MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the book "étale cohomology" by Milne, proposition 2.5 at p.9, it said :

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.

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.

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.

For any maximal ideal $n$ of $B$, the image of $b$ in $B/pB$ is not a zero-divisor, where $p = \phi^{-1} (n)$.

How do you think?

share|cite|improve this question
you are exactly right. cf. the erratum on milne's home page: – LRG Mar 23 '10 at 21:52

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 $B/nB$."

But requiring this for every maximal ideal of $B$ implies that $b$ is a unit! We surely don't want to impose that.

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.)

share|cite|improve this answer
Yes, $\phi : A \rightarrow B$ is the ring homomorphism which defines the $A$-algebra structure of $B$. $p = \phi^{-1} (n)$ is a prime ideal of $A$, but $pB$ is the ideal of $B$ generated by $\phi (p) = \phi \circ \phi^{-1} (n)$, which doesn't equal to $n$ in general. Hence the condition is not as strong as you think. The reason I impose the condition is "flatness is a local property". About the name, I am sorry, is there anyway to change it? – Rothendieck Mar 23 '10 at 13:58
If you were a registered user, you'd go to your profile (click your name where it is displayed at the top of the screen) and click edit where it says "Registered User edit | add openid". I'm not sure if there is a way for an unregistered user to change names. – David Speyer Mar 23 '10 at 14:38
And now I understand where you're coming from mathematically, I'll think about it. – David Speyer Mar 23 '10 at 14:39
@Rothendieck: I've slightly changed your username on the assumption that that's what you wanted to do. Please email me if you have any trouble with your account. – Anton Geraschenko Mar 23 '10 at 20:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.