Mistake in Discriminants, Resultants, and Multidimensional Determinants? I'm studying this proof in the book cited in the title.


The author say that any multiplicative function $\chi\colon\mathsf{GL}(k)\longrightarrow\mathbb{C}^\times$ is a power of the determinant. But what if I take $\chi(g)=\overline{\det g}$? Is this a mistake? If not and I'm missing something where can I find a proof of this fact?
 A: As noted in the comment above, the author is talking about regular functions in algebraic geometry, which does not include complex conjugation.
Here's a proof that over an algebraically closed field that any multiplicative regular function from $GL_n\to \Bbb G_m$ is a power of the determinant.
First, $GL_n$ is irreducible, since it is an open set in an irreducible variety: $GL_n=D(\det)\subset\Bbb A^{n^2}$. Next, the set of matrices with distinct eigenvalues is dense in $GL_n$, as it is the complement of the vanishing locus of the discriminant of the characteristic polynomial. So it suffices to verify the equality $\chi=\det^d$ on the set of matrices with distinct eigenvalues (=the diagonalizable matrices).
A multiplicative function $\chi: GL_n\to\Bbb G_m$ is constant on similarity classes of matrices, as if $A=SBS^{-1}$, then $$\chi(A)=\chi(SBS^{-1})=\chi(S)\chi(B)\chi(S^{-1})=\chi(S)\chi(S^{-1})\chi(B)=\chi(SS^{-1})\chi(B)=\chi(Id)\chi(B)=\chi(B).$$
So we may reduce from diagonalizable matrices to diagonal matrices. $\chi(Id)=1$, so it's enough to calculate on non-identity diagonal matrices. All of these are a product of matrices which have $n-1$ ones and one non-one element $c$ on the diagonal. So it's enough to see what happens to $diag(c,1,\cdots,1)$. Every function on this matrix is a polynomial in $c$. If it's multiplicative, it must be of the form $c^d$ for some integer $d$. But $c$ is the determinant, and we're done.
