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.