Let $L$ be the $2$-dimensional Lie algebra over $\mathbb C$ with basis $x, y$ and $[xy] = x$. Let $V$ be a vector space over $\mathbb C$ with basis $v_1, v_2$. Then $V$ is an $L$-module if we define multiplication by $$xv_1 = 0,\quad xv_2 = v_1,\quad yv_1 = -v_1,\quad yv_2 = v_1$$ Show that $V$ is isomorphic to the adjoint module for $L$.
I have been trying to work with explicit $L$-module isomorphisms such as $$f:V\to L,\ \ \alpha v_1 + \beta v_2\mapsto \alpha x + \beta y$$
but when I try to validate the homomorphism condition $f(lv)=lf(v)$ for $l\in L,\ v\in V$ I always run into a dead end, e.g. $$f(y(\alpha v_1+\beta v_2))=(\beta-\alpha)x\ \ \ \ne\ \ -\alpha x=[y,f(\alpha v_1+\beta v_2)]$$
Any hints as to how I could show isomorphy?
The eigenvectors for $y$ acting on $V$ are $v_1$ with eigenvalue $-1$ and $v_1+v_2$ with eigenvalue $0$. So these have to go (up to scalars) to $x$ (eigenvalue $-1$) and $y$ in the adjoint representation. Now you just have to check what (if any) scalar multiples work to match up the $x$ action.
That is, define a linear map $\phi: V \rightarrow L$ by $\phi(v_1)=ax$ and $\phi(v_1+v_2)=by$ for scalars $a$ and $b$. By design $\phi$ commutes with the action of $y$. Write out the equations $\phi(x v_1)=x \phi(v_1)$ and $\phi(x(v_1+v_2))=x \phi(v_1+v_2)$ to find a choice of $a,b$ that will work.