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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.

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It is not correct that the Hom scheme is projective. It need be neither quasi-compact nor universally closed. Rather it is a countable union of quasi-projective schemes. One example is where both $X$ and $Y$ equal the projective line $\mathbb{P}^1$. Yes, $\textit{Hom}_k(X,Y)_K$ equals $\textit{Hom}_K(X_K,Y_K)$. No, the same is not true for the Kontsevich moduli space. –  Jason Starr Jun 30 '12 at 14:04
    
Apologies, I did not even intend to have the word projective in there, corrected! –  Nadal Jun 30 '12 at 14:55
    

1 Answer 1

up vote 2 down vote accepted

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

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Here is a counterexample for the Kontsevich space. Start with a smooth, plane conic $C$ in $\mathbb{P}^2_k$ such that $C$ has no $k$-rational point, e.g., the conic with defining equation $X_0^2 + X_1^2 + X_2^2$ over $\mathbb{R}$. Let $D$ be an effective, degree $2$, Cartier divisor in $C$, e.g., the common locus of $X_0$ and $X_1^2 + X_2^2$ for the above conic. There is a $k$-point of $\mathcal{M}_{0,0}(\mathbb{P}^2,4)$ "parameterizing" a double cover $f:B\to C$ branched over $D$. However, $f_*\mathcal{O}_B/\mathcal{O}_C$ has degree 1, so $f$ is not defined over $k$. –  Jason Starr Jun 30 '12 at 16:54
    
Typo correction: $f_*\mathcal{O}_B/\mathcal{O}_C$ has degree $-1$, not degree $+1$. The dual has degree $1$ -- this is impossible since $C$ has no $k$-rational point, thus $f$ is not defined over $k$. –  Jason Starr Jun 30 '12 at 17:00
    
Jason, this is not a comment to my trivial, but rather a profound answer by your own. Post it as an answer, please. tea.mathoverflow.net/discussion/1392/… –  Martin Brandenburg Jun 30 '12 at 18:04
    
Great, thank you! –  Nadal Jun 30 '12 at 19:22

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