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

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

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.