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Let $k$ be a field. Let $F$ be a covariant functor on the category of $k$-algebras to the category of sets. Assume that the opposite functor $F^{op}$ on the category of affine $k$-schemes is a sheaf. (This last assumption might not be necessary.)

Now, suppose that $F(R)$ is a finitely generated $R$-module for all $k$-algebras $R$ and that

$F(R) = F(k)\otimes_k R$.

It seems that in this case $F$ is "representable" by the vector space $$F(k).$$

I don't know what this means and I'm probably missing some very standard construction here. It should mean that $F^{op}$ is representable by a finite type $k$-scheme.

I know one can associate to a finite-dimensional vector space $E$ over $k$ the affine variety $\mathrm{Spec} \ \mathrm{Sym}(E).$ Is it clear that in this case the functor $F$ is representable by the affine variety $\mathrm{Spec} \ \mathrm{Sym}(E)$?

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1 Answer 1

up vote 6 down vote accepted

Since $F(k) \cong k^n$ for some $n$, we have $F(R) \cong R^n$, naturally in $R$. But $R^n \cong (\mathbb{A}^n)(R)$, so that $F$ is just the scheme $\mathbb{A}^n$. This is also isomorphic to $\mathrm{Spec}(\mathrm{Sym}(F(k))$.

EDIT: More generally and coordinate free: When $S$ is some base scheme and $\mathcal{E}$ is some quasi-coherent module over $S$, then $\mathbb{V}(\mathcal{E}^\*) := \mathrm{Spec}(\mathrm{Sym}(\mathcal{E}))$ represents the contravariant functor which sends an $S$-scheme $p : T \to S$ the set of $\mathcal{O}_S$-homomorphisms $\mathcal{E} \to p_* \mathcal{O}_T$. When $S=\mathrm{Spec}(k)$ is affine, this is the set of $k$-module homomorphisms $\Gamma(S,\mathcal{E}) \to \Gamma(T,\mathcal{O}_T)$. For example, $\mathbb{V}(\mathcal{O}_S^n) = \mathbb{A}^n_S$.

Now if $F$ is as in your question, we see that $F$ represents the same functor as $\mathbb{V}(F(k))$.

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3  
I think normally we take Spec(Sym(E*)) (E* means E dual). That convention will make morphisms from Speck to this scheme the same as elements of E, if E is a vector space. –  36min May 12 '12 at 16:02
    
Right, I've fixed it. –  Martin Brandenburg May 12 '12 at 17:45

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