Base change is exact for algebraic groups

Question

I need a reference for the following fact:

let $1 \to G' \to G \to G'' \to 1$ be a ses of algebraic groups over $S$. Let $S' \to S$ be a base change. Then $1 \to G'_{S'} \to G_{S'} \to G''_{S'} \to 1$ is still a ses.

Thanks!

Answer 1

Let $S$ be a non-empty scheme. I define the exactness of a sequence of $S$-group schemes $$ 1\to G^{\,\prime}\overset{i}{\to} G\overset{p}{\to} G^{\,\prime\prime}\to 1 $$ to mean that $p$ is faithfully flat and quasi-compact and that $i$ identifies $G^{\,\prime}$ with the scheme-theoretic kernel of $p$, i.e., with $G\times_{\, G^{\,\prime\prime}}S$, where $S\to G^{\,\prime\prime}$ is the unit section of $G^{\,\prime\prime}\to S$. Now let $T\to S$ be any morphism of schemes. To check that $$ 1\to G_{T}^{\,\prime}\overset{i_{T}}{\to} G_{T}\overset{p_{T}}{\to} G_{T}^{\,\prime\prime}\to 1 $$ is an exact sequence of $T$-group schemes, we need to check that $p_{T}$ is faithfully flat and quasi-compact and that $i_{T}$ identifies $G_{T}^{\,\prime}$ with $G_{T}\times_{\, G_{\,T}^{\,\prime\prime}}T$. For the first statement, see EGA IV_2, Corollary 2.2.13(i) and EGA I (1971), Proposition 6.1.5(iii), p.291. For the second, see EGA I (1971), Corollary 1.3.5, p.33.