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Let $k$ be a field and $\mathfrak{g}=kx\oplus ky$ with $[x,y]=y$. Show that $\rho(x)=t\,\frac{d}{dt}$ and $\rho(y)=t\cdot$ (mult. by $t$) define a representation $\rho:\mathfrak{g}\to \mathfrak{gl}(k[t])$. Show that the ideals $(t^n)$ are the only non-zero subrepresentations.

The first part is easy: just check that $\rho$ is a Lie algebra map. I just want to check that I'm thinking about the second part correctly. Checking that each $(t^n)$ is stable under $\rho$ is also easy. Now given a general $\rho$-stable subspace $W\subseteq k[t]$ I want to show that $W=(t^n)$ for some $n$. Now $k[t]$ has basis $\{1,t,t^2,\ldots\}$, so $W$ has a basis coming from some subset of this basis; in particular, $W$ contains $t^n$ for some $n$. So $(t^n)\subseteq W$.

Is this the correct way to think about this? If so, how do I get $W\subseteq (t^n)$?

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up vote 2 down vote accepted

No, you can't assume that the basis for $W$ is a subset of the basis for $k[t]$. For example $k^2$ is a vector space with basis $(0, 1)$ and $(1, 0)$ but we can take any vector, for example $(1, 1)$, and get a $1$ dimensional subspace. We are not just limited to choosing from the basis.

To prove that $W = (t^n)$ first we must show that $t^n \in W$ for some $n$. Choose $f = t^n + a_{n + 1}t^{n + 1} + \cdots + a_mt^m$ in $W$ with the property that $n$ is minimal among all possible choices (why can I assume it's monic?). Now act on $f$ to try and produce $t^n$. For example, start by acting by the element $\frac{1}{n + 1}x$ and then subtracting the result from $f$.

After you've figured that out we have $t^n \in W$ hence $(t^n) \subseteq W$. Now you need to argue that $W \subseteq t^n$. Do that by arguing the contrapositive. Assume $g \notin (t^n)$ and argue that also $g \notin W$. For this step you will use the fact that when we chose $f$ we chose it so that $n$ was minimal.

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Thank you for your help – Bey Feb 12 '13 at 3:08

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