Let $G,H$ be group varieties over a field $k$, where by variety I mean an integral, separated scheme of finite type over $k$. Suppose we have a homomorphism of group varieties $\varphi:G\rightarrow H$, define the kernel of $\varphi$ as the fibre product $ker(\varphi)= G\times_{H}Spec(k)$ where the $H$-scheme structure on $Spec(k)$ is given by the unit $k$-rational point $e:Spec(k)\rightarrow H$.
My question is, is $ker(\varphi)$ a variety? I have a feeling that it ought to be, but I'm not entirely sure. It's obviously a scheme over $k$ which is a good start. As for the rest of the conditions, I am completely stumped.
No, $\ker(\varphi)$ does not have to be integral. For a simple example, let $G=H=\mathbb{G}_m$ and let $\varphi$ be the homomorphism $t\mapsto t^2$. Then intuitively, the kernel of $\varphi$ is not integral since it is $\{1,-1\}$ which is reducible. More formally, the kernel of $\varphi$ is Spec of the ring $k[t]/(t^2-1)$, which is not a domain since $(t+1)(t-1)=0$ in it. (If $k$ has characteristic $\neq 2$ it is isomorphic to $k\times k$, and if $k$ has characteristic $2$ it is isomorphic to $k[s]/(s^2)$ and so the kernel is irreducible but not reduced.)