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

\[ \xymatrix{ R & R' \ar[l]^ t \\ B \ar[u]_ s & B'\ar[u] \ar[l] } \]

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

\[ \xymatrix{ R_{s(g)} = R_{t(f)} & (R')_ f \ar[l]^-t \\ B_ g \ar[u]_ s & (B')_ h \ar[u] \ar[l] } \]

is cartesian too.

**Proof.**
The equality $B' = B \times _ R R'$ tells us that

\[ 0 \to B' \to B \oplus R' \xrightarrow {s, -t} R \]

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

\[ 0 \to B'_ h \to B_ g \oplus R'_ f \xrightarrow {s, -t} R_{s(g)} = R_{t(f)} \]

is an exact sequence. This proves the lemma. $\square$

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