prime avoidance lemma

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 $I\_1,\; I\_2,\; \backslash dots,\; I\_n,\; n\; \backslash ge\; 1$ be ideals such that $I\_i$ are prime ideals for $i\; \backslash ge\; 3$. If E is not contained in any of $I\_i$'s, then E is not contained in the union $\backslash cup\; I\_i$.

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

:$z\_i\; \backslash in\; E\; -\; \backslash cup\_\{j\; \backslash ne\; i\}\; I\_j$

where the set on the right is nonempty by inductive hypothesis. We can assume $z\_i\; \backslash in\; I\_i$ for all i; otherwise, some $z\_i$ avoids all the $I\_i$'s and we are done. Put

:$z\; =\; z\_1\; \backslash dots\; z\_\{n-1\}\; +\; z\_n$.

Then z is in E but not in any of $I\_i$'s. Indeed, if z is in $I\_i$ for some $i\; \backslash le\; n\; -\; 1$, then $z\_n$ is in $I\_i$, a contradiction. Suppose z is in $I\_n$. Then $z\_1\; \backslash dots\; z\_\{n-1\}$ is in $I\_n$. If n is 2, we are done. If n > 2, then, since $I\_n$ is a prime ideal, some is in $I\_n$, a contradiction.

There is the following variant of prime avoidance due to [https://www.genealogy.math.ndsu.nodak.edu/id.php?id=5703 E. Davis].

{{math_theorem|math_statement={{harvnb|Matsumura|1986|loc=Exercise 16.8.}} Let A be a ring, $\backslash mathfrak\{p\}\_1,\; \backslash dots,\; \backslash mathfrak\{p\}\_r$ prime ideals, x an element of A and J an ideal. For the ideal $I\; =\; xA\; +\; J$, if $I\; \backslash not\backslash subset\; \backslash mathfrak\{p\}\_i$ for each i, then there exists some y in J such that $x\; +\; y\; \backslash not\backslash in\; \backslash mathfrak\{p\}\_i$ for each i.}}

Proof:Adapted from the solution to {{harvnb|Matsumura|1986|loc=Exercise 1.6.}} We argue by induction on r. Without loss of generality, we can assume there is no inclusion relation between the $\backslash mathfrak\{p\}\_i$'s; since otherwise we can use the inductive hypothesis.

Also, if $x\; \backslash not\backslash in\; \backslash mathfrak\{p\}\_i$ for each i, then we are done; thus, without loss of generality, we can assume $x\; \backslash in\; \backslash mathfrak\{p\}\_r$. By inductive hypothesis, we find a y in J such that $x\; +\; y\; \backslash in\; I\; -\; \backslash cup\_1^\{r-1\}\; \backslash mathfrak\{p\}\_i$. If $x\; +\; y$ is not in $\backslash mathfrak\{p\}\_r$, we are done. Otherwise, note that $J\; \backslash not\backslash subset\; \backslash mathfrak\{p\}\_r$ (since $x\; \backslash in\; \backslash mathfrak\{p\}\_r$) and since $\backslash mathfrak\{p\}\_r$ is a prime ideal, we have:

:$\backslash mathfrak\{p\}\_r\; \backslash not\backslash supset\; J\; \backslash ,\; \backslash mathfrak\{p\}\_1\; \backslash cdots\; \backslash mathfrak\{p\}\_\{r-1\}$.

Hence, we can choose $y\text{'}$ in $J\; \backslash ,\; \backslash mathfrak\{p\}\_1\; \backslash cdots\; \backslash mathfrak\{p\}\_\{r-1\}$ that is not in $\backslash mathfrak\{p\}\_r$. Then, since $x\; +\; y\; \backslash in\; \backslash mathfrak\{p\}\_r$, the element $x\; +\; y\; +\; y\text{'}$ has the required property. $\backslash square$

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

:$x\_1\; =\; y\_1\; +\; \backslash sum\_\{i\; =\; 2\}^n\; a\_i\; y\_i$

for some $a\_i$ in $A$ such that $x\_1\; \backslash not\backslash in\; \backslash cup\_1^r\; \backslash mathfrak\{p\}\_i$ = the set of zero divisors on M. Now, $I/(x\_1)$ is an ideal of $A/(x\_1)$ generated by $n\; -\; 1$ elements and so, by inductive hypothesis, $\backslash operatorname\{depth\}\_\{A/(x\_1)\}(I/(x\_1),\; M/x\_1M)\; \backslash le\; n\; -\; 1$. The claim now follows.

{{reflist}}

• Mel Hochster, [http://www.math.lsa.umich.edu/~hochster/615W10/supDim.pdf 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

}}

Category:Abstract algebra