Lemma 10.35.24. Let $R$ be a ring. Let $x, y \in R$ be nonzerodivisors. Let $R[x/y] \subset R_{xy}$ be the $R$-subalgebra generated by $x/y$, and similarly for the subalgebras $R[y/x]$ and $R[x/y, y/x]$. If $R$ is integrally closed in $R_ x$ or $R_ y$, then the sequence

\[ 0 \to R \xrightarrow {(-1, 1)} R[x/y] \oplus R[y/x] \xrightarrow {(1, 1)} R[x/y, y/x] \to 0 \]

is a short exact sequence of $R$-modules.

**Proof.**
Since $x/y \cdot y/x = 1$ it is clear that the map $R[x/y] \oplus R[y/x] \to R[x/y, y/x]$ is surjective. Let $\alpha \in R[x/y] \cap R[y/x]$. To show exactness in the middle we have to prove that $\alpha \in R$. By assumption we may write

\[ \alpha = a_0 + a_1 x/y + \ldots + a_ n (x/y)^ n = b_0 + b_1 y/x + \ldots + b_ m(y/x)^ m \]

for some $n, m \geq 0$ and $a_ i, b_ j \in R$. Pick some $N > \max (n, m)$. Consider the finite $R$-submodule $M$ of $R_{xy}$ generated by the elements

\[ (x/y)^ N, (x/y)^{N - 1}, \ldots , x/y, 1, y/x, \ldots , (y/x)^{N - 1}, (y/x)^ N \]

We claim that $\alpha M \subset M$. Namely, it is clear that $(x/y)^ i (b_0 + b_1 y/x + \ldots + b_ m(y/x)^ m) \in M$ for $0 \leq i \leq N$ and that $(y/x)^ i (a_0 + a_1 x/y + \ldots + a_ n(x/y)^ n) \in M$ for $0 \leq i \leq N$. Hence $\alpha $ is integral over $R$ by Lemma 10.35.2. Note that $\alpha \in R_ x$, so if $R$ is integrally closed in $R_ x$ then $\alpha \in R$ as desired.
$\square$

## Comments (1)

Comment #5426 by minsom on