Associated Primes

Prime avoidance lemma

Definition 1 For a ring \(A\) and an \(A\)-module \(M\), a prime ideal of \(A\), \(\mathfrak{p}\), is called an associated prime ideal of \(M\) if for some \(x\in M\) we have \(\mathfrak{p}=\ann(x)\). We write \(\Ass M\) for the collection of associated primes.

In the special case where \(M=\mathfrak{a}\) is an ideal of \(A\), the associated prime ideal of \(\mathfrak{a}\) means by convention not \(\Ass \mathfrak{a}\), but \(\Ass A/\mathfrak{a}\).

Henceforth, whenever we write the annihilator of an element as \(\ann(x)\), we always assume that \(x\neq 0\). In any case, if \(x=0\), then \(\ann(x)=A\), which cannot be a prime ideal, so when dealing with associated prime ideals this convention is merely for notational convenience.

By definition, \(\mathfrak{p}\) being an associated prime of \(M\) is equivalent to \(A/\mathfrak{p}\) being a submodule of \(M\). This follows immediately by taking \(A \rightarrow M\) to be \(1\mapsto x\) and applying the first isomorphism theorem.

In this post we examine various properties of associated prime ideals. In this process, the following lemma plays an important role.

Lemma 2 (Prime avoidance lemma) Given ideals of \(A\), \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n, \mathfrak{b}\), suppose that \(\mathfrak{b}\subseteq \mathfrak{a}_1\cup\cdots\cup \mathfrak{a}_n\). 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 \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n\).

In addition, if \(A\) is graded, \(\mathfrak{b}\) is a homogeneous ideal generated by homogeneous elements of positive degree, and all \(\mathfrak{a}_i\) are prime ideals, the conclusion still holds even if we assume that the homogeneous elements of \(\mathfrak{b}\) belong to \(\mathfrak{a}_1\cup\cdots\cup \mathfrak{a}_n\).

Proof

If \(A\) contains an infinite field \(\mathbb{K}\), viewing each of the ideals as a \(\mathbb{K}\)-vector space we have

\[\mathfrak{b}=\bigcup_{i=1}^n (\mathfrak{b}\cap \mathfrak{a}_i)\]

and the claim is trivial since no \(\mathbb{K}\)-vector space can be expressed as a finite union of its proper subspaces.

The remaining case is proved by induction on \(n\). If \(n=1\), there is nothing to prove.

For larger \(n\), if the inclusion in the condition holds even after removing one of \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n\), this can be resolved by the induction hypothesis, so it suffices to consider only the case where this does not hold. That is, it suffices to assume that satisfying \(x_i\not\in \bigcup_{j\neq i}\mathfrak{a}_j\), there always exists \(x_i\in \mathfrak{b}\), and by the condition on \(x_i\) we necessarily have \(x_i\in \mathfrak{a}_i\).

Now for the case \(n=2\), consider in \(\mathfrak{b}\) the element \(x_1+x_2\). 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, and similarly \(x_1+x_2\) cannot be an element of \(\mathfrak{a}_2\) either. 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 conditions, at least one of \(\mathfrak{a}_1,\ldots, \mathfrak{a}_n\) is a prime ideal, so without loss of generality we may assume that \(\mathfrak{a}_1\) is a prime ideal. Now consider in \(\mathfrak{b}\) the element \(x_1+x_2x_3\cdots x_n\). First, for each \(j\geq 2\) we have \(x_j\not\in \mathfrak{a}_1\), and since \(\mathfrak{a}_1\) is a prime ideal, \(x_2x_3\cdots x_n\not\in \mathfrak{a}_1\); since \(x_1\in \mathfrak{a}_1\), this element does not belong to \(\mathfrak{a}_1\). Also, for each \(j\geq 2\), from \(x_j\in \mathfrak{a}_j\) we have \(x_2x_3\cdots x_n\in \mathfrak{a}_j\), and since \(x_1\not\in \mathfrak{a}_j\), 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 case of a graded ring, it suffices to adjust the degrees by multiplying \(x_i\) repeatedly so that it has the same degree as \(x_2x_3\cdots\).

In particular, the above lemma is used frequently satisfying the latter condition rather than the former condition.

Associated prime ideals

Now we examine \(\Ass M\) in a bit more detail.

Proposition 3 Fix an \(A\)-module \(M\), and suppose \(\mathfrak{a}\) is maximal among ideals for \(x\in M\) of the form \(\ann(x)\). Then \(\mathfrak{a}\) is a prime ideal.

Proof

Suppose \(ab\in \mathfrak{a}\) and \(b\not\in \mathfrak{a}\); we must show \(a\in \mathfrak{a}\). Let \(\mathfrak{a}=\ann(x)\). Then by hypothesis \(abx=0\) and \(bx\neq 0\). Now \(\mathfrak{a}\subseteq\ann(bx)\), so by the maximality of \(\mathfrak{a}\) we have \(\mathfrak{a}=\ann(bx)\). From this, since

\[(a)+\mathfrak{a}\subseteq \ann(bx)=\mathfrak{a}\]

we see that \(a\in \mathfrak{a}\).

The \(\mathfrak{a}\) obtained from the above proposition is a prime ideal and at the same time the annihilator of some element, so by definition it belongs to \(\Ass M\).

Meanwhile, by §Properties of Localization, ⁋Lemma 3, for any \(A\)-module \(M\), that \(x\in M\) is \(0\) is equivalent to its images under the localization \(\epsilon_\mathfrak{m}: M \rightarrow M_\mathfrak{m}\) all being \(0\), and thus to show \(x=0\), it suffices to show that for all prime ideals \(\mathfrak{p}\), \(\epsilon_\mathfrak{p}(x)=0\). The following corollary can also be understood in the same context.

Corollary 4 Over a Noetherian ring \(A\), consider a module \(M\). Then the following hold.

1. For an element of \(M\), \(x\), that \(x=0\) is equivalent to: whenever a maximal associated prime of \(M\), \(\mathfrak{p}\), is given, \(\epsilon_\mathfrak{p}(x)=0\).
2. For a submodule of \(M\), \(L\), having \(L=0\) is equivalent to having, for all \(\mathfrak{p}\in\Ass M\), \(L_\mathfrak{p}=0\).
3. An \(A\)-linear map \(u:M \rightarrow N\) is injective if and only if, for every \(\mathfrak{p}\in \Ass M\), \(u_\mathfrak{p}\) is injective.

Proof

First, in the case of the first result, from the hypothesis that \(A\) is Noetherian, whenever any non-\(0\) element \(x\in M\) is given, among annihilator ideals containing \(\ann(x)\) we can choose a maximal ideal \(\mathfrak{p}\), and by Proposition 3, \(\mathfrak{p}\in \Ass M\). Therefore, in \(M_\mathfrak{p}\), \(x/1\) does not become \(0\). The second result is immediate from the first result, and the third result follows by setting \(L=\ker u\) in the second result.

The goal of this post is to prove Theorem 7. For this, the following two lemmas are needed.

Lemma 5 Suppose given a short exact sequence of \(A\)-modules

\[0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0\]

Then

\[\Ass M'\subseteq \Ass M\subseteq (\Ass M')\cup (\Ass M'')\]

holds.

Proof

The first inclusion is obvious. For the second inclusion, assume \(\mathfrak{p}\in\Ass M\) does not belong to \(\Ass M'\), and let us show that it belongs to \(\Ass M''\). If, for \(x\in M\), we have \(\mathfrak{p}=\ann(x)\), then \(Ax\cong A/\mathfrak{p}\). However, since \(\mathfrak{p}\) is a prime ideal, for any non-\(0\) \(ax\in Ax\), we have

\[a'\in\ann(ax)\iff a'ax=0\iff a'a\in \mathfrak{p}\iff a'\in \mathfrak{p}\]

so that \(\ann(ax)=\mathfrak{p}\). Here, the last equivalence was obtained from the fact that \(ax\neq 0\) and \(Ax\cong A/\mathfrak{p}\) by using \(a\not\in \mathfrak{p}\). Now, this equality shows in particular that any submodule of \(Ax\) other than \(0\) must have \(\mathfrak{p}\) as its annihilator; combining this with the fact that \(\mathfrak{p}\not\in \Ass M'\), we see that \(Ax\cap M'=0\). Therefore, the image of \(Ax\) in \(M''\) is isomorphic to \(Ax\), and from this we see that \(\mathfrak{p}\in \Ass M''\).

Lemma 6 Over a Noetherian ring \(A\), for a finitely generated module \(M\), one can find 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 stated condition.

Proof

First, using Proposition 3, we can find an associated prime of \(M\), \(\mathfrak{p}_1\in\Ass M\), and thus there exists a submodule satisfying \(M_1\cong A/\mathfrak{p}_1\), denoted \(M_1\). Applying the same reasoning to \(M/M_1\), we can obtain \(M_2\), 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.

1. \(\Ass M\) is a nonempty finite set, and each of its elements contains \(\ann M\). Moreover, the minimal prime ideals in the set of prime ideals containing \(\ann M\) are all contained in \(\Ass M\).
2. The union of the associated primes consists of \(0\) and all the zero-divisors of \(M\).

3. For a multiplicative subset \(S\) of \(A\), the equation

   \[\Ass_{S^{-1}A}S^{-1}M=\{\mathfrak{p}S^{-1}A\mid \mathfrak{p}\in\Ass M, \mathfrak{p}\cap S=\emptyset\}\]

   holds.

Proof

First, 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. Meanwhile, by Lemma 5, considering the following 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 that \(\Ass(A/\mathfrak{p})=\{\mathfrak{p}\}\) 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. This is because for any \(p\in \mathfrak{p}\),

\[p(x+\mathfrak{p})=px+\mathfrak{p}=0+\mathfrak{p}\]

holds. If \(\mathfrak{q}\not\subseteq \mathfrak{p}\), there exists \(q\in \mathfrak{q}\setminus \mathfrak{p}\). Then from

\[0=q(x+\mathfrak{p})=qx+\mathfrak{p}\]

we know that \(qx\in \mathfrak{p}\), and then from \(q\not\in \mathfrak{p}\) we see that \(x+\mathfrak{p}\) must have been \(0\). However, this means \(\mathfrak{q}=A\), which is 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 ideal of some nonzero \(x\in M\), we can consider a maximal annihilator ideal containing this annihilator ideal, and since this belongs to \(\Ass M\), it is 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, one can use §Properties of Localization, ⁋Proposition 5 while paying attention to the notation.

Assuming the third result, the remaining part is also obvious. If \(\mathfrak{p}\) is minimal among the prime ideals containing \(\ann M\), then using the third result we can consider the maximal ideal \(\mathfrak{p}\) in the localization \(A_\mathfrak{p}\), and since the only prime ideal containing \(\ann M\) is \(\mathfrak{p}\), we must have \(\mathfrak{p}\in \Ass M\).

Through this, we can obtain 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 the total ring of fractions of \(A\) be \(K\). Then \(K\) is a finite product of fields.

Proof

First, setting \(M=A\), we have \(\ann A=\{0\}\). Therefore, the minimal prime ideals of \(A\), \(\mathfrak{p}_1,\ldots, \mathfrak{p}_k\), all belong to \(\Ass M\), and their union consists of zero-divisors of \(A\).

Moreover, if \(A\) is a reduced ring, their union is precisely the collection of all zero-divisors of \(A\). To verify this, first observe that from the assumption that \(A\) is reduced,

\[(0)=\mathfrak{N}(A)=\bigcap_\text{\scriptsize$\mathfrak{p}$ a prime}\mathfrak{p}= \bigcap_{i=1}^k \mathfrak{p}_i\]

holds. Then for any zero-divisor \(a\neq 0\) and any element with \(ab=0\) satisfying \(b\neq 0\), we must have \(b\not\in \mathfrak{p}_i\) for some \(\mathfrak{p}_i\), and then since \(ab=0\in \mathfrak{p}_i\), we must have \(a\in \mathfrak{p}_i\).

Therefore, the total ring of fractions of \(A\), \(K\), is given for \(S=A\setminus(\mathfrak{p}_1\cup\cdots\cup \mathfrak{p}_k)\) by \(K=S^{-1}A\).

Now by §Localization, ⁋Proposition 8, the prime ideals of \(S^{-1}A\) correspond to the prime ideals disjoint from \(S\) in \(A\), that is, to the prime ideals satisfying \(\mathfrak{p}\subseteq \mathfrak{p}_1\cup\cdots\cup \mathfrak{p}_k\), namely \(\mathfrak{p}\). Applying Lemma 2 (Prime avoidance lemma) to such \(\mathfrak{p}\), since the \(\mathfrak{p}_i\) are all prime ideals, the hypothesis is satisfied and for some \(i\) we have \(\mathfrak{p}\subseteq \mathfrak{p}_i\); since \(\mathfrak{p}_i\) is a minimal prime ideal, we obtain \(\mathfrak{p}=\mathfrak{p}_i\). That is, the prime ideals of \(S^{-1}A\) are precisely \(\mathfrak{p}_1S^{-1}A,\ldots,\mathfrak{p}_kS^{-1}A\), and since the above correspondence preserves inclusion and there are no inclusion relations between distinct minimal prime ideals, each of these is a maximal ideal of \(S^{-1}A\) and therefore they are pairwise comaximal.

On the other hand, since \(A\) is reduced, \(S^{-1}A\) is also reduced: if \((a/s)^n=0\), then for some \(t\in S\) we have \(ta^n=0\), and then since \((ta)^n=t^{n-1}(ta^n)=0\), from \(A\) being reduced we obtain \(ta=0\), that is, \(a/s=0\). 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.

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.

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