# Multiplicative identity of determinant of multiplicative action of a polynomial on a quotient ring (companion matrices)

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.

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.
• Dear Mohan, do you by any chance see an explicit basis of $A[x]/\langle f_1f_2\rangle$ for which the matrix representation of the multiplication map has the asserted block diagonal form? Sep 20 at 21:30
• Also, how does the direct sum decomposition force the multiplication by $g$ in the middle to be blockwise diagonal multiplication by $g$ on the factors? I'd have thought the matrix would at least be block-triangular. Sep 20 at 22:59
• @Arrow Yes, I was sloppy, it is block matrix with zeroes on one block. The basis you need is $1, x,x^2,\ldots, x^{n-1}$, where $\deg f_2=n$ and $f_2,f_2x,\ldots, f_2x^{m-1}$ where $\deg f_1=m$. Sep 20 at 23:35
• @Arrow If $F=M\oplus N$ and $\phi$ is an endomorphism of $F$ such that $\phi(M)\subset M$ then $\phi$ induces an endomorphism of $M$ and $F/M=N$. One can easily check that $\phi$ can be put in the desired form. Sep 21 at 0:02