Consider the following action of the lie algebra $\frak{sl(2;\mathbb{C})}$ on the vector space of complex polynomial.
$$ e\cdot p(z) = z p(z)\quad h\cdot p(z) =2zp'(z)\quad f\cdot p(z)=-zp''(z) $$
It is easy to check that this defines an infinite dimensional $\frak{sl(2;\mathbb{C})}$ module. I want to show it is not irreducible. To do so I must find a non trivial submodule, however I am struggling to even find a subspace of the complex polynomials.
I initially tried the polynomials with integer coefficient, but realised that although they are closed under the $\frak{sl(2;\mathbb{C})}$ action they are not closed under scalar multiplication and so don't even define a subspace. I also tried thinking about even and odd degree polynomials, but $e$ and $f$ alter the degree by $\pm1$ so this cannot be a submodule either.
And help would be very appreciate, especially with finding a subspace.
We see that for all polynomials $p(z)$ the acted versions $e\cdot p$, $h\cdot p$ and $f\cdot p$ vanish at $z=0$.
I am fairly sure that it is the only non-trivial submodule of $\Bbb{C}[z]$. We see that the monomials are weight vectors (=eigenvectors of $h$) with non-negative even integers as weights. Furthermore, the weight vector $p_0(z)=1$ generates the entire module. Therefore $\Bbb{C}[z]$ is like a Verma module of lowest weight $0$. A kind of dual of the usual Verma module, but with the roles of positive/negative roots reversed. In a category containing modules like this, the dot action on weights must be adjusted accordingly. Implying that lowest weights $0$ and $2$ belong to the same block. $V^+$ is then a Verma-module of lowest weight $2$, and is irreducible.