Lemma 10.69.8. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $a \in I$. If $a$ is not contained in any minimal prime of $R$, then $\mathop{\mathrm{Spec}}(R[\frac{I}{a}]) \to \mathop{\mathrm{Spec}}(R)$ has dense image.

Proof. If $a^ k x = 0$ for $x \in R$, then $x$ is contained in all the minimal primes of $R$ and hence nilpotent, see Lemma 10.16.2. Thus the kernel of $R \to R[\frac{I}{a}]$ consists of nilpotent elements. Hence the result follows from Lemma 10.29.6. $\square$

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