We take the polynomial ring $\Bbb C [X]$ and we regard it as a $\Bbb C$-vector space. Then, we have the following linear transformations: $$x:\Bbb C[X]\longrightarrow \Bbb C[X],\ f(X) \longmapsto x(f(X)):=Xf(X)$$ and $$\partial :\Bbb C[X]\longrightarrow \Bbb C[X],\ f(X) \longmapsto \partial(f(X)):=\frac{df}{dX}.$$ We define the first Weyl Algebra as the set $$A_1(\Bbb C):=\{a_m(X)\partial^m+\dotsb+a_1(X)\partial+a_0(X):a_i(X)\in \Bbb C[X], n\in \Bbb N \} \subseteq \mathrm{End}_\Bbb C \Bbb C[X].$$
I am struggling to prove that $A_1(\Bbb C)$ is a ring and a $\Bbb C$- vector space.
Any help please?
It should be clear that $A_1(\mathbb C)$ is closed, as a subset of $\mathrm{End}_{\mathbb C}(\mathbb C[X])$, under addition and under multiplication by scalars in $\mathbb C$. To see that it is closed under composition, it is enough to compute $X^i\partial^m\cdot X^j\partial^n$, since composition in $\mathrm{End}_{\mathbb C}(\mathbb C[X])$ is bilinear. Moreover, $A_1(\mathbb C)$ is clearly closed under multiplication on the left by elements of $\mathbb C[X]$, and under multiplication on the right by powers of $\partial$. Thus we only need to consider a product $\partial^m\cdot X^j$. Finally, composition is associative, so by a double induction we only need to consider the product $\partial\cdot X$. This we now compute: $$ (\partial\cdot X)(X^r) = (r+1)X^r = rX^r + X^r, $$ so that $\partial\cdot X=X\cdot\partial + \mathrm{id}$. The right hand side clearly lies in $A_1(\mathbb C)$, so we are done.
In fact, this gives a presentation of the first Weyl algebra: it is the quotient of the free (non-commutative) polynomial ring $\mathbb C\langle X,Y\rangle$ by the (two-sided) ideal generated by $XY-YX+1$.