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Feb 8, 2016 at 13:02 comment added nfdc23 For each prime ideal $p$ of $B$ and its contraction $q$ in $A$, I claim $A_q \rightarrow B_q$ is an isomorphism. Then the finite presentation hypothesis would provide $a \in A - q$ such that $A[1/a]\rightarrow B[1/a]$ is an isomorphism. The image of $a$ in $B$ lies outside the prime ideal $p$, so such $a$'s would generate 1 in $B$, as desired. So we can assume $A$ is local with maximal ideal $m$ hit by a prime of $B$ (hence $mB \ne B$), so $A \rightarrow B$ is faithfully flat (Theorem 7.2 in Matsumura's "Comm. Ring Theory"). You know the rest. QED (But the "cheat proof" is more illuminating!)
Feb 8, 2016 at 8:26 history asked Zhen Lin CC BY-SA 3.0