Let $A$ be an Artin local ring and let $f:A \to B$ a local ring homomorphism to a Noetherian local one dimensional $A$-algebra $B$.

If $B$ is Cohen--Macaulay and the localization of $B$ at any minimal prime is flat over $A$, then is $B$ flat over $A$?

As far as I can see, this doesn't follow from any of the standard flatness criteria, but I could have missed something simple.

(I am mostly interested in the case that $B$ is the localization of a finite type $A$-algebra at some maximal ideal, but it seems likely that the answer to the question is the same with or without this extra assumption.)


1 Answer 1


That is not true. Let $k$ be a field. A module over the ring of dual numbers, $A=k[\epsilon]/\langle \epsilon^2 \rangle,$ is equivalent to a $k$-vector space $B$ with a square-zero $k$-linear self-map $L_\epsilon:B\to B.$ For every integer $q\geq 1$, the Tor module equals, $$\text{Tor}^A_{q\geq 1}(A/\langle \overline{\epsilon}\rangle,B) = H(L_\epsilon,B) = \text{Ker}(L_\epsilon)/\text{Image}(L_\epsilon).$$ Now let $B$ be the following one-dimensional, local, Noetherian $k$-algebra that is a local complete intersection, hence Cohen-Macaulay, $$B=k[x,y]_{\langle x,y \rangle}/ \langle y^2\rangle,$$ where $L_\epsilon$ is multiplication by $\overline{xy}.$

The localization of $B$ at the unique minimal prime $\mathfrak{p}=\langle \overline{y}\rangle$ is $$B_\mathfrak{p} = k(x)[y]/\langle y^2\rangle.$$ The homology module for this localization is $$H(L_\epsilon,B_{\mathfrak{p}}) = \{0\}.$$ Thus, $B_{\mathfrak{p}}$ is $A$-flat. However, $B$ is not flat since $\text{Tor}_{q\geq 1}^A(B,A/\mathfrak{m})$ equals $$ \text{Ker}(L_{\overline{xy}}:k[x,y]/\langle y^2\rangle \to k[x,y]/\langle y^2\rangle)/\text{Image}(L_{\overline{xy}}:k[x,y]/\langle y^2\rangle \to k[x,y]/\langle y^2\rangle) = $$ $$\langle \overline{y}\rangle / \langle \overline{xy} \rangle = k\cdot \overline{y},$$ which is one-dimensional.

  • $\begingroup$ In the definition of $B$, do you mean the quotient by $y^2$? $\endgroup$
    – Sasha
    Commented Nov 11, 2017 at 13:18
  • $\begingroup$ @Sasha Yes, that was a typo. Thank you for the correction. $\endgroup$ Commented Nov 11, 2017 at 13:38
  • $\begingroup$ @JasonStarr: Thanks! Do you think that there can be a counterexample which has a reduced special fibre? This would suffice for my application (to justify the inclusion on l. 3 of p.12 of this paper). $\endgroup$
    – naf
    Commented Nov 12, 2017 at 5:46
  • $\begingroup$ @ulrich. I do not know whether or not the result is true after imposing a reducedness hypothesis for $B/\mathfrak{m}_A B.$ The result appears to be true if $B/\mathfrak{m}_A B$ is regular. The same argument might apply if $B/\mathfrak{m}_A B$ is a local complete intersection scheme, e.g., the domain curve of a stable map. $\endgroup$ Commented Nov 12, 2017 at 11:02
  • $\begingroup$ @JasonStarr: OK, thanks a lot! I will think some more about the lci case. $\endgroup$
    – naf
    Commented Nov 12, 2017 at 13:20

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.