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Geometrically there is a very big difference between separable and algebraic closures (in the only case where there is a difference at all, i.e., in positive characteristic $p$). Technically, this com …

I shall show that the answer is no when $p=2$ (and it seems to me that a somewhat more involved calculation will work for any $p$). We shall show that there exists a vector bundle $\mathcal E$ such th …

It mostly has to do with finding nice compactifications. Compactifications of varieties are a good thing as they allow us to control what happens at "infinity". If the variety itself is smooth it seem …

I am not going to add any new examples but suggest a systematic way of looking at examples. If one looks at special phenomena in characteristic $2$ one can classify them as follows (though this divisi …

I am currently teaching a course in algebraic geometry where one of the aims is to give an overview of the Enriques-Kodaira classification of surfaces. I am trying to throw in some modern aspects so I …

I don't know how Tate did it but here is one way. Let $\zeta$ be such that $\zeta^{q+1}=-1$ and put $a_j=(0\colon\cdots\colon1\colon\zeta:\cdots\colon0)$, $j=0,\ldots,i$ with the $1$ in coordinate $2 …

The category theoretical definition of stacks (as given for instance in Giraud: Cohomologie non-abélienne) allow for arbitrary categories as targets (the stack condition only involves isomorphisms how …

The automorphism group functor is not representable in general. Consider for instance the case of an algebraically closed field $k$ and $\mathbf A^2=\mathrm{Spec}\ k[x,y]$. Assume the automorphism gro …

Consider the category of algebraic integrable connections on a smooth connected algebraic variety $X$ with a base point $x$ (over a field of characteristic zero). This is a Tannakian category with a f …

Using the Kummer exact sequence $0\rightarrow\mu_n\rightarrow\mathbb{G}_m\rightarrow\mathbb{G}_m\rightarrow0$ we get a long exact sequence $$ 0\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)\rightarrow\mathrm …

The answer is no: Let $X$ the projective plane $\mathbb P^2$ minus a smooth quadric $Q$. Then $[X]=[\mathbb P^2]-[Q]=\mathbb L^2+\mathbb L+1-(\mathbb L+1)=[\mathbb A^2]$ but $X$ is not isomorphic to $ …

[[ Sorry I missed that the question was also concerned with the question in an algebraic topological context. This answer is only concerned with algebraic geometry.]] I think the first question is mu …

Recall that if $G\rightarrow S$ is a flat group scheme, then a $G$-torsor is an $S$-scheme $X\rightarrow S$ with a $G$-action $G\times_SX\rightarrow X$ such that $G\times_SX\rightarrow X\times_SX$ giv …

Essentially because the Tannakian theory gives in the unipotent case (and only in that case) a reasonably sized answer with an easy motivic interpretation. For the size you should be aware that alrea …

It depends on what you mean by "closed subgroup". If you mean a Zariski closed subset which forms a subgroup then the answer is no. If you mean a closed subgroup scheme, then the answer is yes. An exa …