# Deduce Jordan Normal Form from PID finitely generated module structure.

Let $$k$$ be an algebraically closed field. Any $$k$$-linear map $$\phi:k^n\to k^n$$ imposes an extra $$k[x]$$-module structure on $$k^n$$ by defining $$p(x)\cdot v = p(\phi)(v).$$ Clearly then $$k^n$$ is a finitely generated $$k[x]$$-module.

As $$k[x]$$ is a PID, applying the structure theorem gives $$k[x]$$-module isomorphism $$k^n\simeq \bigoplus_{j=1}^{m}\bigoplus_{i=1}^{s_j} k[x]/(x-a_j)^{d_{i,j}}.$$ Note that $$k[x]$$ cannot occur on right hand side as $$k^n$$ is finitely generated. For simplicity let right hand side be W.

Then I asked myself, "what does a $$k[x]$$-module homomorphism $$f: V\to W$$ mean?" Let $$\phi: V\to V$$ and $$\varphi: W\to W$$ be the linear map associated to the $$k[x]$$-module structure of $$V$$ and $$W$$ respectively. Then $$f(\phi(v)) = f(x\cdot v) = x\cdot f(v) = \varphi(f(v))$$ for every $$v\in V$$, in other words, the $$k[x]$$-module homomorphism is the commutative diagram relation $$f\circ \phi = \varphi\circ f$$(where vertical arrows stands for $$f$$ below:)

$$\require{AMScd}$$ $$\begin{CD} V @>{\phi}>> V\\ @VVV @VVV \\ W @>{\varphi}>> W \end{CD}$$

In particular when $$f$$ is an isomorphism, $$\phi$$ and $$\varphi$$ are conjugate, meaning their corresponding matrices are similar.

Back to the question, iff $$\phi$$ stands for the map $$\phi: k^n\to k^n$$ and $$\varphi$$ stands for the linear map $$W\to W$$ from the $$k[x]$$-module structure. Clearly $$\varphi$$, written in matrix form with a suitable basis is the Jordan Normal Form. And the statement is then excatly $$\phi$$, the original matrix, is conjugate(similar) to the Jordan Normal Form of it.

I haven't seen anyone mentioning the "isomorphism implies similar" statement to me. Is this avoidable in other ways? I only did this once by myself.