Prime avoidance lemma explained

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.^[1]

Statement and proof

The following statement and argument are perhaps the most standard.

Theorem (Prime Avoidance Lemma): Let E be a subset of commutative ring R that is an additive subgroup of R and is multiplicatively closed. (In particular, E could be a subring or ideal of R.) Let

I_1,I_2,...,I_n,n\ge1

be ideals such that

I_i

are prime ideals for

i\ge3

. If E is not contained in any of the

I_i

, then E is not contained in the union

\bigcup I_i

Proof by induction on n: The idea is to find an element of R that is in E and not in any of the

I_i

. The base case

n=1

is trivial. Next suppose

n\geq2

. For each i, choose

z_i\inE\setminuscup_jI_j

,where each of the sets on the right is nonempty by the inductive hypothesis. We can assume

z_i\inI_i

for all i; otherwise, there is some

z_k

among them that avoids all of the

I_i

, and we are done. Put

z=z₁ … z_n-1+z_n

.Because E is closed under addition and multiplication, z is in E by construction. We claim that z is not in any of the

I_i

. Indeed, if

z\inI_i

for some

i\len-1

, then

z_n\inI_i

, a contradiction. Next suppose

z\inI_n

. Then

z₁ … z_n-1\inI_n

. If

n=2

, this is already a contradiction. If

n>2

, then, since

I_n

is a prime ideal,

z_i\inI_n

for some

i\leqn-1

, again a contradiction.

\square

E. Davis' prime avoidance

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

Proof:^[2] We argue by induction on r. Without loss of generality, we can assume there is no inclusion relation between the

ak{p}_i

's; since otherwise we can use the inductive hypothesis.

Also, if

x\not\inak{p}_i

for each i, then we are done; thus, without loss of generality, we can assume

x\inak{p}_r

. By inductive hypothesis, we find a y in J such that

x+y\inI-

	r-1
\cup
	1

ak{p}_i

. If

x+y

is not in

ak{p}_r

, we are done. Otherwise, note that

J\not\subsetak{p}_r

(since

x\inak{p}_r

) and since

ak{p}_r

is a prime ideal, we have:

ak{p}_r\not\supsetJak{p}₁ … ak{p}_r-1

.Hence, we can choose

Jak{p}₁ … ak{p}_r-1

that is not in

ak{p}_r

. Then, since

x+y\inak{p}_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\neM

. Also, let

d=\operatorname{depth}_A(I,M)

= the maximal length of M-regular sequences in I = the length of every maximal M-regular sequence in I. Then

d\len

; this estimate can be shown using the above prime avoidance as follows. We argue by induction on n. Let

\{ak{p}_1,...,ak{p}_r\}

be the set of associated primes of M. If

d>0

, then

I\not\subsetak{p}_i

for each i. If

I=(y_1,...,y_n)

, then, by prime avoidance, we can choose

x₁=y₁+

	n
\sum
	i=2

a_iy_i

for some

a_i

such that

x₁\not\in

	r
\cup
	1

ak{p}_i

= the set of zero divisors on M. Now,

I/(x₁₎

is an ideal of

A/(x₁₎

generated by

n-1

elements and so, by inductive hypothesis,

\operatorname{depth}
	A/(x₁₎

(I/(x_1),M/x_1M)\len-1

. The claim now follows.

References

Mel Hochster, Dimension theory and systems of parameters, a supplementary note
Book: Matsumura. Hideyuki. 1986. [{{google books|yJwNrABugDEC|Commutative ring theory|plainurl=yes|page=123}} Commutative ring theory ]. Cambridge Studies in Advanced Mathematics. 8. Cambridge University Press. 0-521-36764-6. 0879273. 0603.13001.

Notes and References

Proof of the fact: suppose the vector space is a finite union of proper subspaces. Consider a finite product of linear functionals, each of which vanishes on a proper subspace that appears in the union; then it is a nonzero polynomial vanishing identically, a contradiction.
Adapted from the solution to