TITLE: The connection between decomposition of $\ker(f_\phi g_\phi)$ and the Chinese remainder theorem
QUESTION [12 upvotes]: Original problem
Suppose $K$ is a field, $f,g\in K[x]$ and $\gcd(f,g)=1$. $V$ is a linear space based on $K$ and $\phi\in\operatorname{End}(V)$.
Notation $f_\phi$ and $g_\phi$ denote linear transformations $f(\phi)$ and $g(\phi)$. Try to show that $$\ker(f_\phi g_\phi)=\ker(f_\phi)\oplus\ker(g_\phi)$$
The sketch of the proof
Notice that sub-ring $K[\phi]\subseteq\operatorname{End}(V)$ is commutative, and $f_\phi,g_\phi\in K[\phi]$, so it's easy to show that $\ker(f_\phi),\ker (g_\phi)\subseteq\ker(f_\phi g_\phi)$.
$\because\gcd(f,g)=1$, $\therefore\exists u,v\in K[x]: uf+vg=1\implies u_\phi f_\phi+v_\phi g_\phi=I_V$, where $I_V$ is the identity operator of $V$.
$\forall x\in\ker(f_\phi g_\phi)$, plug $x$ into the preceding equation, we have $x=u_\phi f_\phi(x)+v_\phi g_\phi(x)$, and it's not quite difficult to show that $u_\phi f_\phi(x)\in\ker(g_\phi)$ and $v_\phi g_\phi(x)\in\ker(f_\phi)$, so $x\in\ker(f_\phi)+\ker(g_\phi)$.
$\forall x\in\ker(f_\phi)\cap\ker(g_\phi)$, plug $x$ into that equation and we have $x=0$, so $\ker(f_\phi)\cap\ker(g_\phi)=0$, Q.E.D
Thoughts
It seems that the proposition is quite similar to Chinese Remainder Theorem, say:
$$\mathbb Z/mn\mathbb Z\cong\mathbb Z/m\mathbb Z\oplus\mathbb Z/n\mathbb Z$$
if $\gcd(m,n)=1$.
Can we find the deeper relationship between these two propositions? Or just an unique aspect?
Thanks!

REPLY [5 votes]: Since my previous answer, I have realised there is a very simple way to link the Chinese remainder theorem for $K[X]$, which is a result for rings, to the kernel decomposition theorem for vector spaces.
The Chinese remainder theorem for a principal ideal domain $R$ says that if $a_1,\ldots,a_k\in R$ are pairwise relatively prime, then $R/a_1\ldots a_kR\cong(R/a_1R)\times\cdots\times(R/a_kR)$ via the map that on component$~i$ of the product is given by reduction modulo$~a_i$ as morphism $R/a_1\ldots a_kR\to R/a_iR$. If $\widehat a_i=a_1\ldots a_k/a_i$ denotes the product of all factors except $a_i$, it essentially states the existence of Bezout coefficients $c_1,\ldots,c_k\in R$ such that $c_1\widehat a_1+\cdots+c_k\widehat a_k=1$, since this requires the class of $c_i\widehat a_i$ to project to the element $e_i=(0,\ldots,0,1,0,\ldots,0)$ of the product ring (the $1$ being in position$~i$), and once these classes are given one can just choose representatives, adapting the last choice so as to make the equality to$~1$ exact (rather than just a congruence modulo $a_1\ldots a_k$).
Applying this for $R=K[X]$, one gets that if $P_1,\ldots,P_k\in K[X]$ are pairwise relatively prime polynomials, then $K[X]/(P_1\ldots P_k)\cong(K[X]/(P_1))\times\cdots\times(K[X]/(P_k))$. Also for any endomorphism$~\phi$, the kernel $\ker(P_1\ldots P_k[\phi])$ is naturally a module over that ring. The final step that I was missing before is the following simple result (it surprises me that I never saw it stated explicitly)
Lemma. Any module $M$ over a product $R=R_1\times\cdots\times R_k$ of (unitary) rings canonically decomposes as direct sum $M=M_1\oplus\cdots\oplus M_k$, where $M_i$ is a module over$~R_i$ (the other factors acting as $0$ on it).
The proof is almost trivial: if $e_i\in R$ is the element with coordinate$~1$ in the factor $R_i$ and$~0$ in the other factors, one has $M_i=e_iM$, and the direct sum decomposition is a consequence of $e_1+\cdots+e_k=1$ together with the relations $e_i^2=e_1$ and $e_ie_j=0$ for $i\neq j$. Not even commutativity of$~R$ is required.
Back to linear algebra, this says that $\ker(P_1\ldots P_k[\phi])$ canonically decomposes as a direct sum of submodules, summand $M_i$ being obtained as image of (multiplication by) the polynomial $C_i[\phi]$ in$~\phi$, where $C_i$ is the Bezout coefficient satisfying $C_1\widehat{P_1}+\cdots+C_k\widehat{P_k}=1$ as above. This submodule can be seen to be equal to the kernel of $P_i[\phi]$, whence the kernel decomposition theorem.