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.
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 $B$.