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How to show $\det(AB) =\det(A)\det(B)$

Consider the map $q:\operatorname{GL}(n,R)→R^*$ given by $q(A)=\det(A)$. I know that $$q(AB)=q(A)q(B)=\det(A)\det(B)$$ does not hold if determinants of $A$ and $B$ are not unity. I would like to know why the map is homomorphic?

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marked as duplicate by Zev Chonoles Dec 28 '11 at 15:33

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the multiplicativity of the determinant holds independently of any other conditions: en.wikipedia.org/wiki/… –  t.b. Dec 28 '11 at 15:03
    
What do you mean by $\text{det}(AB)=\text{det}(A)*\text{det}(B)$ doesn't hold? –  NikolajK Dec 28 '11 at 15:04
    
See the answers to this question. –  Pierre-Yves Gaillard Dec 28 '11 at 15:29
    
@Pierre: Should we close as a duplicate, then? –  Zev Chonoles Dec 28 '11 at 15:32
    
@ZevChonoles: I think so. –  Pierre-Yves Gaillard Dec 28 '11 at 15:33

1 Answer 1

It is a homomorphism because you are incorrect that $\det(AB)\neq\det(A)\det(B)$ unless $\det(A)$ and $\det(B)$ are 1; in fact the statement that $\det(AB)=\det(A)\det(B)$ is always true for matrices over any commutative ring.

This is something I've always just thought, but I had a surprising amount of difficulty coming up with a reference for it, and I couldn't think of a way of proving the general statement myself, which is off-putting. At any rate, here is one proof, which shows that $\det(AB)=\det(A)\det(B)$ holds for matrices over $\mathbb{C}$, then argues that this implies it is true for matrices over any ring $R$ because the determinant is an integer polynomial in the entries of a matrix and we can embed the polynomial ring in $n^2$ variables over $\mathbb{Z}$ into $\mathbb{C}$.

Also, note that one says that a map "is a homomorphism", not "is homomorphic".

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Anyone tried looking at some of Dieudonne's work? –  Geoff Robinson Dec 28 '11 at 15:33
    
I'm afraid I'm not sure what you're referring to, Geoff. –  Zev Chonoles Dec 28 '11 at 15:35

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