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.^[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 ${\displaystyle I_{1},I_{2},\dots ,I_{n},n\geq 1}$ be ideals such that ${\displaystyle I_{i}}$ are prime ideals for ${\displaystyle i\geq 3}$. If E is not contained in any of the ${\displaystyle I_{i}}$, then E is not contained in the union ${\textstyle \bigcup I_{i}}$.

Proof by induction on n: The idea is to find an element of R that is in E and not in any of the ${\displaystyle I_{i}}$. The base case ${\displaystyle n=1}$ is trivial. Next suppose ${\displaystyle n\geq 2}$. For each i, choose

${\displaystyle z_{i}\in E\setminus \bigcup _{j\neq i}I_{j}}$,

where each of the sets on the right is nonempty by the inductive hypothesis. We can assume ${\displaystyle z_{i}\in I_{i}}$ for all i; otherwise, there is some ${\displaystyle z_{k}}$ among them that avoids all of the ${\displaystyle I_{i}}$, and we are done. Put

${\displaystyle z=z_{1}\cdots z_{n-1}+z_{n}}$.

Because E is closed under addition and multiplication, z is in E by construction. We claim that z is not in any of the ${\displaystyle I_{i}}$. Indeed, if ${\displaystyle z\in I_{i}}$ for some ${\displaystyle i\leq n-1}$, then ${\displaystyle z_{n}\in I_{i}}$, a contradiction. Next suppose ${\displaystyle z\in I_{n}}$. Then ${\displaystyle z_{1}\cdots z_{n-1}\in I_{n}}$. If ${\displaystyle n=2}$, this is already a contradiction. If ${\displaystyle n>2}$, then, since ${\displaystyle I_{n}}$ is a prime ideal, ${\displaystyle z_{i}\in I_{n}}$ for some ${\displaystyle i\leq n-1}$, again a contradiction. ${\displaystyle \square }$

E. Davis' prime avoidance

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

Theorem—^[2] Let A be a ring, ${\displaystyle {\mathfrak {p}}_{1},\dots ,{\mathfrak {p}}_{r}}$ prime ideals, x an element of A and J an ideal. For the ideal ${\displaystyle I=xA+J}$, if ${\displaystyle I\not \subset {\mathfrak {p}}_{i}}$ for each i, then there exists some y in J such that ${\displaystyle x+y\not \in {\mathfrak {p}}_{i}}$ for each i.

Proof:^[3] We argue by induction on r. Without loss of generality, we can assume there is no inclusion relation between the ${\displaystyle {\mathfrak {p}}_{i}}$'s; since otherwise we can use the inductive hypothesis.

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

${\displaystyle {\mathfrak {p}}_{r}\not \supset J\,{\mathfrak {p}}_{1}\cdots {\mathfrak {p}}_{r-1}}$.

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

Application

Let A be a Noetherian ring, I an ideal generated by n elements and M a finite A-module such that ${\displaystyle IM\neq M}$. Also, let ${\displaystyle d=\operatorname {depth} _{A}(I,M)}$ = the maximal length of M-regular sequences in I = the length of every maximal M-regular sequence in I. Then ${\displaystyle d\leq n}$; this estimate can be shown using the above prime avoidance as follows. We argue by induction on n. Let ${\displaystyle \{{\mathfrak {p}}_{1},\dots ,{\mathfrak {p}}_{r}\}}$ be the set of associated primes of M. If ${\displaystyle d>0}$, then ${\displaystyle I\not \subset {\mathfrak {p}}_{i}}$ for each i. If ${\displaystyle I=(y_{1},\dots ,y_{n})}$, then, by prime avoidance, we can choose

${\displaystyle x_{1}=y_{1}+\sum _{i=2}^{n}a_{i}y_{i}}$

for some ${\displaystyle a_{i}}$ in ${\displaystyle A}$ such that ${\displaystyle x_{1}\not \in \cup _{1}^{r}{\mathfrak {p}}_{i}}$ = the set of zero divisors on M. Now, ${\displaystyle I/(x_{1})}$ is an ideal of ${\displaystyle A/(x_{1})}$ generated by ${\displaystyle n-1}$ elements and so, by inductive hypothesis, ${\displaystyle \operatorname {depth} _{A/(x_{1})}(I/(x_{1}),M/x_{1}M)\leq n-1}$. The claim now follows.