Is the fiber product of the connected component of a group scheme connected?

Question

Let $G$ be a group scheme over a field $k$. Let $G^0$ be the connected component containing the identity. Is it true that $G^0\times_k G^0$ is connected?
I know that this is true if $G^0$ is geometrically connected, but how to prove it in general?

Answer 1

Thanks to some outside help I figured out the answer to the question. Actually everything needed was here http://stacks.math.columbia.edu/download/varieties.pdf.
Since the identity is a $k$-rational point contained in $G^0$ and this one is connected, we can apply Lemma 5.14. Then $G^0$ is geometrically connected, this allows us to conclude using Lemma 5.4.

Answer 2

Since the numberings in Stacks Project have been updated, for convenience, I rewrite the argument here.

To show that $G^0\times_k G^0$ is connected, by [EGAIV$_2$, 4.5.8], it suffices to show that $G^0$ is geometrically connected. Since $G^0_k$ is connected by definition, and contains a $k$-rational point (which is the neutral section), we apply

[EGAIV$_2$, 4.5.14] Let $X$ be a scheme over a field $k$. If there is a point $x\in X$ such that $\kappa (x)$ is a primary extension of $k$ (for instance, $x$ is a $k$-rational point of $X$), then the connected component of $X$ containing $x$ is geometrically connected.

In conclusion, for every group scheme $G$ over a field $k$, the neutral component $G^0$ is geometrically connected and $G^0\times_k G^0$ is connected.