# Flatness and Cohen-Macaulay rings

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.)

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.
• In the definition of $B$, do you mean the quotient by $y^2$? – Sasha Nov 11 '17 at 13:18
• @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. – Jason Starr Nov 12 '17 at 11:02