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Consider the following: Let $A$ be a commutative ring, let $M$ be an $A$-module. When is the functor from $A$-algebras to Sets given by $R \mapsto R \otimes M$ representable by an $A$-scheme?

Unless I've made a mistake, this is always be an fpqc sheaf. When $M$ is a finitely generated free A-module, then $\mathrm{Spec}( \mathrm{Sym}^\bullet M^*)$ does the trick.

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2 Answers 2

up vote 4 down vote accepted

When $A$ is noetherian and $M$ is finitely generated, Nitin Nitsure showed that the functor is representable if and only if $M$ is projective (see http://arxiv.org/abs/math/0308036).

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Great ! –  Dinakar Muthiah Apr 16 '10 at 19:53

Partial answer: If $M\otimes\kappa(\mathfrak p)$ is infinite-dimensional for some prime $\mathfrak p$ in $A$, then the functor is not representable.

Proof: We may assume that $A=k$ itself is a field and $M=k^{(I)}$ for some infinite set $I$. Suppose there is a representing scheme $X$ (so $X(R)=R^{(I)}$), and let $x$ denote the rational point corresponding to $0\in M$. For a filtered inductive system of rings $R_i$, we may calculate $\varinjlim R_i$ as a inductive limit of abelian groups, so it commutes with the direct sum $R_i^{(I)}$, therefore $X(\varinjlim R_i)=\varinjlim X(R_i)$. Now EGA IV, 8.14.2 tells us that $X$ is locally of finite type over $k$, in particular $\mathcal O_{X,x}$ has finite dimension.

There is an obvious monomorphism $X\to\mathbb A^I=\mathrm{Spec}(k[T_i\mid i\in I])$, so $X$ is separated. Choose an embedding $\mathbb N\to I$. For each $n$, we get a monomorphism $i_n\colon\mathbb A^n\to X$ which is a section to the projection $X\to\mathbb A^n$, so $i_n$ is a closed embedding, whose image contains $x$. This implies that $\mathcal O_{X,x}$ has quotients of arbitrarily large dimension, contradiction.

(EDIT) A different way to conclude is by comparing the finite-dimensional Zariski tangent space with the liftings of $x\in X(k)$ to an $k[\epsilon]/(\epsilon^2)$-valued point, and these liftings are given by $(k\cdot\epsilon)^{(I)}\subset(k[\epsilon]/(\epsilon^2))^{(I)}$. Use the monomorphism to $\mathbb A^I$ to deduce that the two vector space structures agree.

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you seem to use: if $X \to Y$ is a $S$-morphism, which is a monomorphism, where $Y/S$ is separated, then also $X/S$ is separated. is this true? it's well-known that it's true when $X \to Y$ is injective. since the underlying set of a k-scheme can be recovered as the filtered(!) colimit of the K-points, where $K/k$ is a field extension, it's enough to assume that all $X(K) \to Y(K)$ are injective. here, $X(K)=K^{(I)}$ injects to $\mathbb{A}^I(K)=K^I$ and everything is fine. –  Martin Brandenburg Dec 28 '09 at 17:08
    
Monomorphisms are injective (you just gave a proof). –  user2035 Jan 7 '10 at 16:21

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