Proof that Determinant is Multiplicative for Commutative Rings

[Site]

Is there a nice way to show that Det(AB)=Det(A)Det(B) where A and B are n x n matrices over a commutative ring?

I'm hoping there is some analogue to the construction for vector spaces that defines the determinant in a natural way using alternating multilinear mappings...

Otherwise would you just have to bash out the identity using the Leibniz formula for the determinant?

 

[Ben Niehoff]

$\det$ is just a homomorphism from the group of linear maps on $V$ to its representation on the top exterior power $\Lambda^n V$. Taking some $v_i \in V$, we have $\det A : \Lambda^n V \to \Lambda^n V$ given by

$$(\det A)(v_1 \wedge \ldots \wedge v_n) = A(v_1) \wedge \ldots \wedge A(v_n)$$
From here it is easy to show $\det AB = \det A \det B$ by using the usual composition of linear maps on each factor in the wedge product on the right.

 

[Site]

Thanks, Ben. To clarify, do you mean that we set $V=R^n$ so that the group of linear maps on $V$ is the set of $n$x$n$ matrices?

Also, do you know of a textbook that explains exterior algebra (from the module perspective) and its connections to the determinant from the ground up?

 

[Ben Niehoff]

V can be any vector space at all.

 

[Bacle2]

Is there a nice way to show that Det(AB)=Det(A)Det(B) where A and B are n x n matrices over a commutative ring?

I'm hoping there is some analogue to the construction for vector spaces that defines the determinant in a natural way using alternating multilinear mappings...

Otherwise would you just have to bash out the identity using the Leibniz formula for the determinant?

Maybe you can first show (not too hard) , that the determinant is multiplicative for elementary matrices E_i . Then write B as a product of elementary matrices
and rearrange.