Let $X$ be a connected scheme of finite type over a field $k$. I'm trying to understand the following three statements:
(i): If $X$ has a $k$-rational point, then $X_{k'} = X \times_k k'$ is connected for any finite extension of $k$.
(ii): The same as (i), but the conclusion holds for any field extension of $k$.
(iii): If $X$ and $Y$ are geometrically connected schemes of finite type over $k$, then so is $X \times_k Y$.
For (i), a $k$-rational point is a morphism of $k$-schemes $\textrm{Spec } k \rightarrow X$. Equivalently, it is a point of $X$ at which the residue field is $k$.
The projection map $\pi: X_{k'} \rightarrow X$ is surjective and (I think) open and closed. So if $U$ is an open and closed set in $X_{k'}$, then $\pi(U)$ will be an open and closed set in $X$, hence $\pi(U) = X$. Thus every $k$-rational point of $X$ comes from $U$. The same can be said for the complement of $U$ in $X_{k'}$.
In the affine case, here is an equivalent statement for (i): $R$ is a finitely generated $k$-algebra, and there exists a $k'$-algebra homomorphism $R \otimes_k k'\rightarrow k'$ which maps $R$ into $k$. Show that if $R \otimes_k k'$ is isomorphic to a direct product of two nonzero rings, then the same is true for $R$.
I am not sure where to go from here, and would appreciate any hints.
