Representatives of simple modules 
For each of the following rings find a list of representatives of all simple $R$-modules:
1) $R=\mathbb C[x]$
2) $R=\mathbb R[x]$

What I've tried was:
I know that $M$ is a simple $R$-module if and only if $M \cong R/I$ where $I$ is a maximal ideal, so, for each case, first I've tried to find all maximal left ideals.
1) Let $I$ be a maximal left ideal of $\mathbb C[x]$, then $I=<x-\alpha>, \alpha \in \mathbb C$. So every simple module $M$ is isomorphic to a quotient of the form $\mathbb C[x]/<x-\alpha>, \alpha \in \mathbb C$. My question is: How many quotients are up to isomorphisms?, I suspect that $C[x]/<x-\alpha>$ is isomorphic to $\mathbb C$ for any $\alpha$ but I couldn't prove this.
With the same idea, in 2) I've concluded that if $I$ is a maximal ideal in $\mathbb R[x]$, then $I=<x-\lambda>, \lambda \in \mathbb R$ or $I=<(x-z)(x-\overline{z})>, z=a+bi, b \neq 0$. So if $M$ is a simple module, then $M \cong \mathbb R[x] /<x-\lambda>$ or $M \cong  \mathbb R[x] /<(x-z)(x-\overline{z})>$. I have the same problem that in 1), I suspect that $\mathbb R[x]/<x-\lambda>$ is isomorphic to $\mathbb R$ but I don't know how to prove this, as for the quotients of the form $\mathbb R[x]/<(x-z)(x-\overline{z})>$ I don't know how many are there up to isomophism. 
Any help with these points would be greatly appreciated.
 A: You're almost done, everything's fine so far.

My question is: How many quotients are up to isomorphisms?, I suspect that ${\mathbb C}[x]/(x−\alpha)$ is isomorphic to ${\mathbb C}$ for any $\alpha$ but I couldn't prove this.

As complex vector spaces, all the ${\mathbb C}[x]$-modules ${\mathbb C}[x]/(x-\alpha)$ are $1$-dimensional (spanned by the cosets of constant polynomials), hence isomorphic to ${\mathbb C}$; in particular, they are mutually isomorphic. However, as ${\mathbb C}[x]$-modules - and this is what you consider here - they are pairwise non-isomorphic. To prove this, think on how you can recover $\alpha$ from the action of $x$ on ${\mathbb C}[x]/(x-\alpha)$.
The real case is similar. As ${\mathbb R}$-vector spaces, all the modules ${\mathbb R}[x]/(x-\alpha)$ are $1$-dimensional, hence isomorphic to ${\mathbb R}$, but you want to consider them as ${\mathbb R}[x]$-modules. The other family you exhibited consists of $2$-dimensional modules. To show that for $\lambda\neq\mu\in{\mathbb R}$ the ${\mathbb R}[x]$-modules ${\mathbb R}[x]/(x-\lambda)$ and ${\mathbb R}[x]/(x-\mu)$ are not isomorphic, proceed as in the complex case. For $\lambda\notin {\mathbb R}$, again look at the action of $x$ on ${\mathbb R}[x]/(x-\lambda)(x-\overline{\lambda})$ and try to recover $\{\lambda,\overline{\lambda}\}$ from it.
