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Can someone explain to me the proof for "This is the case iff each element $u\in U$ is a non-zero-divisor mod $J$"?

I'm stuck on the forward implication. From what I understand, because we assume that $J = \varphi^{-1}(I)$ we obtain an injective homomorphism $R/J\to R[U^{-1}]/I$ along which the pullback of $U$ from $R[U^{-1}]/I$ into $R/J$ are going to be non-zero divisors? If this is correct, then it makes sense to me, but I'm having a hard time seeing what is happening. Could someone please explain to me what the right intuition is in this situation?

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klein4
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Recall that $\varphi:R\to R[U^{-1}]$ is defined by $x\mapsto x/1$. Let $u\in U$. We need to prove that $u$ is a nonzerodivisor on $R/J$. So assume that $u(x+J)=0$ for some $x\in R$. This means $ux\in J=\varphi^{-1}(I)$, that is, $\varphi(ux)\in I$. Now $\varphi(u)=u/1$ is a unit in $R[U^{-1}]$. Thus we obtain $$\varphi(u)^{-1}(\varphi(ux))=\varphi(u)^{-1}\varphi(u)\varphi(x)=\varphi(x)\in I.$$ In other words, $x\in \varphi^{-1}(I)=J$; hence $x+J=0$ as desired.

cqfd
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