The Stacks project

Lemma 115.5.8. Let $R$ be a ring, let $f \in R$. Suppose we have $S$, $S'$ and the solid arrows forming the following commutative diagram of rings

\[ \xymatrix{ & S'' \ar@{-->}[rd] \ar@{-->}[dd] & \\ R \ar[rr] \ar@{-->}[ru] \ar[d] & & S \ar[d] \\ R_ f \ar[r] & S' \ar[r] & S_ f } \]

Assume that $R_ f \to S'$ is finite. Then we can find a finite ring map $R \to S''$ and dotted arrows as in the diagram such that $S' = (S'')_ f$.

Proof. Namely, suppose that $S'$ is generated by $x_ i$ over $R_ f$, $i = 1, \ldots , w$. Let $P_ i(t) \in R_ f[t]$ be a monic polynomial such that $P_ i(x_ i) = 0$. Say $P_ i$ has degree $d_ i > 0$. Write $P_ i(t) = t^{d_ i} + \sum _{j < d_ i} (a_{ij}/f^ n) t^ j$ for some uniform $n$. Also write the image of $x_ i$ in $S_ f$ as $g_ i / f^ n$ for suitable $g_ i \in S$. Then we know that the element $\xi _ i = f^{nd_ i} g_ i^{d_ i} + \sum _{j < d_ i} f^{n(d_ i - j)} a_{ij} g_ i^ j$ of $S$ is killed by a power of $f$. Hence upon increasing $n$ to $n'$, which replaces $g_ i$ by $f^{n' - n}g_ i$ we may assume $\xi _ i = 0$. Then $S'$ is generated by the elements $f^ n x_ i$, each of which is a zero of the monic polynomial $Q_ i(t) = t^{d_ i} + \sum _{j < d_ i} f^{n(d_ i - j)} a_{ij} t^ j$ with coefficients in $R$. Also, by construction $Q_ i(f^ ng_ i) = 0$ in $S$. Thus we get a finite $R$-algebra $S'' = R[z_1, \ldots , z_ w]/(Q_1(z_1), \ldots , Q_ w(z_ w))$ which fits into a commutative diagram as above. The map $\alpha : S'' \to S$ maps $z_ i$ to $f^ ng_ i$ and the map $\beta : S'' \to S'$ maps $z_ i$ to $f^ nx_ i$. It may not yet be the case that $\beta $ induces an isomorphism $(S'')_ f \cong S'$. For the moment we only know that this map is surjective. The problem is that there could be elements $h/f^ n \in (S'')_ f$ which map to zero in $S'$ but are not zero. In this case $\beta (h)$ is an element of $S$ such that $f^ N \beta (h) = 0$ for some $N$. Thus $f^ N h$ is an element of the ideal $J = \{ h \in S'' \mid \alpha (h) = 0 \text{ and } \beta (h) = 0\} $ of $S''$. OK, and it is easy to see that $S''/J$ does the job. $\square$

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