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.


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).

  • $\begingroup$ Great ! $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Dec 28 '09 at 17:08
  • $\begingroup$ Monomorphisms are injective (you just gave a proof). $\endgroup$ – user2035 Jan 7 '10 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.