# Proof that Determinant is Multiplicative for Commutative Rings

1. Mar 10, 2013

### 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?

2. Mar 10, 2013

### 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.

Last edited: Mar 10, 2013
3. Mar 10, 2013

### 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?

4. Mar 10, 2013

### Ben Niehoff

V can be any vector space at all.

5. Mar 10, 2013

### Bacle2

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