Prove if $f$ and $g$ are coprime then $C(fg) \sim C(f) \oplus C(g)$ In my course we recently learned about rational canonical form. The version I learned in lectures wrote the RCF as the direct product of irreducible polynomials e.g. $C(f_1(x)^{a_1}) \oplus \cdots \oplus C(f_k(x)^{a_k})$ where each $f_i$ is irreducible. $C(f(x))$ denoting the companion matrix of $f(x)$. However I've also come across a RCF theorem that takes the product of coprime polynomials as the blocks.
As I understand it the key lemma that connects these two forms is that if $f$ and $g$ are coprime then $C(fg) \sim C(f) \oplus C(g)$ but I can't seem to come up with or find any simple proof of this.
 A: If you have seen the $k[X]$-module point of view: the result you are looking for is just saying that if $f,g$ are coprime, then we have an isomorphism of $k[X]$-module $k[X]/(fg)\simeq k[X]/(f)\times k[X]/(g)$.
If not, let $u$ be the endomorphism of your $k$-vector space $E$, whose is cyclic with characteristic (or minimal, that the same here) polynomial is $fg$.
Since $f,g$ are coprime and $\chi_u=fg$, we have $E=\ker g(u)\oplus \ker(f(u))$.
Let $v$ be a vector which is $u$-cyclic (meaning  $(v,u(v),\ldots, u^{n-1}(v))$ is a $k$-basis of $E$. Then using a Bézout relation $Uf+Vg=1$, you should be able to prove that
$\ker(g(u))=Im((Uf)(u)), \ker(f(u))=Im((Vg)(u))$, and that
$w=(Vg)(u)(v)$ is  a cyclic vector for the restriction of $u$ to $\ker(f(u))$, and $w'=(Uf)(u)(v)$ is  a cyclic vector for the restriction of $u$ to $\ker(g(u))$ (or something like that, I didn't do the computations...but it is the translation of the $k[X]$-module isomorphism above in terms of endomorphisms, so it should work that way).
Of course, you should also convince yourself along the way that the characteristic polynomials over  each subspace are $f$ and $g$ respectively.
