Definition of Polynomial from Polynomial Ring over Sequences

Theorem
Let $R$ be a ring with unity.

Let $R \left[{x}\right]$ be a polynomial over $R$ in $x$.

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

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

Consider the injection $\phi: R \to P \left[{R}\right]$ defined as:
 * $\forall r \in R: \phi \left({r}\right) = \left \langle{r, 0, 0, \ldots}\right \rangle$

It is easily checked that $\phi$ is a ring monomorphism.

So the set $\left\{{\left \langle{r, 0, 0, \ldots}\right \rangle: r \in R}\right\}$ is a subring of $P \left[{R}\right]$ which is isomorphic to $R$.

So we identify $r \in R$ with the sequence $\left \langle{r, 0, 0, \ldots}\right \rangle$.

Next we note that $P \left[{R}\right]$ contains the element $\left \langle{0, 1, 0, \ldots}\right \rangle$ which we can call $x$.

From the definition of ring product on the polynomial ring over $R$, we have that:
 * $x^2 = \left \langle{0, 1, 0, \ldots}\right \rangle^2 = \left \langle{0, 0, 1, 0, 0, \ldots}\right \rangle$
 * $x^3 = \left \langle{0, 1, 0, \ldots}\right \rangle^3 = \left \langle{0, 0, 0, 1, 0, \ldots}\right \rangle$

... and in general:
 * $x^n = \left \langle{0, 1, 0, \ldots}\right \rangle^n = \left \langle{0, \ldots (n) \ldots, 0, 1, 0, \ldots}\right \rangle$

for all $n \ge 1$.

Hence we see that:

So by construction, $R \left[{x}\right]$ is seen to be equivalent to $P \left[{R}\right]$.

But $R \left[{x}\right]$ is a ring, and so $P \left[{R}\right]$ is one also.

It can also be shown that this proof works for the general ring whether it be a ring with unity or not.