Let $k$ be an arbitrary field of positive characteristic and let $V$ be a smooth projective (irreducible) variety over $k$. Suppose that $K/k$ is a field extension such that $V_K:=V\times_{\text{Spec }k}\text{Spec }K$ remains reduced and irreducible. Can we deduce from this that $V_K$ is smooth over $K$?
If not (and probably this is the case), what minimal assumptions are required to make sure that $V_K$ is smooth? Would it help if we knew that $\text{tr. deg. }K/k=1$ and $k$ is algebraically closed in $K$?
Your question
Given a variety $V$ over a field $k$ and an extension field $k\to K$, the variety $V$ is smooth over $k$ if and only of the variety $V_K$ is smooth over $K$.
Vastly more general result
Smoothness is compatible with base change in all generality: if an arbitrary morphism of schemes $X\to S$ is smooth, it will remain smooth after an arbitrary base change $S'\to S$, i.e. $X\times _S S'\to S'$ will be smooth too: EGA IV,4, Proposition (17.3.3) (iii).
And if $S'\to S$ is faithfully flat, then descent of smoothness holds, namely if $X\times _S S'\to S'$ is smooth, then so was $X\to S$: EGA IV,4, Corollaire (17.7.3) (ii).
Edit
Matt in his comment to the question nailed it: smoothness means that the determinant of some matrix is non zero and this does not depend on the field in which you want to consider that the entries of the matrix belong.
So despite the apparent inpenetrability of the EGA IV reference, this is essentially undergraduate stuff :-)