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 Milne, Etale Cohomology, Proposition 2.5 (§2) is stated as follows:

(All rings noetherian.)

Let $B$ be a flat $A$--algebra, and consider $b \in B$. If the image of $b$ in $B/\mathfrak{m} B$ is not a zero-divisor for any maximal ideal $\mathfrak{m}$ of $A$, then $B/(b)$ is a flat $A$--algebra.

Now consider $A = k[[x_1,x_2]]$, $\mathfrak{m}=(x_1,x_2)$ and $\mathfrak{p} = (x_1)$ as $A$-ideals and $B = A_\mathfrak{p}$. Then $B/\mathfrak{m} B = 0$. Let $b = x_1$. As $B/\mathfrak{m} B$ vanishes, it has no zero-divisors, so $b$ is not a zero-divisor there.

But $B/b B \neq 0$ is obviously not $A$-flat, as it does not preserve the injectiveness of $A \overset{\cdot b}{\hookrightarrow} A$.

Of course, one could object that implicitly all $B \otimes_A k(\mathfrak{m})$ should be taken nonzero, but in this case $B$ would be faithfully flat over $A$ so as to strongly narrow the scope of the proposition.

A comparison with Kurke, Pfister, Roczen, Henselsche Ringe und algebraische Geometrie led me to suppose, that the proposition should be worded slightly different, namely:

"$b$ is not a zero-divisor in $B \otimes_A k(A \cap \mathfrak{n})$ for all $\mathfrak{n} \subseteq B$, maximal"

(see, 1.4.5. Korollar, there)

Is the remark above correct?

share|cite|improve this question

Have a look at Milne's errata and notes page:

share|cite|improve this answer

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.