2
$\begingroup$

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?

$\endgroup$

1 Answer 1

7
$\begingroup$

Have a look at Milne's errata and notes page: http://www.jmilne.org/math/Books/add/ECPUP.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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