambda} \mid \text{ only finitely many } a_\lambda \neq 0\right\}$

Viewing $A^{n}$ as a $A'[\varphi]\text{-module}$, we get:

$\varphi e_{k} = \sum_{i = 1}^{n} a_{ki} e_{i} \Rightarrow \sum_{i = 1}^{n} \varphi \delta_{ik} - a_{ki} e_{i} = 0$

Nakayama's lemma, corollary: If $M = IM$, then $\exists x \in A \text{ such that } x \equiv 1 \text{ mod } I$ and $xM = 0$.

Nakayama's lemma: Let $(A, m)$ be a local ring, $M$ a finite A-module; $M = mM \implies M = 0$; if $M = IM$, with $1 + I \subset A^{\times}$, then $M = 0$.

## Homomorphism theorems of modules

Proofs for these correspond exactly to those in vector spaces.

1. $\text{ker } \varphi \subset M$, and $\text{im } \varphi \subset N$ are submodules.
2. Let $N \subset M$ be a submodule; then $\ni$ a surjective quotient homomorphism $\varphi : M \rightarrow M/N$, ker $\varphi = N$. Elements of $M/N$ can be constructed either as equivalence classes $m \in M \bmod N$, or as cosets $m + N$ in $M$.
3. $M/\text{ker } \varphi \cong \text{im } \varphi$.

## Short exact sequence (s.e.s)

$L$, $M$, $N$ are A-Modules, and $\alpha$, $\beta$ are homomorphisms:

$0 \rightarrow L \xrightarrow[]{\alpha} M \xrightarrow[]{\beta} N \rightarrow 0$

is a split exact sequence if $M \cong L \oplus N$, $L \subset M$ and $N = M/L$, such that $\alpha$ maps $m \mapsto (m, 0)$ and $\beta$ maps $(m, n) \mapsto m$. c Koszul complex of pair ($x$, $y$) is defined as the exact sequence:

$0 \rightarrow A \xrightarrow[]{(-y, x)} A^2 \xrightarrow[]{\begin{pmatrix} x \\ y \end{pmatrix}} I \rightarrow 0$

$S^{-1}$ preserves exactness over the s.e.s:

$L \xrightarrow[]{\alpha} M \xrightarrow[]{\beta} N$ $S^{-1} L \xrightarrow[]{\alpha'} S^{-1} M \xrightarrow{\beta'} S^{-1} N$

## Algebraic dependence and integral dependence

$y$ is algebraic over $k$ if it satisfies the algebraic dependence relation:

$f(y) = a_m y^m + a_{m - 1} y^{m - 1} + ... + a_0$

Over a field, it costs nothing to divide over $a_m$, giving us the monic polynomial:

$f(y) = y^m + a_{m - 1} y^{m - 1} + ... + a_0$

This is termed as integral dependence. $\varphi : A \rightarrow B$ is finite, if $A$ is integral.

C is termed nonsingular if $(\partial f/\partial x, \partial f/\partial y) \neq 0$ for $P \in C$, so that $C$ has a well-defined tangent at every point $P$.

$XY = 1$ is an example of something that is algebraically closed, but not integrally closed. It can be perturbed as $(X + \epsilon Y) Y = 1$ to avoid the unlucky accident of a "missing zero" over $X = 0$.

## The Nullstellansatz

Variety $V(J) = \{P = (a_1, ..., a_n)\; | f(P) = 0\; \forall f \subset J\}$

Tautologically, $X \subset V(I(X))$; $X = V(I(X)) \iff X$ is a variety.

The Nullstellansatz states that:

1. If $J \subsetneq k[X_1 ... X_n]$ then $V(J) \neq 0$
2. $I(V(J)) = \text{rad } J$, the radical of the ideal

A variety is irreducible if it cannot be expressed as the union of two proper subvarieties:

$J(X) = J(X_1) \cup J(X_2) \Rightarrow X = X_1 \text{ or } X_2$; $X$ is a prime ideal.

The following reverse-inclusions are obvious: $X \subset Y \Rightarrow I(X) \supset I(Y)$ $I \subset J \Rightarrow V(I) \supset V(J)$

## Zariski topology

Zariski topology is a topology where the only closed sets are the algebraic ones, the zeros of polynomials. The Zariski topology on a variety is Noetherian.

$V(I) \cup V(J) = V(I \cap J) = V(IJ)$

corresponds exactly to the Zariski topology on $\text{Spec } A$, $\mathcal{V}(I)$:

$\mathcal{V}(I) \cup \mathcal{V}(J) = \mathcal{V}(I \cap J) = \mathcal{V}(IJ)$

where $\mathcal{V}(I) = {P \in \text{Spec A} \mid P \subset I}$.

## Localization

$A_{p}$ is a local ring $\leftrightarrow$ $A_{p}$ has a unique maximal ideal at $p$.

$\mathbb{Z}_{(p)}$, a localization of $\mathbb{Z}$ at p = $\{a/b \text{ with } a, b \in \mathbb{Z}, b \nmid p\}$.

For a general construction, let $S$ be a multiplicative set in $A$, $P$ a prime ideal so that $S = A \backslash P$. Then $A_P = S^{-1} A = A \times S / \sim$, where $\sim$ is an equivalence relation.

$S^{-1}$ is an exact functor in that, if $L \subset M$ and $N = M/L$, then $S^{-1} L \subset S^{-1} M$ and $S^{-1} N = S^{-1} M / S^{-1} L$.

$e : \{\text{ideals of A}\} \to \{\text{ideals of B}\}$ $r : \{\text{ideals of B}\} \to \{\text{ideals of A}\}$

Then, for ideal $J$ in $S^{-1} A$, $e(r(J)) = J$, and for any ideal $I$ of A, $r(e(I)) = \{a \in A \mid as \in I, \text{ for some } s \in S\}$.

These three statements are equivalent:

1. $A$ is local if it has a unique maximal ideal $m$.
2. $m = \{\text{nonunits of A}\}$ is the unique maximal ideal.
3. If $m \subset A$ is a maximal ideal and $x \in A$, then $1 + x$ is unit.

## Support and annihilator

Support of M is defined as $\text{Supp } M = \{ P \subset \text{Spec } A \mid M_p \neq 0\} \subset \text{Spec } A$. Here $M_p = S^{-1} M$, the module of fractions. Assassin $\text{Ass } M \subset \text{Supp } M$. Annihilator of M over A is defined as $\text{Ann } M = \{f \subset A \mid fM = 0\}$.

Example: If $n = p^{\alpha} q^{\beta}$, then $\text{Ass }(\mathbb{Z}/n) = \{(p), (q)\}$. If $m = p^{\alpha - 1} q^{\beta} \text{ mod } n \in \mathbb{Z}/(n)$, then annihilator $\text{Ann } m = p$.

1. If $M$ is generated by a single element $x$, such that $x = \text{Ann } I$, then $\text{Supp } M = \mathcal{V}(I)$.
2. If $L \subset M$ and $N = L/M$, then $\text{Supp } L = \text{Supp } M \cup \text{Supp } N$.
3. If $L \subset M$ and $N = L/M$, then $\text{Ass } L = \text{Ass } M \cup \text{Ass } N$.
4. If $P \subset \text{Supp } M$, then $\mathcal{V}(P) \subset \text{Supp } M$.

In the disjoint union $\mathcal{M} = \sqcup M_{P}$, $M_{P}$ is termed as the stalk of $\mathcal{M}$ over $P$.

## Primary decomposition

Ideal $Q$ of $A$ is primary if $Q \neq A$ and

$fg \in Q \implies f \in Q \text{ or } g^{n} \in Q \text{, for some } n > 0$

A counter-example: $I = (X^{2}, XY)$; $\text{rad}(I) = X$; $I$ is not primary because $XY \in I$, but $X \notin I$ and $Y^{n} \notin I$.

$Q \text{ is P-primary if } Q \text{ is primary and } P = rad(Q)$ $Q \text{ is P-primary} \leftrightarrow \text{Ass}(A/Q) = \{P\}$

In a Noetherian ring $A$, every ideal $I$ has a primary decomposition.

## Discrete valuation rings

$v : K \backslash \{0\} \to \mathbb{Z}$ is a discrete valuation of $K$, a surjective map so that:

1. $v(xy) = v(x) + v(y)$
2. $v(x \pm y) = \text{min}\{v(x), v(y)\} \forall x, y \in K \backslash \{0\}$

By convention, $v(0) = \infty$.

Valuation ring of a discrete valuation $v$ is given by $A = \{x \in K \mid v(x) \geq 0\}$.

For valuation ring $A$:

$A \text{ is a DVR} \leftrightarrow A \text{ is Noetherian}$