가환대수학
Associated Primes of Ideals
Prime avoidance, associated primes, and their properties
This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.
Prime avoidance lemma
Definition 1 For a ring \(A\) and an \(A\)-module \(M\), a prime ideal \(\mathfrak{p}\) of \(A\) is called an associated prime ideal of \(M\) if \(\mathfrak{p}=\ann(x)\) for some \(x\in M\). We write \(\Ass M\) for the set of associated primes.
In the special case where \(M=\mathfrak{a}\) is an ideal of \(A\), the associated primes of \(\mathfrak{a}\) are by convention not \(\Ass \mathfrak{a}\), but \(\Ass A/\mathfrak{a}\).
Henceforth, whenever we write \(\ann(x)\) for the annihilator of an element, we always assume \(x\neq 0\). Since \(\ann(0)=A\), which cannot be a prime ideal, this convention is merely for notational convenience when dealing with associated primes.
By definition, \(\mathfrak{p}\) being an associated prime of \(M\) is equivalent to \(A/\mathfrak{p}\) being a submodule of \(M\). This follows immediately from applying the first isomorphism theorem to the map \(A \rightarrow M\) sending \(1\mapsto x\).
In this post we examine various properties of associated prime ideals. The following lemma plays an important role.
Lemma 2 (Prime avoidance lemma) Let \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n, \mathfrak{b}\) be ideals of \(A\), and suppose \(\mathfrak{b}\subseteq \mathfrak{a}_1\cup\cdots\cup \mathfrak{a}_n\). If \(A\) contains an infinite field, or at most two of the \(\mathfrak{a}_i\) are not prime ideals, then \(\mathfrak{b}\) is contained in one of \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n\).
Moreover, if \(A\) is graded and \(\mathfrak{b}\) is a homogeneous ideal generated by homogeneous elements of positive degree, and all \(\mathfrak{a}_i\) are prime ideals, then the conclusion still holds even if we only assume that the homogeneous elements of \(\mathfrak{b}\) lie in \(\mathfrak{a}_1\cup\cdots\cup \mathfrak{a}_n\).
Proof
If \(A\) contains an infinite field \(\mathbb{K}\), then viewing each ideal as a \(\mathbb{K}\)-vector space we have
\[\mathfrak{b}=\bigcup_{i=1}^n (\mathfrak{b}\cap \mathfrak{a}_i)\]and the claim is obvious because no \(\mathbb{K}\)-vector space can be expressed as a finite union of its proper subspaces.
The remaining case is proved by induction on \(n\). The case \(n=1\) is trivial.
For larger \(n\), if the inclusion in the hypothesis still holds after omitting any one of \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n\), then this is resolved by the inductive hypothesis, so we may assume this is not the case. That is, we may assume there always exists \(x_i\in \mathfrak{b}\) with \(x_i\not\in \bigcup_{j\neq i}\mathfrak{a}_j\), and by the condition on \(x_i\) we necessarily have \(x_i\in \mathfrak{a}_i\).
Now for the case \(n=2\), consider the element \(x_1+x_2\) of \(\mathfrak{b}\). If \(x_1+x_2\in \mathfrak{a}_1\), then \(x_2=(x_1+x_2)-x_1\in \mathfrak{a}_1\), which is a contradiction; similarly \(x_1+x_2\) cannot lie in \(\mathfrak{a}_2\). This contradicts the assumption that \(\mathfrak{b}\subseteq \mathfrak{a}_1\cup \mathfrak{a}_2\).
For \(n\geq 3\) we use a similar idea. From the given hypothesis, at least one of \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n\) is a prime ideal, so without loss of generality we may assume \(\mathfrak{a}_1\) is prime. Now consider the element \(x_1+x_2x_3\cdots x_n\) of \(\mathfrak{b}\). First, for each \(j\geq 2\) we have \(x_j\not\in \mathfrak{a}_1\), and since \(\mathfrak{a}_1\) is prime, \(x_2x_3\cdots x_n\not\in \mathfrak{a}_1\); since \(x_1\in \mathfrak{a}_1\), this element does not belong to \(\mathfrak{a}_1\). Also, for each \(j\geq 2\), since \(x_j\in \mathfrak{a}_j\) we have \(x_2x_3\cdots x_n\in \mathfrak{a}_j\) and \(x_1\not\in \mathfrak{a}_j\), so this element does not belong to \(\mathfrak{a}_j\) either. This contradicts the assumption that \(\mathfrak{b}\subseteq \mathfrak{a}_1\cup\cdots\cup \mathfrak{a}_n\).
In the graded ring case, we simply multiply each \(x_i\) by itself enough times to match the degree of \(x_2x_3\cdots\), adjusting the degree as needed.
In practice, the latter hypothesis is satisfied more often and is used more frequently than the former.
Associated prime ideals
Now we examine \(\Ass M\) in more detail.
Proposition 3 Fix an \(A\)-module \(M\), and suppose \(\mathfrak{a}\) is maximal among ideals of the form \(\ann(x)\) for \(x\in M\). Then \(\mathfrak{a}\) is a prime ideal.
Proof
Suppose \(ab\in \mathfrak{a}\) and \(b\not\in \mathfrak{a}\); we must show \(a\in \mathfrak{a}\). Write \(\mathfrak{a}=\ann(x)\). Then by hypothesis \(abx=0\) and \(bx\neq 0\). Now \(\mathfrak{a}\subseteq\ann(bx)\), so by maximality of \(\mathfrak{a}\) we have \(\mathfrak{a}=\ann(bx)\). Hence
\[(a)+\mathfrak{a}\subseteq \ann(bx)=\mathfrak{a}\]and we conclude \(a\in \mathfrak{a}\).
The \(\mathfrak{a}\) obtained from the above proposition is a prime ideal and also the annihilator of some element, so by definition it belongs to \(\Ass M\).
On the other hand, by §Properties of Localization, ⁋Lemma 3, for any \(A\)-module \(M\), an element \(x\in M\) is zero if and only if its image under the localization map \(\epsilon_\mathfrak{m}: M \rightarrow M_\mathfrak{m}\) is zero for every maximal ideal \(\mathfrak{m}\); thus to show \(x=0\) it suffices to show \(\epsilon_\mathfrak{p}(x)=0\) for all prime ideals \(\mathfrak{p}\). The following corollary can be understood in the same spirit.
Corollary 4 Let \(M\) be a module over a Noetherian ring \(A\). Then the following hold.
- For an element \(x\) of \(M\), we have \(x=0\) if and only if \(\epsilon_\mathfrak{p}(x)=0\) for every maximal associated prime \(\mathfrak{p}\) of \(M\).
- For a submodule \(L\) of \(M\), we have \(L=0\) if and only if \(L_\mathfrak{p}=0\) for all \(\mathfrak{p}\in\Ass M\).
- An \(A\)-linear map \(u:M \rightarrow N\) is injective if and only if \(u_\mathfrak{p}\) is injective for every \(\mathfrak{p}\in \Ass M\).
Proof
For the first result, since \(A\) is Noetherian, for any nonzero \(x\in M\) we can choose an ideal \(\mathfrak{p}\) maximal among annihilator ideals containing \(\ann(x)\), and by Proposition 3 we have \(\mathfrak{p}\in \Ass M\). Hence \(x/1\) is nonzero in \(M_\mathfrak{p}\). The second result is immediate from the first, and the third follows from the second by taking \(L=\ker u\).
The goal of this post is to prove Theorem 7. For this we need the following two lemmas.
Lemma 5 Given a short exact sequence of \(A\)-modules
\[0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0\]we have
\[\Ass M'\subseteq \Ass M\subseteq (\Ass M')\cup (\Ass M'').\]Proof
The first inclusion is obvious. For the second inclusion, assume \(\mathfrak{p}\in\Ass M\) does not belong to \(\Ass M'\); we show it belongs to \(\Ass M''\). If \(\mathfrak{p}=\ann(x)\) for some \(x\in M\), then \(Ax\cong A/\mathfrak{p}\). Since \(\mathfrak{p}\) is prime, for any nonzero \(ax\in Ax\) we have
\[a'\in\ann(ax)\iff a'ax=0\iff a'a\in \mathfrak{p}\iff a'\in \mathfrak{p}\]so \(\ann(ax)=\mathfrak{p}\). Here the last equivalence uses \(ax\neq 0\) and \(Ax\cong A/\mathfrak{p}\) to obtain \(a\not\in \mathfrak{p}\). This equality shows in particular that any nonzero submodule of \(Ax\) must have \(\mathfrak{p}\) as its annihilator; combining this with the fact that \(\mathfrak{p}\not\in \Ass M'\), we see that \(Ax\cap M'=0\). Hence the image of \(Ax\) in \(M''\) is isomorphic to \(Ax\), and we conclude \(\mathfrak{p}\in \Ass M''\).
Lemma 6 For a finitely generated module \(M\) over a Noetherian ring \(A\), there exists a filtration
\[0=M_0\subseteq M_1\subseteq\cdots\subseteq M_n=M,\qquad \text{$M_k/M_{k-1}\cong A/\mathfrak{p}_k$ for some prime $\mathfrak{p}_k$, for all $k$}\]satisfying the above conditions.
Proof
First, using Proposition 3 we can find an associated prime \(\mathfrak{p}_1\in\Ass M\) of \(M\), and thus there exists a submodule \(M_1\) with \(M_1\cong A/\mathfrak{p}_1\). Applying the same argument to \(M/M_1\) yields \(M_2\), and repeating this process, we obtain the desired conclusion from the fact that \(M\) is Noetherian.
Theorem 7 For a nonzero finitely generated module \(M\) over a Noetherian ring \(A\), the following hold.
- \(\Ass M\) is a nonempty finite set, and each of its elements contains \(\ann M\). Moreover, the minimal prime ideals among those containing \(\ann M\) are all contained in \(\Ass M\).
- The union of the associated primes consists of \(0\) together with all zero-divisors of \(M\).
-
For a multiplicative subset \(S\) of \(A\), the following formula holds:
\[\Ass_{S^{-1}A}S^{-1}M=\{\mathfrak{p}S^{-1}A\mid \mathfrak{p}\in\Ass M, \mathfrak{p}\cap S=\emptyset\}.\]
Proof
For the first result, that \(\Ass M\) is nonempty follows from Proposition 3, and that each element of \(\Ass M\) contains \(\ann M\) is obvious. Now, by Lemma 5, considering the short exact sequence
\[0 \rightarrow M_{n-1} \rightarrow M_n \rightarrow M_n/M_{n-1} \rightarrow 0\]we have \(\Ass M_n \subseteq \Ass M_{n-1}\cup \Ass M_n/M_{n-1}=\Ass M_{n-1}\cup \Ass A/\mathfrak{p}_n\).
On the other hand, for any prime ideal \(\mathfrak{p}\) we can show \(\Ass(A/\mathfrak{p})=\{\mathfrak{p}\}\) as follows. Let \(\mathfrak{q}\in \Ass(A/\mathfrak{p})\) and write \(\mathfrak{q}=\ann(x+\mathfrak{p})\). Then first \(\mathfrak{p}\subseteq \mathfrak{q}\) is obvious, because for any \(p\in \mathfrak{p}\),
\[p(x+\mathfrak{p})=px+\mathfrak{p}=0+\mathfrak{p}\]holds. If \(\mathfrak{q}\not\subseteq \mathfrak{p}\), then there exists \(q\in \mathfrak{q}\setminus \mathfrak{p}\). Then from
\[0=q(x+\mathfrak{p})=qx+\mathfrak{p}\]we know \(qx\in \mathfrak{p}\), and since \(q\not\in \mathfrak{p}\) we conclude \(x+\mathfrak{p}\) must have been \(0\). But this means \(\mathfrak{q}=A\), a contradiction.
Therefore, repeating
\[\Ass M \subseteq \Ass M_{n-1}\cup \{ \mathfrak{p}_n\}\subseteq \Ass M_{n-2}\cup \{\mathfrak{p}_n,\mathfrak{p}_{n-1}\}\cdots\]in the above manner yields the finiteness in the first result.
The remaining part of the first result is obtained by proving the third result. Before that, for the second result, if \(a\in A\) belongs to the annihilator of some nonzero \(x\in M\), then considering a maximal annihilator ideal containing this annihilator, which belongs to \(\Ass M\), makes this direction obvious. Conversely, if \(\mathfrak{p}=\ann(x)\in \Ass M\), then any nonzero element of \(\mathfrak{p}\) is a zero-divisor of \(M\) annihilating \(x\). For the third result, paying attention to notation and using §Properties of Localization, ⁋Proposition 5 suffices.
Assuming the third result, the remaining part is also obvious. If \(\mathfrak{p}\) is minimal among prime ideals containing \(\ann M\), then using the third result we can consider the maximal ideal \(\mathfrak{p}\) in the localization \(A_\mathfrak{p}\); since \(\mathfrak{p}\) is the unique prime ideal containing \(\ann M\), we must have \(\mathfrak{p}\in \Ass M\).
From this we can extract much information about \(M\). For example, when \(M=A\) and \(A\) is reduced, we obtain the following corollary.
Corollary 8 For a reduced Noetherian ring \(A\), let \(K\) be the total ring of fractions of \(A\). Then \(K\) is a finite product of fields.
Proof
First, taking \(M=A\) we have \(\ann A=\{0\}\). Hence the minimal prime ideals \(\mathfrak{p}_1,\ldots, \mathfrak{p}_k\) of \(A\) all belong to \(\Ass M\), and their union consists of the zero-divisors of \(A\).
Moreover, if \(A\) is reduced, then their union is exactly the set of all zero-divisors of \(A\). To verify this, first observe from the assumption that \(A\) is reduced that
\[(0)=\mathfrak{N}(A)=\bigcap_\text{\scriptsize$\mathfrak{p}$ a prime}\mathfrak{p}\supseteq \bigcap_{i=1}^k \mathfrak{p}_i\]holds. Then for any zero-divisor \(a\neq 0\) and \(b\neq 0\) with \(ab=0\), there must exist some \(\mathfrak{p}_i\) with \(b\not\in \mathfrak{p}_i\), and then since \(ab=0\in \mathfrak{p}_i\) we must have \(a\in \mathfrak{p}_i\).
Therefore, the total ring of fractions \(K\) of \(A\) is \(K=S^{-1}A\) for \(S=A\setminus(\mathfrak{p}_1\cup\cdots\cup \mathfrak{p}_k)\).
Now by §Localization, ⁋Proposition 8, the prime ideals of \(S^{-1}A\) correspond to the prime ideals of \(A\) disjoint from \(S\), that is, the prime ideals \(\mathfrak{p}\) satisfying \(\mathfrak{p}\subseteq \mathfrak{p}_1\cup\cdots\cup \mathfrak{p}_k\). Applying Lemma 2 (Prime avoidance lemma) to such \(\mathfrak{p}\), since the \(\mathfrak{p}_i\) are all prime ideals the hypothesis is satisfied and we obtain \(\mathfrak{p}\subseteq \mathfrak{p}_i\) for some \(i\); since \(\mathfrak{p}_i\) is a minimal prime ideal we get \(\mathfrak{p}=\mathfrak{p}_i\). Thus the prime ideals of \(S^{-1}A\) are exactly \(\mathfrak{p}_1S^{-1}A,\ldots,\mathfrak{p}_kS^{-1}A\), and since this correspondence preserves inclusion and there are no inclusion relations among distinct minimal prime ideals, each of these is a maximal ideal of \(S^{-1}A\) and hence they are pairwise comaximal.
On the other hand, since \(A\) is reduced, \(S^{-1}A\) is also reduced: if \((a/s)^n=0\) then \(ta^n=0\) for some \(t\in S\), and then \((ta)^n=t^{n-1}(ta^n)=0\) implies \(ta=0\), i.e. \(a/s=0\), because \(A\) is reduced. Then since the nilradical of \(S^{-1}A\) is \(0\) and from the list of prime ideals obtained above we have \(\bigcap_{i=1}^k\mathfrak{p}_iS^{-1}A=0\), by [Ring Theory] §Chinese Remainder Theorem, ⁋Proposition 6 we obtain
\[K=S^{-1}A\cong \prod_{i=1}^k S^{-1}A/\mathfrak{p}_iS^{-1}A\]and since each \(\mathfrak{p}_iS^{-1}A\) is a maximal ideal, the factors on the right are all fields.
References
[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.
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