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
    Nov 11, 2017 at 13:18
  • $\begingroup$ @Sasha Yes, that was a typo. Thank you for the correction. $\endgroup$ 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
    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. $\endgroup$ Nov 12, 2017 at 11:02
  • $\begingroup$ @JasonStarr: OK, thanks a lot! I will think some more about the lci case. $\endgroup$
    – naf
    Nov 12, 2017 at 13:20

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