I was browsing MO and stumbled upon [this post](https://mathoverflow.net/questions/70889/determinant-of-exterior-power), 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 polynomial homomorphism from Gl to the group of units is a power of the determinant
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...


