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If I have an injection $A \to B$ of noetherian reduced rings. Does this in general induce an injection $$ Q(A) \hookrightarrow Q(B) $$ of total rings of fractions?

In the proof of Lemma 2.6 (Greuel et. al.) they say this is clear but I don't see it, since I don't know why a non-zero divisor of $A$ is a non-zero divisor of $B$. Have I overlooked something?

Edit: For an answer to the "induced" question see the comment of user26857 below. Here $\psi(X)$ would be an invertible element under any homomorphism $\psi \colon Q(A) \to Q(B)$. Hence $\psi$ cant be induced by $A \to B$, since $\iota(X)$ is not invertible in $Q(B)$, for $\iota \colon A \to B$.

For the answer why the proof in the paper works, see the answer by Dave.

pyrogen
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    As far as I can see the authors suppose that $A\subset B\subset Q(A)$. Now, if $a\in A$ is a zerodivisor on $B$ it is also a zerodivisor on $Q(A)$, and therefore a zerodivisor on $A$. – user26857 Nov 19 '19 at 09:47
  • Thank you! This makes sense! I read it like it is clear, that Q(A) is included in Q(B) and then they assume as third condition, that B is included in Q(A). But if they mean that they choose B in Q(A), then it makes sense. – pyrogen Nov 19 '19 at 11:04
  • On the other hand, I had the impression that they use the fact, that non-zero divisors are mapped to nonzero divisors for finite monomorphism several times. For example in the proof of Propositions 3.2 in the same paper. Is this true? – pyrogen Nov 19 '19 at 11:08
  • The ring extension $A=K[X]\subset K[X,Y]/(XY,Y^2-Y)=B$ is finite, and $X$ is a zerodivisor on $B$. (Note that $A$, $B$ are reduced.) – user26857 Nov 19 '19 at 21:40
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    Thank you user26857! This answers my question! Sorry that I can't upvote your comment, since I seem to have clicked doubly on the arrow. – pyrogen Nov 20 '19 at 12:43
  • I think the question about non-zerodivisors in finite ring extensions (eventually of reduced rings) deserves a thread of its own. – user26857 Nov 20 '19 at 16:38
  • I made a new thread here: https://math.stackexchange.com/questions/3445174/finite-ring-extensions-doesnt-preserve-non-zero-divisors. Maybe this helps someone – pyrogen Nov 21 '19 at 14:42

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Let $\psi:A\to B$ be an injection, and let $\varphi:Q(A)\to Q(B)$ be the corresponding induced map. We have that $Q(B)=S^{-1}B$, where $S$ is the set of non-zero divisors of $B$.

If $\varphi(\frac{x}{y})=0$ we have that $\psi(x)s=0$ for some $s\in S$. But since $s$ is not a zero divisor we have that $\psi(x)=0$ and so $x=0$ by the injectivity of $\psi$. Then $\frac{x}{y}=0$ and so $\varphi$ is injective.

Edit: Apologies for misinterpreting your question. The authors further suppose in Lemma 2.6 that $B\subseteq Q(\psi(A))$, which guarantees regular elements of $A$ are mapped to regular elements of $B$.

To see this, suppose that $a\in A$ is regular in $A$, and that $\psi(a)b=0$ for some $b\in B$. We have $b=\frac{\psi(r)}{\psi(s)}$ for some $r,s\in A$ by this additional property. Then $t\psi(ar)=0$ for some regular $t\in\psi(A)$, so $\psi(ar)=0$. Then by the injectivity of $\psi$ we have $ar=0$, so by the regularity of $a$ we have $r=0$. Then $b=0$ and so $\psi(a)$ is regular in $B$.

Dave
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  • Thank you Dave! My question was more in regard to why this homomorphism exists in the first place. See the comments above. If any non-zero divisor is mapped to a zero-divisor it would not exist. Do you maybe know if this is true for finite homomorphism of noetherian maps or what I have to assume for this to be true? – pyrogen Nov 19 '19 at 11:11
  • Ah I see what you mean now, sorry! I've updated my answer now I've reread it, I hope this is more helpful – Dave Nov 19 '19 at 13:22