Prime avoidance lemma

Short description: Result concerning ideals of commutative rings

In algebra, the prime avoidance lemma says that if an ideal I in a commutative ring R is contained in a union of finitely many prime ideals P_i's, then it is contained in P_i for some i.

There are many variations of the lemma (cf. Hochster); for example, if the ring R contains an infinite field or a finite field of sufficiently large cardinality, then the statement follows from a fact in linear algebra that a vector space over an infinite field or a finite field of large cardinality is not a finite union of its proper vector subspaces.

Statement and proof

The following statement and argument are perhaps the most standard.

Statement: Let E be a subset of R that is an additive subgroup of R and is multiplicatively closed. Let [math]\displaystyle{ I_1, I_2, \dots, I_n, n \ge 1 }[/math] be ideals such that [math]\displaystyle{ I_i }[/math] are prime ideals for [math]\displaystyle{ i \ge 3 }[/math]. If E is not contained in any of [math]\displaystyle{ I_i }[/math]'s, then E is not contained in the union [math]\displaystyle{ \cup I_i }[/math].

Proof by induction on n: The idea is to find an element that is in E and not in any of [math]\displaystyle{ I_i }[/math]'s. The basic case n = 1 is trivial. Next suppose n ≥ 2. For each i, choose

[math]\displaystyle{ z_i \in E - \cup_{j \ne i} I_j }[/math]

where the set on the right is nonempty by inductive hypothesis. We can assume [math]\displaystyle{ z_i \in I_i }[/math] for all i; otherwise, some [math]\displaystyle{ z_i }[/math] avoids all the [math]\displaystyle{ I_i }[/math]'s and we are done. Put

[math]\displaystyle{ z = z_1 \dots z_{n-1} + z_n }[/math].

Then z is in E but not in any of [math]\displaystyle{ I_i }[/math]'s. Indeed, if z is in [math]\displaystyle{ I_i }[/math] for some [math]\displaystyle{ i \le n - 1 }[/math], then [math]\displaystyle{ z_n }[/math] is in [math]\displaystyle{ I_i }[/math], a contradiction. Suppose z is in [math]\displaystyle{ I_n }[/math]. Then [math]\displaystyle{ z_1 \dots z_{n-1} }[/math] is in [math]\displaystyle{ I_n }[/math]. If n is 2, we are done. If n > 2, then, since [math]\displaystyle{ I_n }[/math] is a prime ideal, some [math]\displaystyle{ z_i, i \lt n }[/math] is in [math]\displaystyle{ I_n }[/math], a contradiction.

E. Davis' prime avoidance

There is the following variant of prime avoidance due to E. Davis.

Theorem —  Let A be a ring, [math]\displaystyle{ \mathfrak{p}_1, \dots, \mathfrak{p}_r }[/math] prime ideals, x an element of A and J an ideal. For the ideal [math]\displaystyle{ I = xA + J }[/math], if [math]\displaystyle{ I \not\subset \mathfrak{p}_i }[/math] for each i, then there exists some y in J such that [math]\displaystyle{ x + y \not\in \mathfrak{p}_i }[/math] for each i.

Proof: We argue by induction on r. Without loss of generality, we can assume there is no inclusion relation between the [math]\displaystyle{ \mathfrak{p}_i }[/math]'s; since otherwise we can use the inductive hypothesis.

Also, if [math]\displaystyle{ x \not\in \mathfrak{p}_i }[/math] for each i, then we are done; thus, without loss of generality, we can assume [math]\displaystyle{ x \in \mathfrak{p}_r }[/math]. By inductive hypothesis, we find a y in J such that [math]\displaystyle{ x + y \in I - \cup_1^{r-1} \mathfrak{p}_i }[/math]. If [math]\displaystyle{ x + y }[/math] is not in [math]\displaystyle{ \mathfrak{p}_r }[/math], we are done. Otherwise, note that [math]\displaystyle{ J \not\subset \mathfrak{p}_r }[/math] (since [math]\displaystyle{ x \in \mathfrak{p}_r }[/math]) and since [math]\displaystyle{ \mathfrak{p}_r }[/math] is a prime ideal, we have:

[math]\displaystyle{ \mathfrak{p}_r \not\supset J \, \mathfrak{p}_1 \cdots \mathfrak{p}_{r-1} }[/math].

Hence, we can choose [math]\displaystyle{ y' }[/math] in [math]\displaystyle{ J \, \mathfrak{p}_1 \cdots \mathfrak{p}_{r-1} }[/math] that is not in [math]\displaystyle{ \mathfrak{p}_r }[/math]. Then, since [math]\displaystyle{ x + y \in \mathfrak{p}_r }[/math], the element [math]\displaystyle{ x + y + y' }[/math] has the required property. [math]\displaystyle{ \square }[/math]

Application

Let A be a Noetherian ring, I an ideal generated by n elements and M a finite A-module such that [math]\displaystyle{ IM \ne M }[/math]. Also, let [math]\displaystyle{ d = \operatorname{depth}_A(I, M) }[/math] = the maximal length of M-regular sequences in I = the length of every maximal M-regular sequence in I. Then [math]\displaystyle{ d \le n }[/math]; this estimate can be shown using the above prime avoidance as follows. We argue by induction on n. Let [math]\displaystyle{ \{ \mathfrak{p}_1, \dots, \mathfrak{p}_r \} }[/math] be the set of associated primes of M. If [math]\displaystyle{ d \gt 0 }[/math], then [math]\displaystyle{ I \not\subset \mathfrak{p}_i }[/math] for each i. If [math]\displaystyle{ I = (y_1, \dots, y_n) }[/math], then, by prime avoidance, we can choose

[math]\displaystyle{ x_1 = y_1 + \sum_{i = 2}^n a_i y_i }[/math]

for some [math]\displaystyle{ a_i }[/math] in [math]\displaystyle{ A }[/math] such that [math]\displaystyle{ x_1 \not\in \cup_1^r \mathfrak{p}_i }[/math] = the set of zero divisors on M. Now, [math]\displaystyle{ I/(x_1) }[/math] is an ideal of [math]\displaystyle{ A/(x_1) }[/math] generated by [math]\displaystyle{ n - 1 }[/math] elements and so, by inductive hypothesis, [math]\displaystyle{ \operatorname{depth}_{A/(x_1)}(I/(x_1), M/x_1M) \le n - 1 }[/math]. The claim now follows.

References

• Mel Hochster, Dimension theory and systems of parameters, a supplementary note
• Matsumura, Hideyuki (1986). Commutative ring theory. Cambridge Studies in Advanced Mathematics. 8. Cambridge University Press. ISBN 0-521-36764-6. https://books.google.com/books?id=yJwNrABugDEC&pg=PA123.