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May 24, 2010 at 11:30 comment added Simon Wadsley Great! Permit me to rephrase this in algebra rather than geometry for the sake of algebraists. Pick $f$ in the Jacobson radical but non-zero and let $P$ be an ideal that is maximal amongst ideals disjoint from the set of powers of $f$. $P$ is prime but cannot be maximal so is (by condition 2) the intersection of the prime ideals that strictly contain it. But by definition of $P$ all these strictly bigger primes conatin $f$ thus $P$ contains $f$ a contradiction. Nice.
May 24, 2010 at 11:27 vote accept Simon Wadsley
May 23, 2010 at 7:23 history answered Kevin Ventullo CC BY-SA 2.5