I‘m currently learning about polynomial rings and am supposed to use them to show that: If the Cayley-Hamilton Theorem holdes over matrices over $\mathbb{C}$ then it holds for matrices with entries in any commutative, unitary ring (in the following just called ring).
Background:
We defined polynomial rings via their unique property (Tried my best to draw a commutative diagramm):
$R[\vec{X}]$, for a ring $R$ is up to unique isomorphism the ring, s.t. for any ring $S$, ring-homomorphism $\varphi: R \rightarrow S$ and $\vec{x} \in S^n$, there exists a unique ring-homomorphism $\varphi_\vec{x}: R[\vec{X}] \rightarrow S$, s.t. the diagram commutes and $\forall i: \varphi_\vec{x}(X_i)=x_i$. ($\iota$ is the inclusion map). $$ \begin{matrix} \ & \varphi & \ & \ \\ R & \rightarrow & S & \ \\ \quad \quad \iota & \searrow & \uparrow & \ \varphi_{\vec{x}} \\ \ & \ & R[\vec{X}] \end{matrix} $$
Furthermore I know:
Therefore, every element $F \in R[\vec{X}]$ induces for every $\varphi: R \rightarrow S$ a function: $$ S^n \rightarrow S, \ \vec{x} \mapsto \varphi_\vec{x}(F) $$ Called the polynomial function associated to the polynomial $F$. Furthermore, if $R$ is a non-finite field, the induced function determines the polynomial uniquely.
I hope I‘m correct this far. So back to Cayley-Hamilton:
We know that for any $A \in \operatorname{Mat}_{n \times n}(\mathbb{C}): \ p_A(A)=0$, with $p_A(\lambda):= \det(A-\lambda\mathbb{1})$. We can therefore define the polynomial function: $$ \mathcal{f}:\mathbb{Q}^{n^2}\rightarrow \mathbb{Q}, \ (x_{i,j})_{i,j} \mapsto p_A(A), \ \operatorname{with:} A_{i,j}:=x_{i,j} $$
Edit: Here there of course is an error, the result of $f$ is a matrix, and each entry of that matrix is itself a polynomial function, the analysis below can then be carried through for all these polynomial functions individually, so the argument stays essentially the same
Because this function maps everything to zero, and $\mathbb{Q}$ is a non-finite field, it‘s associated polynomial $F \in \mathbb{Z}[(x_{i,j})_{i,j}]$ is the zero polynomial. There is a unique ring homomorphism $\varphi: \mathbb{Z} \rightarrow S$, for any ring $S$. So for every matrix $M$ over $S$ there exists a unique $\varphi_{M_{i,j}}: \mathbb{Z}[(x_{i,j})_{i,j}] \rightarrow S$, s.t. $\varphi_{M_{i,j}} \circ \iota = \varphi$ and $\varphi(x_{i,j})=M_{i,j}$, like I explained above (I know this is sloppy notation and I should put the $M_{i,j}$ in a vector $v$ and write $\varphi_v$ but I think it should be somewhat clear).
Thus: $0 = \varphi_{M_{i,j}}(p_A(A))=p_M(M)$?? I‘m really uncertain if I can just do this last step, but unsure how to make it more precise. So I‘d be really happy if someone could elaborate on this last step. Or tell me if I got the entire concept wrong and that there is an error above. Many thanks!