The Stacks project

Proof. If $a$ or $b$ is a unit, then the lemma is true with $R = R'$. Thus we may assume $a, b \in \mathfrak m$. Set $I = (a, b)$. The idea is to blow up $R$ in $I$. Instead of doing the algebraic argument we work geometrically. Let $X = \text{Proj}(\bigoplus _{d \geq 0} I^ d)$. By Divisors, Lemma 31.32.4 the morphism $X \to \mathop{\mathrm{Spec}}(R)$ is an isomorphism over the punctured spectrum $U = \mathop{\mathrm{Spec}}(R) \setminus \{ \mathfrak m\} $. Thus we may and do view $U$ as an open subscheme of $X$. The morphism $X \to \mathop{\mathrm{Spec}}(R)$ is projective by Divisors, Lemma 31.32.13. Also, every generic point of $X$ lies in $U$, for example by Divisors, Lemma 31.32.10. It follows from Varieties, Lemma 33.17.2 that $X \to \mathop{\mathrm{Spec}}(R)$ is finite. Thus $X = \mathop{\mathrm{Spec}}(R')$ is affine and $R \to R'$ is finite. We have $R_ a \cong R'_ a$ as $U = D(a)$. Hence a power of $a$ annihilates the finite $R$-module $R'/R$. As $\mathfrak m = \sqrt{(a)}$ we see that $R'/R$ is annihilated by a power of $\mathfrak m$. By Divisors, Lemma 31.32.4 we see that $IR'$ is a locally principal ideal. Since $R'$ is semi-local we see that $IR'$ is principal, see Algebra, Lemma 10.78.7, say $IR' = (t)$. Then we have $a = a't$ and $b = b't$ and everything is clear. $\square$