Lemma 31.15.10. Let $R$ be a Noetherian UFD. Let $I \subset R$ be an ideal such that $R/I$ has no embedded primes and such that every minimal prime over $I$ has height $1$. Then $I = (f)$ for some $f \in R$.

Proof. By Lemma 31.15.9 the ideal sheaf $\tilde I$ is invertible on $\mathop{\mathrm{Spec}}(R)$. By More on Algebra, Lemma 15.117.3 it is generated by a single element. $\square$

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