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.