Let $A$ be a commutative ring with $f,g\in A[x]$ monics. Consider the $A$-linear endomorphism $\mu\_g^{(f)}\in \mathrm{End}\_A\tfrac{A[x]}{\langle f\rangle}$ given by multiplication by $g$.


For monics $f\_1,f\_2\in A[x]$, how to directly prove that $\det \mu\_g^{(f\_1f\_2)}=\det\mu\_g^{(f\_1)}\det\mu\_g^{(f\_2)}$?


Writing down the matrix representation of $\mu\_g^{(f)}$ w.r.t the monomial basis $1,x,\dots ,x^{\deg f-1}$ is messy and I am unable to see anything through it. On the other hand, $\mu\_g^{(f)}=g(\mu\_x^{(f)})$, and the matrix representation of $\mu\_x^{(f)}$ w.r.t the monomial basis is the companion matrix of $f$, but again I see nothing smart to say about polynomial functions of a companion matrix.



Multiplicative identity of determinant of multiplicative action of a polynomial on a quotient ring (companion matrices)
You have an exact sequence, $0\to A[x]/f\_1\stackrel{f\_2}{\to} A[x]/f\_1f\_2\to A[x]/f\_2\to 0$. This splits as $A$-modules and then multiplication by $g$ in the middle is just the diagonal matrix of multiplication by $g$ in the two factors. So, determinant multiplies.


