Confusion regarding the definition of the Weyl Algebra I've been studying algebraic structures recently; rings, fields, vector spaces and so on. I've recently just started learning what an algebra is, which, from what I can tell is a ring-like structure with scalar multiplication attached, so that the algebra $A$ forms a vector space with the $+$ and scalar multiplication operations.
The first example of an Algebra I encountered was the  First Weyl Algebra over the complex numbers, $A_1(\mathbb{C})$, but I'm confused regarding its definition. The definition I got said to look at the polynomial ring   $\mathbb{C}[x]$, and then regard this as a vector space over $\mathbb{C}$. They go on to define two linear operations $\delta$ and $x$, where $\delta$ is differentiation and $x$ is multiplication. Then $A_1 = {\{\sum_{i = 0}^{n} f_i(x) \delta^i : f_i(x) \in \mathbb{C}[x] }\}$.
$(1)$ How can we just regard the polynomial ring $\mathbb{C}[x]$ as a vector space? Aren't rings and vector spaces fundamentally different?
$(2)$ What exactly is an operator? Can we just think of them as operations like $+$ or multiplication? 
$(3)$ I've seen expressions like $\delta^n \cdot x = n\delta^{n-1} + x\cdot\delta^n$ written, and I don't really understand what this means? Are we differentiating $x$ $n$ times or is this multiplication in the algebra?
Any help would be appreciated, thank you!
 A: (1) Of course we can just regard $\mathbb{C}[x]$ as a vector space. A vector space is an abelian group equipped with the action of a field. A ring is an abelian group equipped with a multiplication, and an algebra is a ring equipped with the action of a field. In particular, $\mathbb{C}[x]$ is a vector space.
(2) An operator on a vector space $V$ is a linear transformation $T:V\to V$. The set of operators on $V$ is usually denoted by $\mathrm{End}(V)$. In particular, the map
$$
\delta: \mathbb{C}[x]\to \mathbb{C}[x]
$$
given by $\delta(f)=f'$ is a linear operator. The map
$$
m_x:\mathbb{C}[x]\to \mathbb{C}[x]
$$
given by $m_x(f)=xf$ is also a linear operator which, by abuse of notation, is referred to as '$x$'. In this context, $A_1\subset \mathrm{End}(V)$ is the algebra generated by $\delta$ and $x$.
(3) Multiplication in $\mathrm{End}(V)$ (and also $A_1$) is given by composition of functions. So, for example,
$$
(\delta\circ x)(f)=\delta(xf)=f + xf'=(1+x\circ\delta)(f)
$$
where the second equality follows by the product rule, and $1\in\mathrm{End}(V)$ is understood to be the identity transformation. The general statement can be proved by induction. Indeed, suppose the formula holds for the $(n-1)$-case. Then
\begin{align}
(\delta^n\circ x)(f)&=\delta((\delta^{n-1}\circ x)(f))\\
&=\delta((n-1)f^{(n-2)}+xf^{(n-1)})\\
&=(n-1)\delta(f^{(n-2)})+\delta(xf^{(n-1)})\\
&=(n-1)f^{(n-1)}+f^{(n-1)}+xf^{(n)}
\end{align}
therefore induction holds.
