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I was wondering if the prime avoidance lemma is very useful or just a nice result. So far I know just only one application: let $R$ be a commutative noetherian ring and $I$ be a proper ideal of $R$. If $I$ consists only of zero divisors of $R$, then $I$ is contained in some associated prime ideal of $(0)$.

So my question is: are there other applications of the prime avoidance lemma in commutative algebra? Thanks in advance for your answers.

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    $\begingroup$ There are many applications of it. Let me give you one. Let $R$ be a Noetherian ring of dimension $d$ and $I$ any ideal. Then $I$ is set-theoretically (that is, upto radicals) generated by $d+1$ elements. $\endgroup$
    – Mohan
    Nov 28, 2016 at 16:02
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    $\begingroup$ en.wikipedia.org/wiki/Prime_avoidance_lemma (I knew this lemma but this particular English name) $\endgroup$
    – YCor
    Nov 28, 2016 at 18:18
  • $\begingroup$ @Ycor I don't understand what you want to say. As far as I know the prime avoidance lemma is stated in Kaplansky's book "Commutative Rings". So I think this result could be called "Kaplansky's lemma'. $\endgroup$
    – Xam
    Nov 28, 2016 at 19:02
  • $\begingroup$ I just gave a link for people to have the statement by a single click (and I meant "but not this particular English name"). $\endgroup$
    – YCor
    Nov 28, 2016 at 19:08
  • $\begingroup$ The application given in the question is missing a noetherian hypothesis. $\endgroup$ Nov 28, 2016 at 19:09

4 Answers 4

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Prime avoidance can be used to show the following fundamental result on regular sequences:

If $R$ is a noetherian ring, $\mathfrak{a}\subseteq R$ is an ideal, and $M$ is an $R$-module of finite type, then every maximal $M$-sequence in $\mathfrak{a}$ has length equal to the $\mathfrak{a}$-depth of $M$.

It can also be used to show the following:

Regular local rings are integral.

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The proof of the First and Second Uniqueness Theorems of Primary Decomposition uses prime avoidance lemma in an essential way.

See Atiyah-Mac Donald, Introduction to Commutative Algebra, Chapter 4 (in that book prime avoidance lemma is referred as Proposition 1.11).

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  • $\begingroup$ I think you are referring to the "dual" statement: if $I_{1}\cap \ldots \cap I_{n}\subseteq P$ for some prime ideal $P$, then there is $j$ such that $I_{j}\subseteq P$. $\endgroup$
    – Xam
    Nov 28, 2016 at 18:28
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    $\begingroup$ The dual statement is used in the First Uniqueness Theorem. For the Second Uniqueness Theorem, it is used the standard form of the result (see Atiyah-Macdonald, computation just before Theorem 4.10). $\endgroup$ Nov 28, 2016 at 18:39
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Theorem. For a given commutative ring $R$, then $Min(R)$, the set of minimal prime ideals of $R$, is a finite set if and only if no minimal prime ideal of $R$ is contained in the union of the remaining minimal primes.

Sketch of Proof. The implication $\Rightarrow$ of the above nice result is deduced from the prime avoidance lemma. The reverse implication is deduced from the fact that $Min(R)$ is quasi-compact with respect to the flat topology.

Remember that the collection of $V(f)=\{\mathfrak{p}\in Spec(R):f\in\mathfrak{p}\}$ with $f\in R$ forms a sub-base for the opens of the flat topology over $Spec(R)$.

For more details on the flat topology see arXiv:1609.00947 or arXiv:1503.04299.

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I just stumbled upon the 4-year-old question. Let me add another application to the list.

Claim. Let $A$ be a commutative Noetherian ring. If primes $\mathfrak p, \mathfrak r$ satisfy $\mathfrak p < \mathfrak r$, then the interval $(\mathfrak p, \mathfrak r)$ contains infinitely many primes or zero primes.

Chat. Here I am working in the ordered set of primes, so $\mathfrak p<\mathfrak r$ means $\mathfrak p\subsetneq \mathfrak r$. Here $(\mathfrak p, \mathfrak r)$ denotes the set of primes strictly intermediate to $\mathfrak p$ and $\mathfrak r$. Finally, this question about the structure of intervals in the ordered set of primes is asked in order to understand Spec($A$).

Reasoning for Claim. Suppose otherwise that $(\mathfrak p, \mathfrak r)$ is finite and nonempty. Shrink this interval to a minimal such one, and we end up with an interval containing some $\mathfrak q\neq \mathfrak p, \mathfrak r$ such that $\mathfrak p\prec \mathfrak q\prec \mathfrak r$ and $(\mathfrak p,\mathfrak r)$ is finite. (The notation $\mathfrak p\prec \mathfrak q$ indicates covering, which means that $\mathfrak p$ is included in $\mathfrak p$ and $(\mathfrak p,\mathfrak q)$ is empty.) When we have shrunk to a minimal such interval, no two distinct primes strictly between $\mathfrak p$ and $\mathfrak r$ will be comparable.

Factor by $\mathfrak p$ to reduce to the case $\mathfrak p=(0)$. Then localize at $\mathfrak r$ to reduce to the case where $\mathfrak r$ is maximal. Now we have $(0)\prec \mathfrak q\prec \mathfrak r$ and that $\mathfrak r$ is maximal. In this reduced case we are trying to show that it is impossible for a Noetherian integral domain of Krull dimension $2$ to have only finitely many primes. This is where Prime Avoidance comes in. Assume that $\mathfrak q_1, \ldots, \mathfrak q_n$ is a complete list of the primes strictly between $(0)$ and $\mathfrak r$ (i.e., height-$1$ primes). By Prime Avoidance, it is impossible to have $\mathfrak r$ contained in $\cup_{i=1}^n \mathfrak q_i$, so choose $a\in \mathfrak r - \cup_{i=1}^n \mathfrak q_i$. By the Krull Principal Ideal Theorem, any prime minimal over $(a)$ must have height $1$, so must be one of the $\mathfrak q_i$'s. But we specifically chose $a$ so that it belongs to none of them, so we are done. $\Box$

The Noetherianness hypothesis is used to allow us to invoke the Krull Principal Ideal Theorem.

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