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The second answer to this question claims: "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 $B$."

The quoted question talks about local Noetherian rings $(A,m_A)$ and $(B,m_B)$ with $f:A \rightarrow B$ a local homomorphism.

Questions: Could one please explain that claim? Are there other ways to prove it? Is the claim talks about general commutative rings, not necessarily local Noetherian?

For a regular sequence of lengh one, for example, take a non-zero divisor $a \in A$, but, unless $B$ is an integral domain, I do not see why $f(a)$ should be a non-zero divisor in $B$ (perhaps this holds in a local homomorphism?).

(Perhaps this question is relevant, especially its second answer).

Thank you very much!

Edit: It seems that the claim is Lemma 10.68.5 (with $M=R$), in case we are dealing with local rings; I still wonder if the above quoted claim talked about a more general case.

user237522
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  • Did you try to tensor the exact sequence $0\to A/(a_1,...,a_i)\stackrel{\cdot a_{i+1}}\to A/(a_1,...,a_i)$ by $B$? – user26857 May 02 '21 at 06:28
  • @user26857, thank you. Truly, no. Just to make sure: $(A/(a_1,\ldots,a_i)) \otimes_A B \cong B/(f(a_1),\ldots,f(a_i))$? So the claim is true for any flat $A \subseteq B$, with $A$ and $B$ not necessarily local Noetherian? – user237522 May 02 '21 at 06:41
  • Correct! (Btw, I think this is proved in Matsumura, CRT) – user26857 May 02 '21 at 08:01
  • Thank you again. The reference is also helpful; I now found this result in CRT, page 132, exercise 16.4, with the additional condition: $(A/(a_1,\ldots,a_n)) \otimes_A B \neq 0$ ($n$ is the length of the given $A$-sequence $a_1,\ldots,a_n$). – user237522 May 02 '21 at 09:30
  • Of course we have to add such condition in order to don't permit the sequence to "explode" in B. – user26857 May 02 '21 at 10:05
  • Thank you for the comment. – user237522 May 02 '21 at 10:07

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