The Stacks project

Proof. Set $I = \mathfrak m_1 \cap \ldots \cap \mathfrak m_ r$. Set $A_ i = A_{\mathfrak m_ i}$ and $A' = \prod A_ i$. Then $IA' = \prod \mathfrak m_ i A_ i$ and $A \to A'$ is a flat ring map such that $A/I \cong A'/IA'$. Thus we may use More on Algebra, Lemma 15.89.16 to see that there exists an $A$-module map $\varphi : A \to B$ with $\varphi _ i$ isomorphic to the localization of $\varphi $ at $\mathfrak m_ i$. Then we can use the discussion in More on Algebra, Remark 15.89.19 to endow $B$ with an $A$-algebra structure matching the given $A$-algebra structure on $B_ i$. The final statement of the lemma follows easily from the fact that $\mathop{\mathrm{Ker}}(\varphi )_{\mathfrak m_ i} \cong \mathop{\mathrm{Ker}}(\varphi _ i)$ and $\mathop{\mathrm{Coker}}(\varphi )_{\mathfrak m_ i} \cong \mathop{\mathrm{Coker}}(\varphi _ i)$. $\square$

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. $\square$

Proof. Choose a minimal prime $\mathfrak q \subset R$. Observe that $\dim (R/\mathfrak q) = 1$, in particular $R/\mathfrak q$ is not a field. We can choose a discrete valuation ring $A$ dominating $R/\mathfrak q$ with the same fraction field, see Algebra, Lemma 10.119.1. Observe that $a$ and $b$ map to nonzero elements of $A$ as nonzerodivisors in $R$ are not contained in $\mathfrak q$. Let $v$ be the discrete valuation on $A$. Then $v(a) > 0$ as $a \in \mathfrak m$. We claim $n = v(b)/v(a)$ works.

Let $R \subset R'$ be given. Set $A' = A \otimes _ R R'$. Since $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is surjective (Algebra, Lemma 10.36.17) also $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ is surjective (Algebra, Lemma 10.30.3). Pick a prime $\mathfrak q' \subset A'$ lying over $(0) \subset A$. Then $A \subset A'' = A'/\mathfrak q'$ is a finite extension of rings (again inducing a surjection on spectra). Pick a maximal ideal $\mathfrak m'' \subset A''$ lying over the maximal ideal of $A$ and a discrete valuation ring $A'''$ dominating $A''_{\mathfrak m''}$ (see lemma cited above). Then $A \to A'''$ is an extension of discrete valuation rings and we have $b = a^ m c$ in $A'''$. Thus $v'''(b) \geq mv'''(a)$. Since $v''' = ev$ where $e$ is the ramification index of $A'''/A$, we find that $m \leq n$ as desired. $\square$

Proof. Since at least one $a_ i$ is not a unit we find that $\mathfrak m$ is not an associated prime of $A$. Moreover, for any $A \subset B$ as in the statement $\mathfrak m$ is not an associated prime of $B$ and $\mathfrak m_ j$ is not an associate prime of $B_ j$. Keeping this in mind will help check the arguments below.

First, we claim that it suffices to prove the lemma for $r = 2$. We will argue this by induction on $r$; we suggest the reader skip the proof. Suppose we are given $A \subset B$ and $\pi _ j$ in $B_ j = B_{\mathfrak m_ j}$ and factorizations $a_ i = u_{i, j} \pi _ j^{e_{i, j}}$ for $i = 1, \ldots , r - 1$ in $B_ j$ with $u_{i, j} \in B_ j$ units and $e_{i, j} \geq 0$. Then by the case $r = 2$ for $\pi _ j$ and $a_ r$ in $B_ j$ we can find extensions $B_ j \subset C_ j$ and for every maximal ideal $\mathfrak m_{j, k}$ of $C_ j$ a nonzerodivisor $\pi _{j, k} \in C_{j, k} = (C_ j)_{\mathfrak m_{j, k}}$ and factorizations

as in the lemma. There exists a unique finite extension $B \subset C$ with $C/B$ annihilated by a power of $\mathfrak m$ such that $C_ j \cong C_{\mathfrak m_ j}$ for all $j$, see Lemma 42.4.1. The maximal ideals of $C$ correspond $1$-to-$1$ to the maximal ideals $\mathfrak m_{j, k}$ in the localizations and in these localizations we have

for $i \leq r - 1$. Since $a_ r$ factors correctly too the proof of the induction step is complete.

Proof of the case $r = 2$. We will use induction on

If $\ell = 0$, then either $a_1$ or $a_2$ is a unit and the lemma holds with $A = B$. Thus we may and do assume $\ell > 0$.

Suppose we have a finite extension of rings $A \subset A'$ such that $A'/A$ is annihilated by a power of $\mathfrak m$ and such that $\mathfrak m$ is not an associated prime of $A'$. Let $\mathfrak m_1, \ldots , \mathfrak m_ r \subset A'$ be the maximal ideals and set $A'_ i = A'_{\mathfrak m_ i}$. If we can solve the problem for $a_1, a_2$ in each $A'_ i$, then we can apply Lemma 42.4.1 to produce a solution for $a_1, a_2$ in $A$. Choose $x \in \{ a_1, a_2\} $ such that $\ell = \text{length}_ A(A/xA)$. By Lemma 42.2.5 and (42.2.2.1) we have $\text{length}_ A(A/xA) = \text{length}_ A(A'/xA')$. On the other hand, we have

by Algebra, Lemma 10.52.12. Since $x \in \mathfrak m$ we see that each term on the right hand side is positive. We conclude that the induction hypothesis applies to $a_1, a_2$ in each $A'_ i$ if $r > 1$ or if $r = 1$ and $[\kappa (\mathfrak m_1) : \kappa (\mathfrak m)] > 1$. We conclude that we may assume each $A'$ as above is local with the same residue field as $A$.

Applying the discussion of the previous paragraph, we may replace $A$ by the ring constructed in Lemma 42.4.2 for $a_1, a_2 \in A$. Then since $A$ is local we find, after possibly switching $a_1$ and $a_2$, that $a_2 \in (a_1)$. Write $a_2 = a_1^ m c$ with $m > 0$ maximal. In fact, by Lemma 42.4.3 we may assume $m$ is maximal even after replacing $A$ by any finite extension $A \subset A'$ as in the previous paragraph. If $c$ is a unit, then we are done. If not, then we replace $A$ by the ring constructed in Lemma 42.4.2 for $a_1, c \in A$. Then either (1) $c = a_1 c'$ or (2) $a_1 = c a'_1$. The first case cannot happen since it would give $a_2 = a_1^{m + 1} c'$ contradicting the maximality of $m$. In the second case we get $a_1 = c a'_1$ and $a_2 = c^{m + 1} (a'_1)^ m$. Then it suffices to prove the lemma for $A$ and $c, a'_1$. If $a'_1$ is a unit we're done and if not, then $\text{length}_ A(A/cA) < \ell $ because $cA$ is a strictly bigger ideal than $a_1A$. Thus we win by induction hypothesis. $\square$