# Generalizing miracle flatness (Matsumura 23.1) via finite Tor-dimension

Let $(A,m_A)$ and $(B,m_B)$ be noetherian local rings and $f:A\rightarrow B$ a local homomorphism. Let $F = B/m_AB$ be the fiber ring and assume that $$\mathrm{dim}(B) = \mathrm{dim}(A) + \mathrm{dim}(F).$$

The following Theorem (23.1 in Matsumura's CRT) is really quite a miracle:

Theorem: If $A$ is regular and $B$ is Cohen-Macaulay then $f$ is flat.

I am wondering to what extent this theorem can be generalized. What I have in mind is a statement of the type:

"Theorem": If $A$ is $X$ and $B$ is $Y$ then $f$ is of finite Tor-dimension (i.e. $\mathrm{Tor}^i_A(B,A)=0$ for all $i$ sufficiently large).

Here, $X$ and $Y$ are ring-theoretic conditions which should be strictly weaker than "regular" and "CM" respectively. Is the "Theorem" above true just requiring $A$ and $B$ to be normal? How about both CM? Or maybe CM plus finitely many $(R_i)$?

Any thoughts/ counterexamples?

• I think you want "finite flat dimension" or something similar. That would be equivalent to $\mathrm{Tor}_A^i(B,M)=0$ for all ${}_A M$ and all $i \gg 0$. All those $\mathrm{Tor}(B,A)$ in the question vanish already, since $A$ is $A$-flat. Dec 1 '09 at 14:40
• Yes, of course, this is what I meant to say (I think finite flat dimension is simply another name for finite tor dimension...) Dec 1 '09 at 17:56

The "Theorem" isn't true with both rings just normal, or just CM, or even normal and CM. Let $A = k[[x,y,z]]/(xz-y^2) \cong k[[a^2,ab,b^2]]$ and let $B = k[[a,b]]$, with $f$ the natural inclusion. The dimensions add up as they must, since $f$ is module-finite. In this case finite flat dimension is the same as finite projective dimension, but $B$ does not have finite projective dimension over $A$.

I don't expect that any addition of assumptions $(R_i)$ would help.

• Of course, as you say, requiring $A$ to be $R_i$ for all $i$ (even just all $i\le \mathrm{dim}(A)$ helps. One might speculate that we could get away with assuming $R_i$ for $i< \mathrm{dim}(A)$, though maybe your nice example already kills this too...(I didn't check). This example also shows that even if one assumes $A$ and $B$ are local complete intersections, there is still no "Theorem". Given the hierarchy regular implies lci implies gorenstein implies CM, it may be too much to expect anything short of regular... Dec 1 '09 at 20:35
• Sorry, I don't really understand your comment. Assuming $A$ has $R_i$ only makes sense for $i \leq \mathrm{dim}A$; $i =\mathrm{dim}A$ is taken care of by my first answer (then you get a "Theorem"), and the second shows that you don't get a "Theorem" even if you assume $R_i$ for all $i < \mathrm{dim}A$. (The example is a normal domain, so $R_1$ and $S_2$.) I think Long's answer is a better avenue to pursue in looking for a "Theorem". Dec 2 '09 at 0:52
• Yes, you have just rephrased exactly what I said in my above comment. In any case, thanks for the good answer and comments! Dec 2 '09 at 7:12

I post this answer to give some intuition about what is really happening behind the scene in the theorem mentioned. If $f:A\rightarrow B$ is flat, then obviously the image of any $A$-regular sequence under $f$ is a $B$-regular sequence. This can be seen by tensoring the $A$-Koszul complex on an $A$-regular sequence, by $B$.

Now let's ask this question: Suppose a map $f:A\rightarrow B$ has the property that it maps any $A$-regular sequence to a $B$-regular sequence. Is $f$ flat then? The answer is no. As an example, you can consider the Frobenius endomorphism $F:A\rightarrow A$ of a local ring of characteristic $p>0$. Obviously it maps every regular sequence to a regular sequence, but $F$ is not flat, unless $A$ is regular, by a theorem of Kunz. Another example is any endomorphism $f$ of a local Cohen-Macaulay ring $(A,\mathfrak{m}_A)$ for which $f(\mathfrak{m}_A)A$ is $\mathfrak{m}_A$-primary. One can see quickly that the image of any regular sequence is a regular sequence, but in general $f$ need not be flat.

The reason for this failure is existence of modules of infinite projective dimension. The condition that $f$ sends any regular sequence to a regular sequence only guarantees that finite free resolutions of $A$-modules stay exact after tensoring by $B$. This quickly follows from Buhsbaum-Eisenbud exactness criterion. (cf. p. 37, Corollary 6.6 in Topics in the homological theory of modules over commutative rings, M. Hochster.)

When $A$ is regular, however, every finite $A$-module has finite projective dimension. That's why in this case the condition that every $A$-regular sequence will be mapped to a $B$ regular sequence by $f$ is equivalent to flatness! (keep in mind that flatness only needs to be checked on finite modules.) The conditions $B$ Cohen-Macaulay and $\dim B=\dim A+\dim F$ are just meant to guarantee that any $A$-regular sequence is mapped to a $B$-regular sequence, as you can check quickly. To check this, take an $A$-regular sequence $x_1,\ldots,x_t$, extend it to a maximal regular sequence $\underline{x}:=(x_1,\ldots,x_d)$ in $A$, then use the dimension assumption and the fact that $B$ is Cohen-Macaulat to show that $f(\underline{x}):=(f(x_1),\ldots,f(x_d))$ is a regular sequence in $B$.

(Note that on one hand, the inclusion $f(\underline{x})B\subseteq\mathfrak{m}_AB$ gives $\dim B/f(\underline{x})B\geq\dim B/\mathfrak{m}_AB$. On the other hand the map $A/\underline{x}\rightarrow B/f(\underline{x})B$ gives $\dim B/f(\underline{x})B\leq \dim A/\underline{x}+\dim B/\mathfrak{m}_AB=0+\dim B/\mathfrak{m}_AB$. Hence $\dim B/f(\underline{x})B=\dim B-\dim A$.)

Hi Bryden,

I agree with Graham that it would be hard to have a generalization in the sense you want. As Graham pointed out you already have finite Tor-dimension if $A$ is regular. In general, finite Tor dimension are much more miraculous. If $A$ is even an hypersurface of isolated singularity of any dimensions, then one can still cook up CM extensions with infinite Tor-dimensions.

However, if you want something like: "Assume $f:A\to B$ has finite Tor-dimension, and assume $A$ is $X$ and $B$ is $Y$, then $f$ is flat", then there is much better chance. For example, one can get results with $X,Y=normal$ plus some low codimension conditions:

http://www.ams.org/proc/1999-127-01/S0002-9939-99-04501-3/home.html

Also, these papers may be worth a look, but you probably already knew them:

Kollar, "Flatness criteria", J. Algebra 175, 712-727.

Cutkosky, "Purity of branch locus and Lefschetz theorems", Compositio Math. 96, (1995) 173-195.

• Hi Long, and thanks for the interesting references. In fact, I want a criterion for finite Tor dimension, so I don't want to assume it! Looks pretty hopeless, though.... Dec 2 '09 at 7:10
• May I ask what do you need finite Tor-dimension for? Dec 2 '09 at 7:45
• Yes, sure! I'll post this as another question tomorrow.... thanks for the interest! Dec 2 '09 at 20:27

If you are willing to impose instead a condition on the closed fiber, namely, that it be regular, then you only need to assume something like $A$ being an excellent normal domain with perfect residue field to get flatness (this is Theorem 3.3.27 in my "Ultraproducts" book).

If $A$ is still regular and $B$ is anything at all, then $B$ has finite flat dimension over $A$. So this is strictly weaker on one ring, though not on both.