I was thinking about factorization of elements in an arbitrary ring recently and wondered whether there was always a largest $k$ such that $a^k | b$, where $a$ and $b$ are arbitrary ring elements and $k$ is a natural number.
I then came up with a slightly extended polynomial ring $R[x, x^\infty]$ that permits infinite terms as well. In this ring, $x^\infty$ is a bizarre sort of quasi-annihilator that erases the distinction between powers of $x$.
It's easy to check that the ring axioms still hold:
Let $p, q, r$ be polynomials. Let $a, b, c$ be $R$-elements. Let $p''$ be the sum of the coefficients of a polynomial $p$.
$$ (p+ax^\infty)-(q+bx^\infty) = (p-q) + (a-b)x^\infty \\ (p + ax^\infty) * (q + bx^\infty) = pq + (bp'' + aq'' + ab)x^\infty $$
I will verify the right distributive law as an example showing that the ring laws are satisfied.
$$ ((p+ax^\infty) + (q+bx^\infty))*(r + cx^\infty) = \\ ((p+q) + (a+b)x^\infty) * (r + cx^\infty) = \\ (pr + qr) + (ar'' + br'' + p''c + q''c + p''r'' + q''r'')x^\infty = \\ (pr + (ar'' + p''c + p''r)x^\infty) + (qr + br'' + q''c + q''r'')x^\infty = \\ (p+ax^\infty)(r+cx^\infty) + (q + bx^\infty)(r+ cx^\infty) $$
$(\mathbb{N} \cup \{\infty\}, 0, +)$ is a perfectly nice commutative monoid with identity $0$.
This makes me wonder whether we can replace the exponents of a polynomial with any commutative monoid $M$ and give us back $R[x^M]$ or something and, if so, what it's called.