Isomorphic $\mathbb{C}[X]$-modules

Question

My question is :

It is true that $\mathbb{C}[X]/(x-c)$ and $\mathbb{C}[X]/(x-d)$ are isomorphic $\mathbb{C}[X]$-modules if and only if $c=d$?

I have the feeling that the answer is simple but I haven't found it.

Answer 1

Let $f:\mathbb C[x]/(x-c)\to \mathbb C[x]/(x-d)$ a linear map of vector spaces such that $xf(P)=f(xP)$.

Then, $0=(x-d)f(P)=f((x-d)P)=f((x-c+c-d)P)=f((c-d)P)=(c-d)f(P)$, for all $P$.

So, $d=c$ if $f$ is not the $0$ map.

Answer 2

A $\mathbb{C}[X]$-module is the same thing as a $\mathbb{C}$-vector space $V$ with an endomorphism $u$, and a morphism of $\mathbb{C}[X]$-modules $(V,u)\to (V',u')$ is a linear map $f:V\to V'$ such that $u'\circ f=f\circ u$.

Since $\mathbb{C}$ is algebraically closed, $u$ always has an eigenvector, so $V$ has a 1-dimensional subspace which is stable by $u$. This implies the fact that you mentioned: simple $\mathbb{C}[X]$-modules are 1-dimensional, and $u$ acts by a constant $c\in \mathbb{C}$, so this module is isomorphic to $\mathbb{C}[X]/(X-c)$.

You see that if $(V,u)$ and $(V',u')$ are 1-dimensional and isomorphic, then $u$ must act on $V$ by the same constant as $u'$ on $V'$. So indeed $\mathbb{C}[X]/(X-c)$ is isomorphic to $\mathbb{C}[X]/(X-d)$ iff $c=d$.