The following is a confusion I can't get rid of concerning group-schemes.
Let $G$ be a group scheme over some base $S$: for any scheme $T$ over $S$, the set of $T$-valued points $G(T)$ has a group structure that I denote multiplicatively, and such that for any $T'\rightarrow T \rightarrow S$, the natural map $G(T)\rightarrow G(T')$ is a group homomorphism.
Let me fix $T\rightarrow S$ and consider the base change $G_T=G\times _S T$ of $G$ to $T$. I know that $G_T$ still is a group scheme over $T$, but how can I describe the multiplication in $G_T(T')$ whith respect to that in $G(T')$, for $T'$ a scheme over $T$ ?
That is, elements of $G_T(T')$ are pairs $(g,t)\in G(T')\times T(T')$ of points which agree when seen as points in $S(T')$. If I want to multiply $(g,t)$ with $(\hat g,\hat t)$, on the first component I could see $g\hat g$, but what about the second component ? How can I take both $t$ and $\hat t$ into account ?
In particular, I wonder what happens when $T$ is itself another group scheme over $S$. In this case, could the group law on $G\times _S T$ be $(g,t)(\hat g,\hat t)=(g\hat g,t\hat t)$ ?
EDIT: Actually I may have understood the origin of my confusion. The description I give of $G_T(T')$ is wrong : it would be true if I considered $T'$ as a scheme over $S$. But here, I'm considering it over $T$. I am given a canonical $t\in T(T')$ which is the structure morphism, and $G_T(T')$ consist of the pairs $(g,t)$ which $g\in G(T')$. The second component is always the fixed point $t$. Then, I have simply $(g,t)(\hat g,t)=(g\hat g,t)$. In general, I can't give $G_T$ the structure of a group scheme over $S$.
However, when $T$ is also a group scheme over $S$, then $G_T$ can be given a structure of group scheme over $S$, via $(g,t)(\hat g,\hat t)=(g\hat g,t\hat t)$. In this situation, it's better to denote it by $G\times _S T$ rather than $G_T$, to remove ambiguity.
I ended up answering this by myself, but I would still appreciate any confirmation whether my way of thinking is correct. Thanks !