Is it a Ring? Is it PID?

Question

Let $H$ be a Hilbert separable space, $f: H \to H$ a linear operator and

$$ A_f := \{ (a_i)_{i\in \omega +1 } \in \mathbb{C}^{\omega + 1}: \forall v \in H \ [ \sum_{i\in \omega}a_if^i+a_\omega \lim_{m \rightarrow \infty }f^m](v) \text{ is an element of } H \}$$

I can think an element $z$ of $A_f$ as a power series, i.e.

$$ z=\sum_{i\in \omega}a_ix^i + a_\omega x^\omega$$

So I may try to make $A_f$ a ring with following multiplication.

$$\begin{eqnarray*}&&\left(\sum_{i\in \omega}a_ix^i + a_\omega x^\omega\right) * \left(\sum_{i\in \omega}b_ix^i + b_\omega x^\omega\right)\\ &:=& \left(\sum_{i\in \omega}a_ix^i\right)\left(\sum_{i\in \omega}b_i x^i \right)+ \left(b_\omega\sum_{i\in \omega}a_i + a_\omega\sum_{i\in \omega}b_i +a_\omega b_\omega\right)x^\omega\end{eqnarray*} $$

Is this working, is $A_f$ a ring this way? Is it PID?

Answer 1

The question as it is stated has answer "no", since your wannabe-product $A_f\times A_f\to A_f$ is not a total operation on $A_f$: take $f=0$, then every $(a_i)$ is allowed since every $v\in H$ is sent to 0, for any choice of a net $(a_i)$. Now, how do you cope with non-convergent $(a_i)$'s?

Fortunately, if you restrict to those $(a_i)$ which have well-defined sums in $\mathbb{C}$[1], your operation is (bilinear and) associative. Let's have fun with notations:

1. Let $\mathfrak a,\mathfrak b,\mathfrak c$ be three "admissible" nets in $A_0$;
2. let $\mathfrak a_<$ be the truncation of the net $\mathfrak a$ to indices $<\omega$, and similarly for $\mathfrak b_<, \mathfrak c_<$;
3. let $\cdot$ denote the Cauchy product of series;
4. let $A=\sum \mathfrak a_<$ be the sum of the series $\mathfrak a_<$, and similarly for $B,C$;

then on the one side $$ (\mathfrak a * \mathfrak b) * \mathfrak c = (\mathfrak a_<\cdot \mathfrak b_<)\cdot \mathfrak c_< + (c_\omega AB + \ell_\omega(C+c_\omega))x^\omega $$ where $\ell_\omega = a_\omega B + b_\omega A + a_\omega b_\omega$; on the other hand, $$ \mathfrak a * (\mathfrak b * \mathfrak c) = \mathfrak a_<\cdot(\mathfrak b_<\cdot\mathfrak c_<) + (a_\omega BC + p_\omega (A + a_\omega)) $$ where $p_\omega = b_\omega C + c_\omega B + b_\omega c_\omega$.

Now it's easy to conclude, please don't let me TeX you that both sides are equal to $$ \mathfrak a_<\cdot \mathfrak b_<\cdot\mathfrak c_< + \Big(a_\omega BC + A b_\omega C + AB c_\omega + A b_\omega c_\omega + a_\omega B c_\omega + a_\omega b_\omega C + a_\omega b_\omega c_\omega\Big)x^\omega $$ ...too late. []

EDIT: actually I think that you are simply generalizing Cauchy product a bit: consider that the coefficient of $x^\omega$ is nothing more than $\sum_{|i|+|j|=|\omega|}a_i b_j$, where $|X|$ is the cardinality of a set. But maybe it's only a late-night mirage, I'm putting this here to check again tomorrow.

[1] it's enough that $\sum_{i<\omega}a_i<\infty$.