ac.commutative algebra - Generalizing miracle flatness (Matsumura 23.1) via finite Tor-dimension

I post this answer to give some intuition about what is really happening behind the scene in the theorem mentioned. If $f:Arightarrow B$ 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 $f:Arightarrow B$ 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 $F:Arightarrow A$ 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 $f(underline{x})Bsubseteqmathfrak{m}_AB$ gives $dim B/f(underline{x})Bgeqdim B/mathfrak{m}_AB$. On the other hand the map $A/underline{x}rightarrow B/f(underline{x})B$ gives $dim B/f(underline{x})Bleq dim A/underline{x}+dim B/mathfrak{m}_AB=0+dim B/mathfrak{m}_AB$. Hence $dim B/f(underline{x})B=dim B-dim A$.)

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