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.

  • 1
    $\begingroup$ To write ideals using angle brackets use \langle and \rangle. $\endgroup$ – Mariano Suárez-Álvarez Nov 3 '14 at 7:40

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.

  • $\begingroup$ Thanks for your answer. Let me see if I could partially follow you: consider $M=\mathbb C[x]/ \langle x-\alpha \rangle$ and $N=\mathbb C[x]/ \langle x-\lambda \rangle$, $\alpha \neq \lambda$. In $M$, $x.z=(x-\alpha+\alpha).z=(x-\alpha).z+\alpha.z=\alpha z$. Similarly, in $N$ we have $x.z=\lambda z$. This means that the ring acts differently in each module if $\lambda \neq \alpha$. So, my question may be silly but how can I assure that this implies these two modules can't be isomorphic? $\endgroup$ – user16924 Nov 3 '14 at 11:59

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