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Is it possible to embedd an elliptic curve $E:\;\; y^2=x^3+ax+b$, defined over an algebraically closed field $k$, into some $GL_n(k)$ ?

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Presumably you want to do this as a group as well as as a variety? The first problem that comes to mind is the point at infinity. The group of such an elliptic curve is a projective variety, and it is IMHO impossible to embed a projective variety into an affine one. One reason that comes to mind is that an affine variety has plenty of globally defined rational functions, but a projective variety has only the constants. –  Jyrki Lahtonen Jul 13 '12 at 22:36
    
Isn't a projective variety as zero-set of (homogeneous) polynomials automatically also an affine variety ? –  tj_ Jul 13 '12 at 22:43
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@tj_ The zero-set of a homogeneous polynomial is not the same as the projective variety defined by the homogeneous polynomial. The difference is that in the first case, you are working in affine space, whereas in the second, you are working in projective space (so you have an equivalence relation on the points). –  M Turgeon Jul 13 '12 at 23:04

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up vote 6 down vote accepted

If $E\hookrightarrow\mathrm{GL}_n$ is a closed immersion of $k$-schemes ($k$ any field) inducing an isomorphism with the closed subscheme $Z$ of $\mathrm{GL}_n$, then $Z$ is necessarily affine, and, being isomorphic over $k$ to $E$, is also proper over $k$. It is therefore finite over $k$. But this implies that $E$ is finite over $k$, which is not the case.

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Yep. My answer was overkill. –  M Turgeon Jul 13 '12 at 23:36
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In fact, any immersion of $E$ into a separated algebraic variety is a closed immersion by properness of $E$. –  user18119 Jul 14 '12 at 9:04

By the Chevalley's theorem on algebraic groups (see for example this modern proof of B. Conrad), there exists a unique normal linear algebraic closed subgroup of your elliptic curve $E$ such that the quotient is an abelian variety (at least if you assume your base field is perfect, which is trivially the case in your situation). In particular, since the identity subgroup satisfies the requirement, we can conclude that an elliptic curve cannot embed into some $\mathrm{GL}_n$, since otherwise $E$ would be that unique normal subgroup.

Remark: This is probably overkill, but at least it answers the question.

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