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I was browsing MO and stumbled upon this post, and I got very curious. I searched for about half an hour and could not find a proof for the statement that any polynomial group homomorphism $\mathrm{Gl}_n(\Bbbk)\to\Bbbk^\times$ is a power of the determinant. Now this is certainly classic, so could someone point me to a good reference? I would prefer a proof that can be presented to graduate students with a mild background in classical algebraic geometry.

Edit: I was also wondering about the following: If $G$ is an affine algebraic group (i.e. an affine $\Bbbk$-variety with compatible group structure), then there is a closed immersion of $\iota:G\hookrightarrow\mathrm{Gl}_n(\Bbbk)$. From such an immersion, I get a "determinant" on $G$, namely $\iota^\sharp(\det)$. Is this, by any chance, independent of the embedding up to taking powers?

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No. You can modify the resulting "determinant" by any character $\chi:G\to \text{GL}_1(\bf k)$, simply by enlarging $n$ by one and using the extra entry on the diagonal as target for $\chi$. –  Wilberd van der Kallen Jul 13 '12 at 8:06
    
Your argument seems to be kindof circular - the group homomorphisms $G\to\Bbbk^\times$ are the characters, and if all of them are powers of a determinant, then so is the modification you propose, because the extra entry on the diagonal will mean multiplication with some power of the determinant. –  Jesko Hüttenhain Jul 13 '12 at 8:19
    
@Jesko -- As I believe others point out below, van der Kallen's argument is completely correct and certainly is not circular. If you take an algebraic group like $\mathbf{G}_m\times \mathbf{G}_m$, a product of two copies of the multiplicative group, then the group of characters has rank $>1$, i.e., it is not the case that all characters are related to, i.e., commensurate with, a single "determinant" character. Thus if you modify your linear representation as van der Kallen suggests by these characters, the associated determinants will also be non-commensurate. –  Jason Starr Jul 13 '12 at 17:13
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4 Answers

up vote 5 down vote accepted

Since ${\mathbb k}^\times$ is abelian, any group hom $\phi : GL_n ({\mathbb k}) \rightarrow{\mathbb k}^\times$ factors through $\phi : GL_n ({\mathbb k}) \rightarrow GL_n({\mathbb k})/[GL_n({\mathbb k}),GL_n({\mathbb k})] \rightarrow{\mathbb k}^\times$. It is a nice easy exercise that $[GL_n({\mathbb k}),GL_n({\mathbb k})] = SL_n({\mathbb k})$. Thus, the factorization turns out to be $\phi : GL_n ({\mathbb k}) \xrightarrow{\det} {\mathbb k}^\times \rightarrow{\mathbb k}^\times$.

If you want your hom to be polynomial the corresponding endomorphism of the multiplicative group ought to be polynomial. Notice that I am not making any assumptions about the field...

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That is, indeed, very nice and brief. Do you have any ideas about my edit? –  Jesko Hüttenhain Jul 13 '12 at 9:31
    
Off course, it is not independent. Take $G$ to be the Klein-4-group. It will have 3 diagonal immersions into $GL(2)$, each immersion different elements will have determinant 1. –  Bugs Bunny Jul 13 '12 at 9:50
    
I don't fully understand: What exactly are these diagonal immersions? What do you mean by your last statement? –  Jesko Hüttenhain Jul 13 '12 at 10:54
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Technical note: $SL_n(k) = [GL_n(k), GL_n(k)]$ in all cases EXCEPT $n=2$ and $k=\mathbb{F}_2$ or $\mathbb{F}_3$. Doesn't effect the end result, because you could always extend the base field before running your argument. See Lang's <i>Algebra</i>, Chapter XIII. –  David Speyer Jul 13 '12 at 11:51
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David, except $n=2$ and $k=F_2$ only, which is a very special case as $k^\times$ is trivial:-)) So you dont need any extensions. If $k=F_3$, $SL_2$ is not perfect but still the commutant of $GL_2$ –  Bugs Bunny Jul 13 '12 at 12:48
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I can't tell if you want a reference to a proof or just a proof, but here's a proof.

By multiplying by a suitable power of the determinant we may assume that such a homomorphism extends to a polynomial monoid homomorphism $\mathcal{M}_n(k) \to k$, and since it is polynomial it extends to a homomorphism $\phi : \mathcal{M}_n(\bar{k}) \to \bar{k}$. Since $\bar{k}$ is abelian, $\phi$ is conjugation-invariant. Restricted to diagonal matrices $\phi$ must be a symmetric polynomial in their diagonal entries, and since the diagonalizable matrices are Zariski dense in $\mathcal{M}_n(\bar{k})$, $\phi$ must be a symmetric polynomial in the eigenvalues (hence a polynomial in the coefficients of the characteristic polynomial) identically. By scaling entries of elements of $\mathcal{M}_n(\bar{k})$, $\phi$ must be a homogeneous polynomial.

Assume WLOG that $\phi$ is not divisible by the determinant. If $\phi$ has degree $0$ as a homogeneous polynomial, then it must be the trivial homomorphism. Otherwise, $\phi$ is a homogeneous polynomial of positive degree, so in particular $\phi(0) = 0$. Now, if $D_1$ is a diagonal matrix with a zero entry, then we can find conjugates $D_2, ... D_n$ of $D_1$ which are also diagonal but with the zero entry at every possible other location. It follows that $D_1 ... D_n = 0$, so $\phi(D_1 ... D_n) = \phi(D_1)^n = 0$, so $\phi(D_1) = 0$. Hence $\phi(M)$ vanishes if any of the eigenvalues of $M$ vanish, so $\phi$ is divisible by the determinant; contradiction.

Edit: Your proposed extension is false. Take $G = \text{GL}_1 \times \text{GL}_1$ and embed using different powers for each factor...

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A simple proof in case that $\mathbb{K}$ is infinite can be found in my answer here:

determinants and polynomials in matrices

It is based on the fact that $SL_n(\mathbb{K})$ is perfect. This is just Theorem 14.8 in this paper. The proof is short and elementary.

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I think it is even simpler than that. I will have a go... –  Bugs Bunny Jul 13 '12 at 9:10
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As people have pointed out, the question raised in the Edit has a negative answer. To clarify this, start with $G = \mathrm{GL}_n(K)$ (for an algebraically closed field $K$) and embed $G$ (or any closed subgroup) into $\mathrm{GL}_{n+1}(K)$ as block diagonal matrices $(A, 1/\det A)$ with zero entries elsewhere. In this way $G$ actually embeds in the special linear group, so applying $\det$ gives the trivial map on the embedded copy of $G$.

This is just a variant of the standard procedure used to see that $G$ (which at first looks like a Zariski-open subset of affine $n^2$-space) can in fact be viewed concretely as the zero set of some polynomials in an affine space of dimension $n^2+1$.

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