Associated prime ideals and regular elements in polynomial rings

This post is here to record a fact that ought to be better known and for which I ought to have known a proper proof beforehand.

Let's start by defining two thirds of the terms of the title.

By polynomial ring, I simply mean a ring of polynomials $A[T_1,T_2,\dots]$ over a commutative ring $A$, in any number of indeterminates. The most important case here will however be that of a ring of polynomials in one single indeterminate, everything would follow from that one by “standard” arguments. For simplicity, I'll write $A[T]$ for such  a ring of polynomials.

Regular elements in a ring $R$ are those which are not zero divisors. Explicitly, $a\in R$ is regular if $ab=0$ implies $ b=0$ for any $b\in R$. Or $a$ is a zero divisor if there exists a nonzero $b\in R$ such that $ab=0$.

The classic result linking these two notions that I have in mind is the following result.

Proposition. — Let $A$ be a commutative ring. If $P\in A[T]$ is not regular, then there exists a nonzero $a\in A$ such that $aP=0$.

There is a classic, rather elementary, proof of that result which I recall first. Let us first treat the case where is only one indeterminate. By assumption, there exists a nonzero polynomial $Q\in Q[T]$ such that $PQ=0$, and we may choose $Q$ to be of minimal degree. Write $P=\sum_{k=0}^m a_k T^k$ and $Q=\sum_{k=0}^n b_k T^k$, where $m$ and $n$ are the degrees of $P$ and $Q$, so that $a_m\neq 0$ and $b_n\neq 0$.

Lemma. — One has $a_k Q=0$ for all $k$.

The result being obvious for $k>m$, we may argue by decreasing induction on $k$, assuming that $a_\ell Q=0$ for all $\ell>k$. Then, in the product $PQ$, the term of largest degree is $a_k b_n T^{k+n}$, so that $a_k b_n=0$. In particular, the polynomial $a_kQ$ has degree $<n$. Since $P(a_k Q)=a_k PQ=0$, and by the choice of $Q$, we have $a_kQ=0$, as claimed.

Given the lemma, we see that $a_k b_\ell=0$ for all integers $k$ and $\ell$. Choosing any nonzero coefficient $b_\ell$ of $Q$, we obtain $b_\ell P=0$. This concludes the proof of the proposition when there is only one indeterminate.

To treat the general case, it is probably possible to make an induction on the number of indeterminates that appear in an equality $PQ=0$, where $Q$ is a nonzero element of $A[T]$. Another possibility, closer to the above proof, would be to choose a graded monomial ordering, for example the `deglex' order (we first order monomials by total degree, and then by lexicographic order with respect to some linear ordering of the indeterminates). I leave the details to later thought.

The last concept of the title is that of associated prime ideal. There are in fact two different notions for this, and in absence of noetherian hypotheses, it is better to consider the one coined as “weakly associated” by N. Bourbaki (Algèbre commutative) or D. Lazard (“Autour de la platitude”, Bulletin de la S. M. F., tome 97 (1969), p. 81-128). This is also the choice I made in my book, (Mostly) commutative algebra, Universitext, 2021. And so let me skip the adverb “weakly”, defining a prime ideal $P$ of $A$ to be an associated prime to an $A$-module $M$ if there exists $m\in M$ such that $P$ is minimal among the set of prime ideals containing the annihilator $\operatorname{ann}_A(m)$ of $m$. (When $A$ is noetherian, more generally when $P$ is finitely generated, one can prove that there exists $m$ such that $P$ equals the annihilator of $m$, and this is the — sometimes — more classic definition of an associated prime ideal.) The set of associated prime ideals of $M$ is denoted by $\operatorname{Ass}_A(M)$ (or even $\operatorname{Ass}(A)$ when $M=A$), where the abbreviation stands for assassin — In mathematics, forcing something to vanish is often, too often, related to the vocabulary of murder…

This looks like it is a strange notion, but it has some algebraic content, as summaried by the following proposition for which proof I refer to Theorem 6.5.8 of my above mentioned book.

Proposition — Let $A$ be a commutative ring and let $M$ be an $A$-module.

1) An element $a\in A$ belongs to some associated prime ideal of $M$ if and only if there exists a nonzero $m\in M$ such that $am=0$.

2) An element $a$ belongs to every associated prime ideal of $M$ if and only if the localized module $M_a$ vanishes.

Associated primes also have some geometric meaning when $M=A/I$, for some ideal $I$ of $A$. Indeed, the minimal prime ideals of $A$ containing $I$ are associated, and they represent the irreducible components of the locus $\mathrm V(I)$ in $\operatorname{Spec}(A)$. When the ideal $I$ is equal to its radical, this is all there is, but in general there are more associated prime ideals, and the corresponding closed subsets of $\mathrm V(I)$ are called “embedded” components.

It is now a natural question to elucidate associate prime ideal in the context of polynomial rings. The answer is given by a theorem of J. W. Brewer and W. J. Heinzer in their paper “Associated primes of principal ideals”, Duke Math. Journal, 1974.

Theorem (Brewer & Heinzer). — Let $A$ be a commutative ring. The associated prime ideals of the polynomial ring $A[T]$ are the ideals of the form $p[T]$, where $p$ is an associated prime ideal of $A$.

This theorem implies the above proposition. Indeed, if $P\in A[T]$ is a zero divisor, there exists an associated prime ideal $q\in A[T]$ such that $P\in q$, and $q$ is of the form $p[T]$ for some associated prime ideal $p$ of $A$. By definition, there exists an element $a\in A$ such that $p$ is minimal among the prime ideals annihilating $a$. Then $ap=0$, which implies that $aP=0$. Since $p$ is prime, one has $a\neq 0$, hence the proposition.

It doesn't look so clear to me how to immediately deduce the theorem from the proposition, although that might well be possible, and even easy. Anyway, let us describe the proof given by Brewer and Heinzer (this is theorem 7 in their paper), adjusting the notation and terminology. In fact, I cannot really prove the theorem without using the proposition, but I might have overlooked some easy argument.

We consider an associated prime ideal $q$ of $A[T]$ and let $p=A \cap q$ be its trace on $A$. Note that $p$ is a prime ideal of $A$.

Lemma. — One has $q=p[T]$.

The inclusion $p[T]\subseteq q$ is obvious; let us show the other one. Let $f\in q$ and let us prove $f\in p[T]$. Localizing at $p$, we recall that $A_p$ is a local ring with maximal ideal $pA_p$: in that case, elements of $pA_p$ are zero divisors (because $pA_p$ is associated), and elements outside of $pA_p$ are units. In particular, elements of $A_p[T]$ outside of $pA_p[T]$ have at least one coefficient which is a unit, hence can't be zero-divisors in $A_p[T]$ (by the proposition). This implies that $qA_p=pA_p[T]$ after localization, hence there exists $a\notin p$ such that $af\in p[T]$. Using that $p$ is prime, we get $f\in p[T]$, as was to be shown.

Let us now show that $q$ is an associated prime ideal of $A$. We still assume that $A$ is local with maximal ideal $p$. Let $f\in A[T]$ be such that $q$ is a minimal prime ideal among those containing $\operatorname{ann}(f)$.

Lemma. — One has $q=\sqrt{\operatorname{ann}(f)}$.

The radical of an ideal is the intersection of the minimal prime ideals that contain it, and $q$ is among this set of prime ideals, so that it suffices to prove that $q$ is the only minimal prime ideal of $A[T]$ that contains $\operatorname{ann}(f)$. Let then $q'$ be such a prime ideal. By definition, it is an associated prime ideal of $A[T]$ so that, by what has already been proved of the theorem, there exists a prime ideal $p'$ of $A$ such that $q'=p'[T]$. Since $A$ is local with maximal ideal $p$, one has $p'\subseteq p$, hence $q'=p'[T]\subseteq p[T]=q$. Since $q$ was assumed to be a minimal prime ideal containing $\operatorname{ann}(f)$, we get $q'=q$, as claimed.

From this lemma, one deduces that $p=\sqrt{\operatorname{ann}(f)\cap A}$.

Let then $a$ be a nonzero coefficient of $f$. By definition, one has $\operatorname{ann}_{A[T]}(f)\cap A \subseteq \operatorname{ann}_A(a)$, so that $p = \sqrt{\operatorname{ann}_{A[T]}(f)\cap A} \subseteq \sqrt{\operatorname{ann}_A(a)}$. Since Since $A$ is local with maximal ideal $p$, this implies that $p=\sqrt{\operatorname{ann}_A(a)}$. Consequently, $p$ is the only prime ideal containing $\operatorname{ann}_A(a)$, which implies that is is an associated prime ideal of $A$, concluding the proof of the theorem of Brewer and Heinzer.

On the other hand, there is a slightly different approach to the theorem of Brewer and Heinzer, and I want to sketch it now. This will build on proposition II/3.2 of Lazard's paper mentioned above.

Proposition (Lazard). — Let $B$ be a flat $A$-algebra and let $M$ be a $B$-module. Then $\operatorname{Ass}_A(M)$ equals the image of $\operatorname{Ass}_B(M)$ under the natural continuous map from $\operatorname{Spec}(B)$ to $\operatorname{Spec}(A)$.

Now specialize that proposition to the case where $B=A[T]$ and $M=B$. This says that the image of $\operatorname{Ass}_B(B)$ (the associated prime ideals of $B$) in $\operatorname{Spec}(A)$ is exactly $\operatorname{Ass}_A(A[T])$. On the other hand, the freeness of $A[T]$ as an $A$-module implies that $\operatorname{Ass}_A(A[T])=\operatorname{Ass}_A(A)$. Indeed, one reduces to the case of finite free modules, and then apply the classic inclusion $$ \operatorname{Ass}_A(N) \subseteq \operatorname{Ass}_A(M) \subseteq \operatorname{Ass}_A(N) \cup \operatorname{Ass}_A(M/N) $$ for every submodule $N$ of $M$, which will show by induction that $\operatorname{Ass}_A(M)=\operatorname{Ass}_A(A)$ when $M$ is free. On the other hand, we have seen in the first (and easiest) part of the proof of the theorem of Brewer and Heinzer that every associated prime ideal of $A[T]$ is of the form $q=p[T]$ for some prime ideal $p$ of $A$. Since $p=q\cap A$, this shows that $p$ is an associated prime of $A$ and gives the alternative proof.

Lazard's proof is very similar to (but predates by 5 years) the arguments given by Brewer and Heinzer. I don't really know why they didn't use it in their paper.