Lemma 15.5.3. Suppose given a cartesian diagram of rings
i.e., $B' = B \times _ R R'$. If $h \in B'$ corresponds to $g \in B$ and $f \in R'$ such that $s(g) = t(f)$, then the diagram
is cartesian too.
Lemma 15.5.3. Suppose given a cartesian diagram of rings
i.e., $B' = B \times _ R R'$. If $h \in B'$ corresponds to $g \in B$ and $f \in R'$ such that $s(g) = t(f)$, then the diagram
is cartesian too.
Proof. The equality $B' = B \times _ R R'$ tells us that
is an exact sequence of $B'$-modules. We have $B_ g = B_ h$, $R'_ f = R'_ h$, and $R_{s(g)} = R_{t(f)} = R_ h$ as $B'$-modules. By exactness of localization (Algebra, Proposition 10.9.12) we find that
is an exact sequence. This proves the lemma. $\square$
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