Let $k$ be a field and $V$ a vector space over $k$. Does anybody know a proof of the fact that the functor of $k$algebras $$ R \longmapsto V \otimes_k R $$ is not representable by a scheme when $V$ has infinite dimension?

1$\begingroup$ This argument is also the proof of Claim 3.1 in my paper at the following URL: arxiv.org/pdf/math/0602646.pdf $\endgroup$ – Jason Starr Apr 6 '17 at 12:31

1$\begingroup$ Here is a proof when the dimension of $V$ is countably infinite. For $R=k[\epsilon]/\langle \epsilon^2\rangle$, consider the subspace $V\otimes_k \epsilon R $ of $V\otimes_k R$. If the functor is representable by a $k$scheme $X$, for the stalk of the structure sheaf $(\mathcal{O},\mathfrak{m})$ at the $k$point of $X$ representing the zero vector in $V$, the vector space above equals the vector space $\text{Hom}_k(\mathfrak{m}/\mathfrak{m}^2,k)$. This vector space is uncountably generated if $\mathfrak{m}/\mathfrak{m}^2$ has infinite dimension. $\endgroup$ – Jason Starr Apr 6 '17 at 12:32

4$\begingroup$ Another proof: let $F$ be the functor in question. Then the natural map $F(k[[t]])\to\varprojlim_n F\left(k[[t]]\,/\,(t^{n+1})\right)$ is not surjective. It would be bijective if $F$ were representable. $\endgroup$ – Laurent MoretBailly Apr 6 '17 at 12:46

1$\begingroup$ This has been discussed, see mathoverflow.net/questions/6764/… $\endgroup$ – Anton Fonarev Apr 6 '17 at 14:32