The Stacks project

Proof. We will use the results of Lemma 15.90.7 without further mention. The functor $M \otimes _ R -$ is right exact (Algebra, Lemma 10.12.10) hence we get (1).

The kernel of $M \to M \otimes _ R R_ f = M_ f$ is $M[f^\infty ]$. Thus (2) follows.

If the sequence is exact in the middle, then elements of the form $(x, 0)$ with $x \in (M \otimes _ R R')[f^\infty ]$ are in the image of the first arrow. This implies that $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is surjective. Conversely, assume that $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is surjective. Let $(x, y)$ be an element in the middle which maps to zero on the right. Write $y = y'/f^ n$ for some $y' \in M$. Then we see that $f^ n x - y'$ is annihilated by some power of $f$ in $M \otimes _ R R'$. By assumption we can write $f^ nx - y' = z$ for some $z \in M[f^\infty ]$. Then $y = y''/f^ n$ where $y'' = y' + z$ is in the kernel of $M \to M/f^ nM$. Hence we see that $y$ can be represented as $y'''/1$ for some $y''' \in M$. Then $x - y'''$ is in $(M \otimes _ R R')[f^\infty ]$. Thus $x - y''' = z' \in M[f^\infty ]$. Then $(x, y'''/1) = (y''' + z', (y''' + z')/1)$ as desired.

If $(R \to R', f)$ is a glueing pair, then (15.90.10.1) is exact in the middle for any $M$ by Algebra, Lemma 10.12.10. This gives the penultimate statement of the lemma. The final statement of the lemma follows from this and the fact that $(R, f)$ is a glueing pair if and only if $(R \to R^\wedge , f)$ is a glueing pair. $\square$