# is a cartesian square of a group scheme with $\mathbb{G}_a^n$ fibres reduced?

Let $G$ be a group scheme over $S$ where $S$ is a reduced scheme of finite type over a field $k$ of characteristic 0, and let every fibre $G_s$ over a closed point of $S$ be isomorphic to $\mathbb{G}_a^n$ for some $n$ that varies with $s$.

Is it true then that $G$ is reduced?

Suppose that $G$ is reduced. Is it true that $G \times_S G$ is reduced?

(This question is a continuation of this one; the motivation comes from this question)

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Is $n$ independent of $s$? – Laurent Moret-Bailly Feb 15 '12 at 21:01
No. Perhaps I should have mentioned explicitly that $n$ varies with $s$. – Dima Sustretov Feb 15 '12 at 21:17
Why "tensor square"? Isn't it just a cartesian square (over $S$)? – Qfwfq Feb 15 '12 at 23:02
@Qfwfq: corrected – Dima Sustretov Feb 16 '12 at 12:12

Take $S=\mathrm{Spec}(R)$ where $R$ is, say, a noetherian domain. Let $t\in S$ be nonzero and noninvertible. Let $G$ be the kernel of $f:\mathbb{G}_{a,S}^2\to\mathbb{G}_{a,S}$ sending $(x,y)$ to $t^2\,y$. For each $s\in S$, the fiber $G_s$ is equal to $\mathbb{G}_{a,s}^2$ if $t$ is zero at $s$, and to $\mathbb{G}_{a,s}\times\{0\}$ otherwise. But $G$ is (in general) not reduced since the function $ty$ is nonzero, with square zero.
Oh, of course, you are right; I have asked a question that is too weak. I am sorry. In the comments to David Speyer's answer to the referenced question I have asked if a product of such $G$ is reduced if $G$ is reduced. I wanted to formulate some condition on a group scheme that would presumably imply reducedness. I will update the question – Dima Sustretov Feb 15 '12 at 22:47