[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 \(\mathfrak{p}\) 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 \(f\).
  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 \(\mathfrak{p}\) 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.
  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 \(\mathfrak{p}\) 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:

  1. \(N = 0\).
  2. \(N_\mathfrak{p} = 0\), for all prime ideals \(\mathfrak{p}\) of \(A\).
  3. \(N_\mathfrak{m} = 0\), 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:

  1. \(\varphi\) is injective.
  2. \(\varphi_M : S_\mathfrak{p} \rightarrow T_\mathfrak{p}\) is injective for each prime ideal \(\mathfrak{p}\).
  3. \(\varphi_M : S_\mathfrak{m} \rightarrow T_\mathfrak{m}\) is injective for each maximal ideal \(\mathfrak{m}\).

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

  1. \(N\) is a flat \(A\)-module.
  2. \(N_\mathfrak{p}\) is flat for each prime ideal \(\mathfrak{p}\).
  3. \(N_\mathfrak{m}\) 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`.


  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

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:

  1. \(x \in B\) is integral over \(A\).
  2. \(A[x]\) is a finitely generated \(A\)-module.
  3. \(A\) is contained in a subring \(C\) of \(B\), such that \(C\) is a finitely generated \(A\)-module.
  4. 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


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 \(P\).