Talk:Algebraic group

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 Field: Algebra

Semi-Direct Product Versus Extensions[edit]

Every algebraic group G contains a linear subgroup H such that G/H is an abelian variety, but the sequence 1 -> H -> G -> G/H -> 1 is in general not split. Hence it is wrong to say that G is a semidirect product of H and G/H. Joerg Winkelmann 11:53, 1 May 2006 (UTC)

Dear Joerg,

I think this has been fixed in the current version.

DeaconJohnFairfax (talk) 23:51, 30 June 2008 (UTC)


The text says that each finite group is also algebraic. I think that needs further explaining: why is that? Why is even the set of zeros of some polynomials (over which field? The group-algebra?). And why are inversion and multiplication morphismns? —Preceding unsigned comment added by (talk) 16:03, 24 March 2008 (UTC)

Dear Anonymous User,

I think you're right about the finite group example needing more explanation.

Regarding inversion and multiplication being given by morphisms, nobody said they were. In fact, inversion is and multiplication is not (at least not with the natural definition x -> xy where y is fixed; i.e., xz -> xyxz = xzy is not true in general.)

The article said they were given by regular functions. For matrix inversion in Gln(C), that is Cramer's rule. The variety is defined by det(X) != 0. I'm not absolutely sure how matrix multiplication is given by a regular function, or by regular functions, but I think it is just by the inner product of a row and a column giving a coordinate of the product of the two matricies. I'm pretty sure the coordinate functions on an algebraic variety are regular functions (that's what the previous statement boils down to.)

For elliptic curves, the regular functions are given in the ususal proof that the product of two rational points on the curve is rational.

So, if you replace "morphism" by "regular function" in your request, then, I agree that it is a good suggestion.

DeaconJohnFairfax (talk) 23:49, 30 June 2008 (UTC)