# Applications of the prime avoidance lemma

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.

• 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. Nov 28, 2016 at 16:02
• en.wikipedia.org/wiki/Prime_avoidance_lemma (I knew this lemma but this particular English name)
– YCor
Nov 28, 2016 at 18:18
• @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'.
– Xam
Nov 28, 2016 at 19:02
• I just gave a link for people to have the statement by a single click (and I meant "but not this particular English name").
– YCor
Nov 28, 2016 at 19:08
• The application given in the question is missing a noetherian hypothesis. Nov 28, 2016 at 19:09

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.

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).

• 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 $I_{j}\subseteq P$.
– Xam
Nov 28, 2016 at 18:28
• 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). Nov 28, 2016 at 18:39

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.

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.