a question about flatness - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:57:01Z http://mathoverflow.net/feeds/question/19107 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19107/a-question-about-flatness a question about flatness Rothendieck 2010-03-23T13:03:33Z 2010-03-23T13:36:45Z <p>In the book "étale cohomology" by Milne, proposition 2.5 at p.9, it said :</p> <p>Let $B$ be a flat $A-$algebra where $A$ and $B$ are noetherian rings, and consider $b \in B$. If the image of $b$ in $B/mB$ is not a zero-divisor for any maximal ideal $m$ of $A$, then $B/(b)$ is a flat $A-$ algebra.</p> <p>At the beginning of the proof, he said we can reduce to the case where $\phi : A \rightarrow B$ is a local homomorphism of local noetherian rings. The proof in this case uses the fact that it's a local homomorphsim.</p> <p>But I think that in order to reduce the general case to the local case, we need the following condition, which I can't get from the original condition.</p> <p>For any maximal ideal $n$ of $B$, the image of $b$ in $B/pB$ is not a zero-divisor, where $p = \phi^{-1} (n)$. </p> <p>How do you think?</p> http://mathoverflow.net/questions/19107/a-question-about-flatness/19111#19111 Answer by David Speyer for a question about flatness David Speyer 2010-03-23T13:36:45Z 2010-03-23T13:36:45Z <p>I am not sure about Milne's reduction, but your fix is too strong. First off all, I don't understand why you write $B/p B$ with $p = \phi^{-1}(n)$. I assume $\phi$ is the map $A \to B$. But then $\phi(n)$ is an ideal of $A$, not $B$, and $b$ is an element of $B$. I am going to assume you meant "$b$ is not a zero divisor in $B/nB$."</p> <p>But requiring this for every maximal ideal of $B$ implies that $b$ is a unit! We surely don't want to impose that.</p> <p>Let me add that, in my opinion, it is rude to use the name of a living person as a pseudonym. I do not think that Grothendieck would appreciate other people posting under his name. If you want to honor him, why not name yourself for one of his theorems or definitions, as fpqc does? (Of course, if your name is Grothendieck, I apologize.)</p>