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I am trying to solve following exercise from Humphrey's Lie algebra:

Ler $R=F[X,Y]$ be polynomial ring in two variables over field $F$. Then $L=\mathfrak{sl}(2,F)$ acts naturally on subspace $\{mX+nY: m,n\in F\}$ it:

$$\begin{bmatrix} X \\ Y\end{bmatrix} \mapsto \begin{bmatrix} a & b\\ c & -a\end{bmatrix} \begin{bmatrix} X \\ Y\end{bmatrix}.$$ Extend this action to the ring $R=F[X,Y]$ by the derivation rule: $z.fg=(z.f)g+f(z.g)$ for $z\in L$ and $f,g\in F[X,Y]$.

Show that this extension is well defined and becomes an $L$-module.

I didn't get any intuition to prove wee-definedness; can you help me?

This looks something simple exercise, but, I was confused in understanding- prove that it is well-defined. What should I do actually to prove well-definedness?

What I did: consider $x=\begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}$, $y=\begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}$ and $h=\begin{bmatrix} 1 & 0\\ 0 & -1\end{bmatrix}$. Then

$$x.X^n=0, \,\,\,\, x.Y^n=nY^{n-1}X, \,\,\,\, y.X^n=nX^{n-1}Y, \,\,\,\, y.Y^n=0.$$ After this, what to do?

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Consider $x=\pmatrix{a &b\cr c&d}$, you have $x.\pmatrix{X\cr Y}=\pmatrix{aX+bY\cr cX+dY}$. This means that $x.X=aX+bY$ and $x.Y=cX+dY$.

You can define recursively $x.X^n=x.(XX^{n-1})=(x.X)X^{n-1}+X(x.X^{n-1})=(aX+bY)X^{n-1}+X(x.X^{n-1})$.

Similar with $x.Y^n$

show $[x,y].P=x.(y.P)-y.(x.P)$.

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    We have relation $[h,x]=2x$, $[h,y]=-2y$ and $[x,y]=h$. So for each of these relations, should we verify that action of LHS of each relation is equal to the action of RHS of that relation; does this completes the proof that action is well defined? – 2017-01-12
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    yes, you can show it by like that. – 2017-01-12
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    But one point also: if we write $X^n$ as $X^k.X^{n-k}$ and also as $X^l.X^{n-l}$ then should we also verify that action on each of these is same? This is a point confusing me, what exactly should we do to show well-definedness – 2017-01-12
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    $x.(PQ)=(x.P)Q+P(x.Q)=(x.Q)P+Q(x.P)=x.(QP)$. – 2017-01-12