## Rings of fractions

Rings of fractions, and the associated localization process, correspond to the [algebro-geometric picture](/ag) of studying behavior of an open set or near a point. Using the way we construct $\mathbb{Q}$ from $\mathbb{Z}$, we can construct equivalence classes of $a, s$, where $(a, s) \in A, s \neq 0$, defined as follows: $$(a, s) \equiv (b, t) \iff at - bs = 0$$

but this definition only works if $A$ is an integral domain. A more general definition of equivalence involves using a `multiplicative subset` $S$ of $A$ and defining $\equiv$ over $A \times S$: $$(a, s) \equiv (b, t) \iff \exists u \in S : u(at - bs) = 0$$

Let $a/s$ denote the equivalence class of $(a, s)$ and $S^{-1} A$ the set of all equivalence classes. We can endow $S^{-1} A$ with a ring structure by defining addition and multiplication as operations on fractions, as in elementry algebra: $$a/s + b/t = (at + bs)/st$$ $$(a/s)(b/t) = ab/st$$

We also have the ring homomorphism $A \rightarrow S^{-1} A$ defined by $x \mapsto x/1$. It is not injective in general. $S^{-1} A$ is termed the `ring of fractions` of $A$ with respect to $S$.

Let $g : A \rightarrow B$ be a ring homomorphism where $g(s)$ is unit in $B$ for all $s \in S$. Then there exists a unique ring homomorphism $h : S^{-1} A \rightarrow B$ such that $g = h \circ f$.

The ring $S^{-1} A$ and the homomorphism $f : A \rightarrow S^{-1} A$ have the following properties:

- $s \in S \implies f(s)$ is unit in $S^{-1} A$.
- $f(a) = 0 \implies as = 0$ for some $s \in S$.
- Every element of $S^{-1} A$ has the form $f(a)f(s)^{-1}$ for some $a \in A, s \in S$.

Conversely, these properties determine $S^{-1} A$ uniquely upto isomorphism.

Some examples follow:

- Let $\mathfrak{p}$ be a prime ideal in $A$. Then $A - \mathfrak{p}$ is multiplicatively closed and is written as $S^{-1} A = A_\mathfrak{p}$, the localization of $A$ at $\mathfrak{p}$. $a/s$ with $a \in \mathfrak{p}$ form an ideal $M$ in $A_\mathfrak{p}$. If $b/t \neq M$, and $b \notin \mathfrak{p}$, then $b/t$ is unit. $M$ is the only maximal ideal in $\mathfrak{p}$, and $A_\mathfrak{p}$ is a local ring. If $A = \mathbb{Z}, \mathfrak{p}$ is the set of prime numbers, then $A_\mathfrak{p}$ is the set of all rationals with denominators as powers of $f$.
- $S^{-1} A$ is the zero ring $\iff 0 \in S$.
- Let $f \in A$ $S^{-1} = (1/f, 1/f^2, \ldots, 1/f^n)$. Then we write $A_f$ for $S^{-1} A$ in this case. Let $A = k[t_1, \ldots, t_n]$ be a polynomial ring where $k$ is a field, and $\mathfrak{p}$ a prime ideal in $A$. Then, $A_\mathfrak{p}$ is the ring of all rational functions $g/h, g \notin \mathfrak{p}$. Let $V$ be a variety defined by $\mathfrak{p}$, that is to say the set $(x_1, \ldots, x_n) \in k^n$ such that $f(x) = 0$ whenever $f \in \mathfrak{p}$. Then, $A_\mathfrak{p}$ can be identified with the ring of all rational functions in $k^n$ which are defined at almost all points of $V$; it is the local ring of $k^n$ defined along variety $V$. This is the prototype of local rings, as they arise in algebraic geometry.
- Let $I$ be an ideal in $A$, such that $S = 1 + I$. Then, $S$ is multiplicatively closed.

## Modules of fractions

The same construction involving $S^{-1}$ can be carried out in modules instead of rings. Let $u : M \rightarrow N$ be an $A$-module homomorphism. This gives rise to an $S^{-1} A$-module homomorphism $S^{-1} M \rightarrow S^{-1} N$, such that $S^{-1} u$ maps $m/s$ to $u(m)/s$. We have $S^{-1} (v \circ u) = S^{-1}v \circ S^{-1}u$.

$S^{-1}$ preserves exactness, in that if the following sequence is exact at $M$: $$L \xrightarrow{f} M \xrightarrow{g} N$$

then, this sequence is exact at $S^{-1} M$: $$S^{-1} L \xrightarrow{S^{-1} f} S^{-1} M \xrightarrow{S^{-1} g} S^{-1} N$$

This can proved by noticing that if $g \circ f = 0$, then so is $S^{-1} g \circ S^{-1} f = S^{-1} (0)$. Hence, $\text{Im}(S^{-1} f) \subseteq \text{Ker}(S^{-1} g)$. To prove inclusion in the other direction, let $m/s \in S^{-1} g$, then $g(m)/s \in S^{-1} N$, hence $\exists t : t g(m) = 0$ in $N$. But $t g(m) = g(tm)$, since $g$ is an $A$-module homomorphism, hence $t(m) \in \text{Ker}(g) = \text{Im}(f)$. Therefore, $\exists m' \in M : t(m) = f(m')$. Hence, in $S^{-1} M$, we have $m/s = f(m')/st = S^{-1} f(m'/st) \in \text{Im}(S^{-1} f)$. Hence, $\text{Ker}(S^{-1} g) \subseteq \text{Im}(S^{-1} f)$.

If $N$ is a submodule of $M$, then the mapping $S^{-1} M \rightarrow S^{-1} N$ is injective, and $S^{-1} N$ is a submodule of $S^{-1} M$.

We have the following properties of fractions of modules, given that $M, N$ are submodules of $A$-module $M$:

- $S^{-1}(N + P) = S^{-1} N + S^{-1} P$.
- $S^{-1}(N \cap N) = S^{-1} N \cap S^{-1} P$.
- $S^{-1}(M/N) \cong (S^{-1} M)/(S^{-1} N)$.

The first is obvious. To prove (ii), consider $y/s = z/t, y \in N, z \in P, s, t \in S$. Then, $\exists u \in S : u(ty - zt) = 0$. Hence, $w = uty = usz \in N \cap P$, and therefore, $y/s = w/stu \in S^{-1} N \cap P$. Consequently, $S^{-1} N \cap S^{-1} P \subseteq S^{-1}(N \cap P)$. The reverse inclusion is obvious.

To prove (iii), apply $S^{-1}$ to the exact sequence $0 \rightarrow M \rightarrow N \rightarrow M/N \rightarrow 0$.

Let $M$ be an $A$-module. Then, there exists a unique isomorphism $f : S^{-1} A \otimes_A M \rightarrow S^{-1} M$ such that: $$f((a/s) \otimes m) = am/s \quad \forall a \in A, m \in M, s \in S$$

To prove this, notice that the mapping $S^{-1} A \rightarrow S^{-1} M$ given by $(a/s, m) \mapsto am/s$ is $A$-bilinear, and by universal property of the tensor product, induces the $A$-homomorphism: $$f : S^{-1} A \otimes_A M \rightarrow S^{-1} M$$

$f$ is clearly surjective, and it remains to be proved that it is injective as well. Notice that every element of $S^{-1} A \otimes M$ is of the form $(1/s) \otimes m$. Suppose that $(1/s) \otimes m = 0$, then $m/s = 0$ and hence $\exists t : tm = 0$. Therefore: $$\frac{1}{s} \otimes m = \frac{t}{ts} \otimes m = \frac{1}{s} \otimes tm = 0$$

Hence, $f$ is injective, and therefore an isomorphism.

$S^{-1} A$ is a [flat](/ac/1#exactness-of-the-tensor-product) $A$-module, and this is immediately obvious.

If $M, N$ are $A$-modules, there is an isomorphism of $S^{-1} A$-modules $f : S^{-1} M \otimes_{S^{-1} A} S^{-1} N \rightarrow S^{-1}(M \otimes_A N)$ such that: $$f((m/s) \otimes (n/t)) = (m \otimes n)/st$$

In particular, if $\mathfrak{p}$ is a prime ideal: $$M_\mathfrak{p} \otimes_{A_\mathfrak{p}} N_\mathfrak{p} \cong (M \otimes_A N)_\mathfrak{p}$$

## Local properties

A ring $A$, or a module $N$ is said to have a `local property` $P$ if, $A$ has $\mathfrak{p} \iff A_\mathfrak{p}$ has $\mathfrak{p}$, or $N$ has $\mathfrak{p} \iff N_\mathfrak{p}$ has $\mathfrak{p}$, for prime ideal $\mathfrak{p}$. Now, for example, the following statements are equivalent:

- $N = 0$.
- $N_\mathfrak{p} = 0$, for all prime ideals $\mathfrak{p}$ of $A$.
- $N_\mathfrak{m} = 0$, for all maximal ideals $\mathfrak{m}$ of $A$.

Since (i) $\implies$ (ii) $\implies$ (iii) is obvious, let us prove (i), given (iii). Suppose $N \neq 0$, then $x \in N$, and let $I = \text{Ann}(x)$. $I$ is an ideal, and therefore must be contained within some maximal ideal $\mathfrak{m}$. Let $x/1 \in N_\mathfrak{m}$. Since $N_\mathfrak{m} = 0$, $x/1 = 0$, and hence, $x$ is killed by some element of $N - \mathfrak{m}$. But this is impossible since $\text{Ann}(x) \subseteq \mathfrak{m}$.

Let $\varphi: S \rightarrow T$ be an $A$-module homomorphism. Then, the following statements are equivalent:

- $\varphi$ is injective.
- $\varphi_M : S_\mathfrak{p} \rightarrow T_\mathfrak{p}$ is injective for each prime ideal $\mathfrak{p}$.
- $\varphi_M : S_\mathfrak{m} \rightarrow T_\mathfrak{m}$ is injective for each maximal ideal $\mathfrak{m}$.

Flatness is a local property. For any $A$-module $N$, the following statements are equivalent:

- $N$ is a flat $A$-module.
- $N_\mathfrak{p}$ is flat for each prime ideal $\mathfrak{p}$.
- $N_\mathfrak{m}$ is flat for each maximal ideal $\mathfrak{m}$.

## Primary decomposition

Just as prime ideals are generalizations of primes in $\mathbb{Z}$, in some sense, powers of primes can be generalized to primary ideals: ideal $Q$ of $A$ is primary if $Q \neq A$ and $$fg \in Q \implies f \in Q \text{ or } g^{n} \in Q \text{, for some } n > 0$$

In other words, for $Q$ to be primary, $$A/Q \neq 0 \text{, and every zerodivisor in } A/Q \text{ is nilpotent}$$

Clearly, every prime ideal is primary. Also, the contraction of a primary ideal is primary, for if $f : A \rightarrow B$ and $Q$ is primary in $B$, $A/Q^c \cong B/Q$.

Let $Q$ be a primary ideal in $A$. Then $\sqrt{Q}$ is the smallest prime ideal containing $Q$. To prove this, it suffices to show that $\sqrt{Q}$ is prime. Let $xy \in \sqrt{Q}$. Then $(xy)^m \in Q, m \gt 0$, which implies that either $x^m \in Q$ or $y^{mn} \in Q, n \gt 0$; i.e. $x \in \sqrt{Q}$ or $y \in \sqrt{Q}$. If $P = \sqrt{Q}$, then $Q$ is said to be `$P$-primary`.

Examples:

- $0$ and $P^n$ are primary in $\mathbb{Z}$, where $P$ is a prime number.
- Let $A = k[x, y]$ and $Q = (x, y^2)$. Then, $A/Q \cong k[y]/(y^2)$, in which all zero-divisors are multiples of $y$, hence nilpotent. Hence, $Q$ is primary and its radical $P = (x, y)$ is such that $P^2 \subset Q \subset P$, so that a primary ideal is not necessarily a prime power.
- Let $I = (x^{2}, xy), \sqrt{I} = x$. $I$ is not primary because $xy \in I$, but $x \notin I$ and $y^{n} \notin I$.

If $\sqrt{P}$ is maximal, then $P$ is primary. In particular, the powers of a maximal ideal $M$ are $M$-primary. To prove this, let $\sqrt{I} = M$. The image of $M$ in $A/I$ is the nilradical of $A/I$, and hence $A/I$ only has one prime ideal. Every element of $A$ is either unit or nilpotent, so every zero-divisior of $A$ is nilpotent.

If $Q_i$ are $P$-primary, then so is $\bigcap_i Q_i$. To prove this, notice that $\sqrt{Q} = \sqrt{\bigcap_i Q_i} = \bigcap_i \sqrt{Q_i} = P$. Let $xy \in Q$, $y \notin Q$. Then, for some $i$, we have $xy \in Q_i$ and $y \notin Q_i$. Hence, $x \in P$ since $Q_i$ is primary.

Let $Q$ be $P$-primary, and $x \in A$. Then:

- If $x \in Q$, then the [ideal quotient](/ac/1#operations-on-ideals) $(Q : x) = (1)$.
- If $x \notin Q$ then $(Q : x)$ is $P$-primary, and hence $\sqrt{(Q : x)} = P$.
- If $x \notin P$, then $(Q : x) = Q$.

Let us prove (iii), since (i) and (ii) are obvious. Let $y \in (Q : x)$ and $xy \in Q$, and hence, since $x \notin Q$, $y \in P$. Hence, $Q \subseteq (Q : x) \subseteq P$; taking radicals, we get $\sqrt{(Q : x)} = P$. Let $yz \in (Q : x)$ so that $y \notin P$; then $xyz \in Q$, hence $xz \in Q$ and $z \in (Q : x)$.

The `primary decomposition` of ideal $I$ is given by finite intersection of primary ideals: $$I = \bigcap_i Q_i$$

An ideal is said to be `decomposable` if such a decomposition exists.

`First uniqueness theorem`: Let $I$ be decomposable, and $I = \bigcap_i Q_i$ be its minimal primary decomposition. Let $P_i = \sqrt{Q_i}$; then prime ideals $P_i$ are precisely the ones that occur in the set $\sqrt{(I : x)}$, for $x \in A$, and are hence independent of a particular primary decomposition of $I$. To prove this, let $x \in A$, so that $(I : x) = \bigcap_i (Q_i : x)$. Hence $\sqrt{(I : x)} = \sqrt{\bigcap_i (Q_i : x)} = \bigcap_{x \notin Q_j} P_j$. Suppose $\sqrt{(I : x)}$ is prime; then $\sqrt{(I : x)} = P_j$, for some $j$. Hence, every prime ideal of the form $\sqrt{(I : x)}$ is one of $P_j$. Conversely, for each $i$, there exists $x_i \notin Q_i$, $x_i = \bigcap_{i \neq j} Q_j$, since the decomposition is minimal. Hence, $\sqrt{(I : x_i)} = P_i$.

Let $I$ be a decomposable ideal. Then, any prime ideal $P \supseteq I$ contains a minimal prime ideal belonging to $I$, and thus the minimal prime ideals of $I$ are exactly the minimal elements in the set of all prime ideals containing $I$.

Let $I = \bigcap_i Q_i$ be a minimal primary primary decomposition, and $\sqrt{Q_i} = P_i$. Then, $$\bigcup_i P_i = \{x \in A \mid (I : x) \neq I\}$$

In particular, if the zero ideal is decomposable, then the set $D$ of zero-divisors of $A$ is the union of prime ideals belonging to $0$.

Let $S$ be a multiplicatively closed subset of $A$, and let $Q$ be $P$-primary. Then:

- If $S \cap P \neq \phi$, then $S^{-1} Q = S^{-1} A$.
- If $S \cap P = \phi$, then $S^{-1} Q$ is $S^{-1} P$-primary, and its contraction in $A$ is $Q$.

Let $S$ be a multiplicatively closed subset of $A$, and $I = \bigcap_i Q_i$ be the minimal primary decomposition of $I$ in $A$. Let $P_i = \sqrt{Q_i}$, and suppose $Q_i$ is numbered so that $S$ meets $P_{m + 1} \ldots P_n$, but not $P_1 \ldots P_m$. Then: $$S^{-1} I = \bigcap_{i = 1}^m S^{-1} Q_i, \quad S(I) = \bigcap_{i = 1}^m Q_i$$

Moreover, these are minimal primary decompositions.

The set $\Sigma$ of prime ideals belonging to $I$ is said to be `isolated` if it satisfies the following condition: if $P'$ is a prime ideal belonging to $I$, and $P' \supseteq P$ for some $P \in \Sigma$, then $P' \in \Sigma$.

`Second uniqueness theorem`: Let $I = \bigcap_i Q_i$ be a minimal primary decomposition of ideal $I$, and let $\{P_{i_1} \ldots P_{i_n}\}$ be an isolated set of prime ideals of $I$. Then, $Q_1 \cap \ldots \cap Q_n$ is independent of the decomposition. In particular, isolated primary components are uniquely determined by $I$.

## Integral dependence

Let $B$ be a ring, and $A$ a subring of $B$. Then, element $x$ of $B$ is said to be `integral over $A$` if $x$ is the root of the polynomial with coefficients in $A$, that is, if $x$ satisfies: $$x^n + a_1 x^{n - 1} + \ldots + a_n = 0$$

Let $A = \mathbb{Z}, B = \mathbb{Q}$. If a rational number $x = r/s$ is integral over $\mathbb{Z}$, where $r, s$ have no common factor, we have: $$r^n + a_1 r^{n - 1} s + \ldots + a_n s^n = 0$$

the $a_i$ being rational integers. Hence, $s$ divides $r^n$, $s = \pm 1$, and $x \in \mathbb{Z}$.

The following are equivalent:

- $x \in B$ is integral over $A$.
- $A[x]$ is a finitely generated $A$-module.
- $A$ is contained in a subring $C$ of $B$, such that $C$ is a finitely generated $A$-module.
- There exists a faithful $A[x]$-module $M$ which is finitely generated as an $A$-module.

To prove (i) $\implies$ (ii), we have: $$x^{n + r} = -(a_1 x^{n + r - 1} + \ldots + a_n x^n)$$

for all $r \geq 0$. Hence, by induction, all positive powers of $x$ lie in the $A$-module generated by $1, x, \ldots, x^{n - 1}$. Hence, $A[x]$ lie in the $A$-module generated by $1, x, \ldots, x^{n - 1}$. Hence, $A[x]$ is generated by $1, x, \ldots, x^{n - 1}$.

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 going-up theorem

## The going-down theorem

## Valuations

## Chain conditions

## Noetherian rings

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

## Artin rings

## Discrete valuation rings

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

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

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

## Dedekind domains

## Completions

## Graded rings and modules

## Associated graded ring

## The Nullstellansatz

Variety $V(J) = \{P = (a_1, \ldots, a_n) \mid f(P) = 0\, \forall f \subset J\}$

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

The Nullstellansatz states that:

- If $J \subsetneq k[X_1 \ldots X_n]$ 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: $$X \subset Y \Rightarrow I(X) \supset I(Y)$$ $$I \subset J \Rightarrow V(I) \supset V(J)$$

## 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:

- $A$ 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.

## 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 $\text{Ann } m = p$.

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