TITLE: Is this action of $\mathbb F[[x]]$ on $\bigoplus_{i=0}^{\infty}\mathbb F$ natural?
QUESTION [6 upvotes]: The title of my question has a field $\mathbb F$ in it, but to make sure I'm not losing anything, I would like to introduce my question in full generality. But still, I will be happy with an answer in fields, which is why I chose this title. Please feel free to consider $R$ a field and all modules to be vector spaces.
Notation
I'm assuming $\mathbb N=\{0,1,\ldots\}$ in this question.
Let $R$ be any unital ring. $R[x]$ denotes the ring of polynomials over $R$. $R[[x]]$ denotes the ring of formal power series over $R$. (The linked article requires that $R$ be commutative. I don't.) $M_n(R)$ will denote the ring of all square matrices over $R$ indexed by the set $\{0,1,\ldots,n-1\}\times\{0,1,\ldots,n-1\}.$
$M_{\infty}(R)$ denotes the set of all $\mathbb N\times\mathbb N$-matrices over $R$ whose columns have finitely many non-zero coefficients. This restriction allows the usual multiplication of such matrices, and so they form a ring.
I will also be using the ring of endomorphisms $E(R)$ of the free $R$-module $V(R)=\bigoplus_{i=0}^{\infty}R,$ with a fixed basis $e_0,e_1,\ldots$ 
Background
The rings $M_\infty (R)$ and $E(R)$ are isomorphic. Just like for finite matrices over fields, the $i$-th column of a matrix $A$ represents the element $y$ of $V(R)$ to which $e_i$ is mapped by a corresponding endomorphisms $\alpha.$ To be precise, the column contains the coefficients of $y$ in the basis $e_0,e_1,\ldots$ This explains the requirement that there be only finitely many non-zero elements in the columns, because every $y\in V(R)$ can be written uniquely as a linear combination of the basis, and linear combinations are finite sums.
Let now $\phi:R[[x]]\longrightarrow M_\infty(R)$ such that 
$$\sum_{i=0}^\infty a_ix^i\mapsto 
\begin{pmatrix}
a_0    & a_1    & a_2    & a_3    & \ldots\\
0      & a_0    & a_1    & a_2    & \ldots\\
0      & 0      & a_0    & a_1    & \ldots\\
0      & 0      & 0      & a_0    & \ldots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}
$$
$\phi$ is a ring isomorpisms onto its image $M_\triangledown(R).$ I found it very interesting that formal power series (and therefore also polynomials) can be embedded in matrices. It turns out to be useful. I'm reading a 1974 paper by Jan Krempa which uses it. The title is On the Jacobson Radical of Polynomial rings.
Very similarly, we can define the map $\psi:R[x]/\langle x^n\rangle\longrightarrow M_n(R)$ such that 
$$
\left(\sum_{i=0}^{n-1}a_ix^i\right)+\langle x^n\rangle\mapsto
\begin{pmatrix}
a_0    & a_1    & a_2    & \ldots & a_{n-1}\\
0      & a_0    & a_1    & \ldots & a_{n-2}\\
0      & 0      & a_0    & \ldots & a_{n-3}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0      & 0      & 0      & \ldots & a_0
\end{pmatrix}$$
Question
Let $f=\sum_{i=0}^\infty a_ix^i\in R[[x]].$ Now $\phi(f)$ is a matrix over $R$ and as such can be identified with an endomorphism $\alpha_f$ of $V(R).$ As an endomorphism, $\alpha_f$ can take arguments from $V(R)$ and send them to some other elements of $V(R).$ However, the original power series $f$ is not usually thought to have such a capability. We can define for any $f\in R[[x]]$ and $y\in V(R)$ $$fy:=\alpha_f(y).$$
I would like to know if thinking of formal power series acting in this way on $V(R)$ is natural. I was trying to see some natural way in which a power series can be thought of as an endomorphism of a free module or vector space but I failed.
It's easy to write what $fy$ actually is. Let $y\neq 0$ for simplicity. Let $\{y_i\}_{i=1}^\infty$ be the coefficients of $y$ in the basis $e_0,e_1,\ldots$ Let $n$ be the greatest index such that $y_n=0$ but $y_{n-1}\neq 0$. Now calculating $fy$ is essentially the same as calculating 
$$
\begin{pmatrix}
a_0    & a_1    & a_2    & \ldots & a_{n-1} & \ldots\\
0      & a_0    & a_1    & \ldots & a_{n-2} & \ldots\\
0      & 0      & a_0    & \ldots & a_{n-3} & \ldots\\
\vdots & \vdots & \vdots & \ddots & \vdots  & \ldots\\
0      & 0      & 0      & \ldots & a_0     & \ldots\\
\vdots & \vdots & \vdots & \vdots & \vdots  & \ddots
\end{pmatrix}
\begin{pmatrix}
y_0 \\ y_1 \\ y_2 \\ \vdots \\ y_{n-1} \\ \vdots
\end{pmatrix}
=
\begin{pmatrix}
a_0y_0+a_1y_1+a_2y_2 +\cdots +a_{n-1}y_{n-1}\\
a_0y_1+a_1y_2+\cdots +a_{n-2}y_{n-1}\\
a_0y_2+\cdots +a_{n-3}y_{n-1}\\
\vdots\\
a_0y_{n-1}\\
\vdots
\end{pmatrix}
$$
I've looked at these formulas for quite some time now and they still tell me nothing. Could please tell me if you recognize them? 
An analogous question can be asked about the finite-dimensional case of $R[x]/\langle x^n\rangle$ and the map $\psi$. I hereby ask it then, as this post is already so long and latex-laden that I have to wait one or two seconds for a reaction to my typing on the screen.

REPLY [1 votes]: First take the case of a field, $\mathbb F$, of characteristic zero, then let $p_n(y) = \frac{y^n}{n!}$.
Then the $p_n$ form a basis for $\mathbb F[y]$ as a vector space over $\mathbb F$.
What is the operation of $F[[x]]$ on $\mathbb F[y]$?  It turns out, it is the operation that sends $p_n(y)$ to $p_{n-1}(y)$.  So if an element of $\mathbb F[y]$ is written as $\sum {a_n p_n(y)}$, the result of having $x$ act on $\mathbb F[y]$ is the result of differentiating the polynomial.  In other words, $x$ acts as $D_y$.  So we can think of:
$$\mathbb F[[x]] \cong \mathbb F[[D_y]]$$
Essentially, $(\sum a_i x^i)\cdot p(y) = \sum a_i p^{(i)}(y)$
So this is action is differentiation in the case of a field of characteristic zero. 
You can extend this idea to a ring $R$ if $\mathbb Q\subset R$.
Now if $R$ is commutative of arbitrary characteristic, then you can define a ring on $\oplus_{i\in\mathbb N} R$ by defining $(a_i)\cdot(b_i) = (c_i)$ where:
$$c_n = \sum_{i=0}^n \binom{n}{i} a_i b_{n-i}$$
This is well-defined because $\binom{n}{i}$ is an integer.
Now you can verify the rule: $$x\cdot((a_i)\cdot(b_i)) = (x\cdot(a_i))\cdot (b_i) + (a_i)\cdot (x\cdot (b_i))$$
That is, we write $D=x$, $a=(a_i)$, and $b=(b_i)$ we get:
$$D(ab) = (Da)b + a(Db)$$
That might also be true if $R$ is non-commutative.
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