[ac/2] Modules of fractions, Primary decomposition

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:

  1. \(s \in S \implies f(s)\) is unit in \(S^{-1} A\).
  2. \(f(a) = 0 \implies as = 0\) for some \(s \in S\).
  3. 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:

  1. Let \(P\) be a prime ideal in \(A\). Then \(A - P\) is multiplicatively closed and is written as \(S^{-1} A = A_P\), the localization of \(A\) at \(P\). \(a/s\) with \(a \in P\) form an ideal \(M\) in \(A_P\). If \(b/t \neq M\), and \(b \notin P\), then \(b/t\) is unit. \(M\) is the only maximal ideal in \(P\), and \(A_P\) is a local ring. If \(A = \mathbb{Z}, P\) is the set of prime numbers, and \(A_P\) is the set of all rationals with denominators as powers of \(f\).
  2. \(S^{-1} A\) is the zero ring \(\iff 0 \in S\).
  3. 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 \(P\) a prime ideal in \(A\). Then, \(A_P\) is the ring of all rational functions \(g/h, g \notin P\). Let \(V\) be a variety defined by \(P\), that is to say the set \((x_1, \ldots, x_n) \in k^n\) such that \(f(x) = 0\) whenever \(f \in P\). Then, \(A_P\) 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](/ag).
  4. 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 \(M\):

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

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 \(P\) is a prime ideal: \[M_P \otimes_{A_P} N_P \cong (M \otimes_A N)_P\]

Local properties

A ring \(A\), or a module \(N\) is said to have a `local property` \(P\) if, \(A\) has \(P \iff A_P\) has \(P\), or \(N\) has \(P \iff N_P\) has \(P\), for prime ideal \(P\). Now, for example, the following statements are equivalent:

  1. \(N = 0\).
  2. \(N_P = 0\), for all prime ideals \(P\) of \(A\).
  3. \(N_M = 0\), for all maximal ideals \(M\) 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 \(M\). Let \(x/1 \in N_M\). Since \(N_M = 0\), \(x/1 = 0\), and hence, \(x\) is killed by some element of \(N - M\). But this is impossible since \(\text{Ann}(x) \subseteq M\).

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

  1. \(\varphi\) is injective.
  2. \(\varphi_M : S_P \rightarrow T_P\) is injective for each prime ideal \(P\).
  3. \(\varphi_M : S_M \rightarrow T_M\) is injective for each maximal ideal \(M\).

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

  1. \(N\) is a flat \(A\)-module.
  2. \(N_P\) is flat for each prime ideal \(P\).
  3. \(N_M\) is flat for each maximal ideal \(M\).

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`.


  1. \(0\) and \(P^n\) are primary in \(\mathbb{Z}\), where \(P\) is a prime number.
  2. 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.
  3. 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:

  1. If \(x \in Q\), then the [ideal quotient](/ac/1#operations-on-ideals) \((Q : x) = (1)\).
  2. If \(x \notin Q\) then \((Q : x)\) is \(P\)-primary, and hence \(\sqrt{(Q : x)} = P\).
  3. 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:

  1. If \(S \cap P \neq \phi\), then \(S^{-1} Q = S^{-1} A\).
  2. 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

\(y\) is algebraic over \(k\) if it satisfies the algebraic dependence relation: \[f(y) = a_m y^m + a_{m - 1} y^{m - 1} + \ldots + a_0\]

Over a field, it costs nothing to divide over \(a_m\), giving us the monic polynomial: \[f(y) = y^m + a_{m - 1} y^{m - 1} + \ldots + a_0 \]

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 going-up theorem

The going-down theorem


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:

  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 \(A\): \[A \text{ is a DVR} \leftrightarrow A \text{ is Noetherian}\]

Dedekind domains


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:

  1. If \(J \subsetneq k[X_1 \ldots X_n]\) 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 \(\mathcal{V}(I) = {P \in \text{Spec A} \mid P \subset I}\).


\(A_{p}\) 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 \(P\).