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 Pi's, then it is contained in Pi 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.


Statement: Let E be a subset of R that is an additive subgroup of R and is multiplicatively closed. Let I1,I2,…,In,n≥1{displaystyle I_{1},I_{2},dots ,I_{n},ngeq 1}I_{1},I_{2},dots ,I_{n},ngeq 1 be ideals such that Ii{displaystyle I_{i}}I_{i} are prime ideals for i≥3{displaystyle igeq 3}igeq 3. If E is not contained in any of Ii{displaystyle I_{i}}I_{i}'s, then E is not contained in the union ∪Ii{displaystyle cup I_{i}}cup I_{i}.


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


zi∈E−∪j≠iIj{displaystyle z_{i}in E-cup _{jneq i}I_{j}}z_{i}in E-cup _{jneq i}I_{j}

where the set on the right is nonempty by inductive hypothesis. We can assume zi∈Ii{displaystyle z_{i}in I_{i}}z_{i}in I_{i} for all i; otherwise, some zi{displaystyle z_{i}}z_{i} avoids all the Ii{displaystyle I_{i}}I_{i}'s and we are done. Put



z=z1…zn−1+zn{displaystyle z=z_{1}dots z_{n-1}+z_{n}}z=z_{1}dots z_{n-1}+z_{n}.

Then z is in E but not in any of Ii{displaystyle I_{i}}I_{i}'s. Indeed, if z is in Ii{displaystyle I_{i}}I_{i} for some i≤n−1{displaystyle ileq n-1}ileq n-1, then zn{displaystyle z_{n}}z_{n} is in Ii{displaystyle I_{i}}I_{i}, a contradiction. Suppose z is in In{displaystyle I_{n}}I_{n}. Then z1…zn−1{displaystyle z_{1}dots z_{n-1}}z_{1}dots z_{n-1} is in In{displaystyle I_{n}}I_{n}. If n is 2, we are done. If n > 2, then, since In{displaystyle I_{n}}I_{n} is a prime ideal, some zi,i<n{displaystyle z_{i},i<n}z_{i},i<n is in In{displaystyle I_{n}}I_{n}, a contradiction.



Notes





  1. ^ 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.




References



  • Mel Hochster, Dimension theory and systems of parameters, a supplementary note









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