# Commutative Algebra

Last updated: Tue, 16 Oct 2018

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. $$L$$, $$M$$, $$N$$ are modules. $$\varphi$$ is a homomorphism.

In Miles Reid, 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.

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

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

Maximal ideal $$\Rightarrow$$ Prime ideal

$$A$$ 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 idempotent 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: $\begin{cases} I + J = I \cap J \\ A/IJ \cong A/I \times A/J \end{cases}$

For instance, $$\sin$$ and $$\cos$$.

## 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.

Theory of modules $$\leftrightarrow$$ Theory of linear algebra over $$A$$:

$AM = \sum a_i m_i\; \colon a_i \subset A, m_i \subset M$

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:

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

## Homomorphism theorems of modules

Proofs for these correspond exactly to those in vector spaces.

(i) ker $$\varphi \subset M$$, and im $$\varphi \subset N$$ are submodules.

(ii) 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 $$M$$.

(iii) $$M/$$ ker $$\varphi \cong$$ im $$\varphi$$.

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

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

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

$$M \cong L \oplus N$$, $$L \subset M$$ and $$N = M/L$$.

Koszul complex of pair ($$x$$, $$y$$) is defined as the exact sequence:

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

## 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.

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:

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

The Nullstellansatz states that:

• If $$J \subsetneq k[X_1 ... X_n]$$ then $$V(J) \neq 0$$
• $$I(V(J)) = \text{rad } J$$, 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: $X \subset Y \Rightarrow I(X) \supset I(Y)$ $I \subset J \Rightarrow V(I) \supset V(J)$

## 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:

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

## Primary decomposition

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.

Annihilator of M over A is defined as $$\text{Ann } M = \{f \subset A \mid fM = 0\}$$