A polynomial homomorphism from Gl to the group of units is a power of the determinant 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? 
 A: 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... 
A: 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...
A: 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. 
A: 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 $n^2+1$.
