Product rule for determinant over a commutative ring.

Question

Can someone help me show that $\det(AB)=\det(A)\det(B)$ where $A$ and $B$ are matrices over a commutative ring. In the proofs that I have seen of the corresponding result about matrices over a field they use the fact that we can perform Gaussian elimination.

Answer 1

It true for matrices $A$ and $B$ over a general ring, if it true for the matrices $A'=(X_{i,j})$ and $B'=(Y_{i,j})$ over the particular ring $S=\Bbb Z[X_{i,j},Y_{i,j}]$, the polynomial ring in $2n^2$ generators over the integers. This is because for a general ring $R$ and general matrices $A=(a_{i,j})$ and $B=(b_{i,j})$ there is a ring homomorphism $S\to R$ taking $X_{i,j}$ to $a_{i,j}$ and $Y_{i,j}$ to $b_{i,j}$. Applying this ring homomorphism to $\det(A'B')=\det(A')\det(B')$ in $S$ gives $\det(AB)=\det(A)\det(B)$ in $R$.

But $S$ is an integral domain, and so a subring of a field, so from the determinant identity for fields, $\det(A'B')=\det(A')\det(B')$ does hold in $S$.

Answer 2

Any commutative ring is $\mathbb{Z}$-algebra, thus a quotient ring of some polynomial ring over $\mathbb{Z}$. Then considering in the quotient field, we have $\det AB=\det A\det B$.