Prime avoidance lemma

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 $I_1, I_2, \dots, I_n, n \ge 1$ be ideals such that $I_i$ are prime ideals for $i \ge 3$. If ''E'' is not contained in any of $I_i$'s, then ''E'' is not contained in the union $\cup I_i$.

Proof by induction on ''n'': The idea is to find an element that is in ''E'' and not in any of $I_i$'s. The basic case ''n'' = 1 is trivial. Next suppose ''n'' ≥ 2. For each ''i'', choose
:$z_i \in E - \cup_ I_j$
where the set on the right is nonempty by inductive hypothesis. We can assume $z_i \in I_i$ for all ''i''; otherwise, some $z_i$ avoids all the $I_i$'s and we are done. Put
:$z = z_1 \dots z_ + z_n$.
Then ''z'' is in ''E'' but not in any of $I_i$'s. Indeed, if ''z'' is in $I_i$ for some $i \le n - 1$, then $z_n$ is in $I_i$, a contradiction. Suppose ''z'' is in $I_n$. Then $z_1 \dots z_$ is in $I_n$. If ''n'' is 2, we are done. If ''n'' > 2, then, since $I_n$ is a prime ideal, some $z_i, i < n$ is in $I_n$, a contradiction.

E. Davis' prime avoidance

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

Proof:Adapted from the solution to We argue by induction on ''r''. Without loss of generality, we can assume there is no inclusion relation between the $\mathfrak_i$'s; since otherwise we can use the inductive hypothesis.

Also, if $x \not\in \mathfrak_i$ for each ''i'', then we are done; thus, without loss of generality, we can assume $x \in \mathfrak_r$. By inductive hypothesis, we find a ''y'' in ''J'' such that $x + y \in I - \cup_1^ \mathfrak_i$. If $x + y$ is not in $\mathfrak_r$, we are done. Otherwise, note that $J \not\subset \mathfrak_r$ (since $x \in \mathfrak_r$) and since $\mathfrak_r$ is a prime ideal, we have:
:$\mathfrak_r \not\supset J \, \mathfrak_1 \cdots \mathfrak_$.
Hence, we can choose $y'$ in $J \, \mathfrak_1 \cdots \mathfrak_$ that is not in $\mathfrak_r$. Then, since $x + y \in \mathfrak_r$, the element $x + y + y'$ has the required property. $\square$

Application

Let ''A'' be a Noetherian ring, ''I'' an ideal generated by ''n'' elements and ''M'' a finite ''A''-module such that $IM \ne M$. Also, let $d = \operatorname_A(I, M)$ = the maximal length of ''M''-regular sequences in ''I'' = the length of ''every'' maximal ''M''-regular sequence in ''I''. Then $d \le n$; this estimate can be shown using the above prime avoidance as follows. We argue by induction on ''n''. Let $\$ be the set of associated primes of ''M''. If $d > 0$, then $I \not\subset \mathfrak_i$ for each ''i''. If $I = (y_1, \dots, y_n)$, then, by prime avoidance, we can choose
:$x_1 = y_1 + \sum_^n a_i y_i$
for some $a_i$ in $A$ such that $x_1 \not\in \cup_1^r \mathfrak_i$ = the set of zerodivisors on ''M''. Now, $I/(x_1)$ is an ideal of $A/(x_1)$ generated by $n - 1$ elements and so, by inductive hypothesis, $\operatorname_(I/(x_1), M/x_1M) \le n - 1$. The claim now follows.

Notes

References

* Mel Hochster
Dimension theory and systems of parameters
a supplementary note
*{{cite book
, last1 = Matsumura
, first1 = Hideyuki
, year = 1986
, title = Commutative ring theory
, series = Cambridge Studies in Advanced Mathematics
, volume = 8
, url = {{google books, yJwNrABugDEC, Commutative ring theory, plainurl=yes, page=123
, publisher = Cambridge University Press
, isbn = 0-521-36764-6
, mr = 0879273
, zbl = 0603.13001

Algebra