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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?

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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. –  Graham Leuschke 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...) –  B. Cais Dec 1 '09 at 17:56
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5 Answers

up vote 4 down vote accepted

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.

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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... –  B. Cais 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". –  Graham Leuschke 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! –  B. Cais Dec 2 '09 at 7:12
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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:


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.

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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.... –  B. Cais Dec 2 '09 at 7:10
May I ask what do you need finite Tor-dimension for? –  Hailong Dao Dec 2 '09 at 7:45
Yes, sure! I'll post this as another question tomorrow.... thanks for the interest! –  B. Cais Dec 2 '09 at 20:27
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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 $\dim B/f(\underline{x})B=\dim B-\dim A$.)

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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).

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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.

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