Product of varieties in is a variety? I know that the question may look similar to this: Is fibre product of varieties irreducible (integral)?, but I am forced by the context to use a different definition for variety.
Definition. Let $K$ be a field (not necessarly algebraically closed), and $X$ a scheme over $K$. Then $X$ is a variety if it is separated, of finite type, and geometrically integral.
Assume that $X$ and $Y$ are two varieties. Is it true that $X\times_K Y$ (the scheme-theoretical fiber product) is a variety? (Of course in the sense of the previous definition).
If you are able to find a clear reference for this, it would be better than an explicit argument.
 A: Here is a reference: on page 136 of the book of Görtz-Wedhorn, Proposition 5.51 (ii) says that a $k$-scheme $X$ is geometrically integral over $k$ if and only if for every integral $k$-scheme $Y$, $X\times_kY$ is integral. This doesn't quite give you what you want immediately, in that there is still an argument to be made, but this is the key ingredient. The fact that $X\times_kY$ is separated and of finite type over $k$ is standard, and presumably not the main point of your question. By the aforementioned result, it is integral. Why is it geometrically integral? Remember we are assuming $Y$ as well as $X$ is geometrically integral over $k$. For any extension $K/k$, $(X\times_kY)\times_k\mathrm{Spec}(K)=(X\times_k\mathrm{Spec}(K))\times_k(Y\times_k\mathrm{Spec}(K))$. Since $X$ is geometrically integral, $X\times_k\mathrm{Spec}(K)$ is as well (geometric integrality is a definition which is obviously insensitive to an extension of the base field), and since $Y$ is geometrically integral, $Y\times_k\mathrm{Spec}(K)$ is in particular integral, so, invoking the result again, we get that $(X\times_kY)\times_k\mathrm{Spec}(K)$ is integral. As $K$ was arbitrary, we win: $X\times_kY$ is geometrically integral, and hence a variety in the sense given. 
