Polynomial Ring of Sequences is Ring

Theorem
Let $R$ be a ring.

Let $P \left[{R}\right]$ be the polynomial ring over $R$.

Then $P \left[{R}\right]$ is itself a ring.

Proof
We have by definition of polynomial ring that:
 * $P \left[{R}\right] = \left\{{\left \langle {r_0, r_1, r_2, \ldots}\right \rangle}\right\}$

where each $r_i \in R$, and all but a finite number of terms is zero.

Proof that Operations are Closed
We need to ensure that the operations as defined are closed.

Let $r = \left \langle {r_0, r_1, r_2, \ldots}\right \rangle, s = \left \langle {s_0, s_1, s_2, \ldots}\right \rangle \in P \left[{R}\right]$.

As all but a finite number of terms of $r$ and $s$ are zero, there exist $m, n \ge 0$ such that:
 * $\forall i > m: r_i = 0$
 * $\forall j > n: s_j = 0$

Let $l = \max \left\{{m, n}\right\}$.

We can express the operations on $P \left[{R}\right]$ as:

We have that:
 * $\forall i > l: r_i + s_i = 0$

and so
 * $r + s \in P \left[{R}\right]$

Equally clearly:
 * $-r \in P \left[{R}\right]$

Now consider:
 * $\displaystyle \left({r s}\right)_i = \sum_{j \mathop + k \mathop = i}r_j s_k$

Let $i > m + n$.

Then in any $r_j s_k$ such that $j + k = i$, either $j > m$ or k > n.

In the first case $r_j = 0$ and in the second $s_j = 0$.

In either case $r_j s_k = 0$.

So:
 * $\displaystyle \forall i > m + n: \left({r s}\right)_i = \sum_{j \mathop + k \mathop = i}r_j s_k = 0$

So:
 * $r s \in P \left[{R}\right]$

Proof of Additive Group
The addition operation $r + s$ is clearly commutative and associative, and:
 * $\left \langle {0, 0, \ldots}\right \rangle = 0_{P \left[{R}\right]}$

Equally clearly:
 * $\forall r \in P \left[{R}\right]: r + \left({-r}\right) = \left \langle {r_i + \left({-r_i}\right)}\right \rangle = \left \langle {0, 0, \ldots}\right \rangle = 0_{P \left[{R}\right]}$

and so $-r$ is the inverse of $r$ for addition.

So $\left({P \left[{R}\right], +}\right)$ is an abelian group, as it needs to be for $P \left[{R}\right]$ to be a ring.

Proof of Ring Product
We need to establish that the ring product is associative.

Let $r = \left \langle {r_0, r_1, \ldots}\right \rangle, s = \left \langle {s_0, s_1, \ldots}\right \rangle, t = \left \langle {t_0, t_1, \ldots}\right \rangle \in P \left[{R}\right]$.

Then:

Similarly for $\left({r \left({s t}\right)}\right)_n$.

So the ring product is associative, and so forms a semigroup.

Proof of Distributivity
Finally we need to show that the ring product is distributive over ring addition.

Similarly for $\left({t \left({r + s}\right)}\right)_n$.

Hence the result.