It is known that in a commutative ring $R$ an ideal contained in a finite union of prime ideals $P_i , ( i=1,...,n)$ is a subset of one of them (prime avoidance theorem). Now, if $P_i$'s are arbitrary ideals and $R$ contains an infinite field I want to prove the same conclusion. Thanks to anybody helping me!

  • $\begingroup$ In another direction: two of the ideals can be non-prime. $\endgroup$
    – Hoot
    Commented Jan 16, 2015 at 17:57

1 Answer 1


Since ideals are vector subspaces, it is enough to prove that in a vector space $V$ over an infinite field a vector subspace $I\subset P_1\cup\dots\cup P_n$ included in the union of finitely many vector subspaces $P_i$ is already included in one of them: $I\subset P_{i_0}$ for some $i_0$ .
If this were false, we would have $V_j:=I\cap P_j\subsetneq I $ for all $j$ and also $\cup V_j=I$.
But this is absurd : over an infinite field a vector space $I$ cannot be the union of finitely many strict subspaces $V_j$.


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