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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 …

To determine which potential zeta functions are actual zeta functions of curves is very difficult. The zeta function of a curve is determined by the zeta function of its Jacobian so one could instead …

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 …

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 …

If all Tate modules (i.e., for all $\ell$) are isomorphic then they differ by the twist by a locally free rank $1$ module over the endomorphism ring of one of them. This is true for all abelian variet …

I wanted to add some things to the comments I had already made but the list of comments have become very large and the comments I have already made are becoming more and more difficult to follow so I' …

Starting, generally, with a commutative ring $R$ and a rank $2$ projective module $P$ given with a quadratic form $\varphi$ we can form its Clifford algebra $C(P)$. Its even part $S:=C^+(P)$ is then a …

Also for the étale fundamental group there is in fact always some universal cover. However, in the abstract way that Grothendieck formulated the theory of coverings a universal cover would only exist …

If you rewrite the equation as $b^2=-a^2/(a^2+1)$ as Will did you can continue as follows: The equation can be rewritten as $$\left(\frac{b}{a}\right)^2 = -(a^{-2}+1)$$ (excepting $a=b=0$). Putting $x …

The key to solving both problems is the use of the following two facts: 1) Any closed subgroup of $T^n$ is the intersection of the kernels of characters of $T^n$, i.e., continuous group homomorphisms …

As $\pi_1(B)$ is the same over $\overline{\mathbb Q}$ as over $\mathbb C$ we may embed $\overline{\mathbb Q}$ in $\mathbb C$ and then extend scalars to $\mathbb C$ to reduce to the case when $\mathbb …

It seems likely (and may be known) that it does depend on the permutation representation. However, it is a subtle question as the stable isomorphism class (of any faithful permutation, or in fact arbi …

I think that this MR0277590 (43 #3323) André, M. Hopf algebras with divided powers. J. Algebra 18 1971 19--50 may be relevant. It says that a graded commutative divided power Hopf algebra is the co-en …

Just some comments that are well-known in the theory of toric varieties (and no doubt to other areas as well). What we are asked to determine is membership in a finitely generated submonoid $\Gamma$ o …

The condition $\mathbb Q\Lambda_1=\mathbb Q\Lambda_2=:X$ means that we have $\Lambda_1,\Lambda_2\subseteq X$ and then we have $\Lambda_1,\Lambda_2\subseteq \Lambda_1+\Lambda_2\subseteq X$. This mean …