One really stupid, trivial question: A Quasi coherent sheaf $F$ on an affine group scheme(Spec R) is simply an Rmodule. What happens in case R is a Hopf algebra? Will the Q.coherent sheaf $F$ be an algebra in this case?

1$\begingroup$ A quasicoherent sheaf on $Spec(R)$ corresponds to a $R$module; this has nothing to do with extra structures like group schemes. Or what condition do you impose on the sheaf in this situation? $\endgroup$ – Martin Brandenburg Oct 10 '10 at 21:29

1$\begingroup$ No. Quasicoherent sheaves correspond to $Rmodules, and nothing more. $\endgroup$ – Donu Arapura Oct 10 '10 at 21:33

1$\begingroup$ To get extra structure on your module from the extra structure from the extra structure on your scheme, you need some compatibility condition. For example, in your case, you have a group scheme and if you demand that F is an equivariant sheaf, you should get an equivariant module under the group scheme. But there is no way to get an algebra structure, unless you impose an algebra structure. $\endgroup$ – Lennart Meier Oct 11 '10 at 8:07

$\begingroup$ Thanks to all : Martin Brandenburg, Donu Arapura, Lennart Meier,Scott Carnahan. So I understand that the quasi coherent sheaf $F$ need not be an algebra, even if we start with a Hopf Algebra H. But when would $F$ be an algebra, What extra conditions would be required for defining multiplication? $\endgroup$ – Neha Oct 11 '10 at 10:50

1$\begingroup$ Neha, I think I'm starting to understand what you are after. Perhaps the following may help. Given a scheme $X$ (not necessarily a group scheme), quasicoherent sheaves of commutative algebras on $X$ correspond to affine schemes over $X$. $\endgroup$ – Donu Arapura Oct 11 '10 at 12:28
No. Consider the case of the trivial group scheme over a field $k$ (so $R=k$). In this situation, a quasicoherent sheaf is just a $k$vector space. As Lennart Meier said in a comment, you need additional structure to get an algebra, e.g., a multiplication map.
Added: If you just want a "pointwise" multiplication operation on your module $M$, you should ask for an $R$linear map $M \otimes M \to M$. This does not use the group law $m: G \times G \to G$ (i.e., the coalgebra structure on $R$). It sounds like you might be looking for some kind of convolution product that uses the group law, for example, a map $F \boxtimes F \to m^*F$ on $G \times G$. Again, this is not a condition, but an extra structure. If we return to the case where $G$ is trivial, we see that the only conditions on a $k$vector space that endow it with a canonical algebra structure are those conditions that imply the vector space is zero.

$\begingroup$ Thanks Scott. So I understand that the quasi coherent sheaf F need not be an algebra, even if we start with a Hopf Algebra H. But when would F be an algebra, What extra conditions would be required for defining multiplication? $\endgroup$ – Neha Oct 11 '10 at 10:58

$\begingroup$ Thanks so much. So what I understand finally is that the quasi coherent sheaf $F$ over an affine scheme, Spec $R$ is simply an Rmodule. But if F is a quasi coherent sheaf of algebras, then $F$ would be an $R$algebra, say $M$ ?!? Do I think of it as a constant sheaf with all its sections as $M$ ?!? $\endgroup$ – Neha Oct 12 '10 at 10:56

$\begingroup$ In the notation of EGA1, 1.3.4, $F$ has the form $\tilde{M}$ for an $R$algebra $M$. It is not a constant sheaf in any natural sense of the word (although you could say it is constant if Spec $R$ only has one point). $\endgroup$ – S. Carnahan♦ Oct 12 '10 at 16:52