Question

Let $K/k$ be an arbitrary field extension and $X$, $Y$ varieties over $k$ (lets assume projective and perhaps smooth to avoid technicalities). There is a fine moduli space of morphisms between $X$ and $Y$ parametrized by a scheme $Hom(X,Y)$ over $k$. My question is whether $Hom(X,Y)_K$ is isomorphic to $Hom(X_K, Y_K)$ where ${}_K$ denotes tensoring with $Spec K$. From the universal property of $Hom(X,Y)$ we have a bijection between maps $Spec K\to Hom(X,Y)$ (i.e. $K$-rational points, as a set contained in $Hom(X,Y)_K$) and maps $Y_K\to X_K$. I seem to be a bit confused as to whether this is enough to conclude.

Finally, could we get a similar thing to work for the coarse Kontsevich moduli space $\mathcal{M}_{g,n}(X,\beta)$ even though there is no universal family? Thanks.

Answer 1

If $X,Y$ are $S$-schemes, then $\underline{\hom}_S(X,Y)$ denotes the sheaf on $S$-schemes defined by $T \mapsto \hom_T(X_T,Y_T)$. Now, if $S'/S$ is any base change, we have for every $S'$-scheme $T$:

$$\hom_{S'}(T,\underline{\hom}_S(X,Y)_{S'}) = \hom_S(T,\underline{\hom}_S(X,Y)) = \hom_T(X_T,Y_T)$$

$$ = \hom_T((X_{S'})_T,(Y_{S'})_T)) = \hom_{S'}(T,\underline{\hom}_{S'}(X_{S'},Y_{S'}))$$

Hence, $\underline{\hom}_S(X,Y)_{S'} = \underline{\hom}_{S'}(X_{S'},Y_{S'})$.