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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.

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Arrow
    Sep 20 at 21:30
  • $\begingroup$ 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. $\endgroup$
    – Arrow
    Sep 20 at 22:59
  • $\begingroup$ @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 $\deg f_1=m$. $\endgroup$
    – Mohan
    Sep 20 at 23:35
  • $\begingroup$ Dear @Mohan, is there a way to see the block triangular decomposition from the splitting of the exact sequence? I do not see a way to circumvent the use of the basis above. $\endgroup$
    – Arrow
    Sep 20 at 23:40
  • $\begingroup$ @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. $\endgroup$
    – Mohan
    Sep 21 at 0:02

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