Definition of Polynomial from Polynomial Ring

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Theorem

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

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

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:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \left \langle{r_0, r_1, \ldots, r_n, 0, \ldots \ldots}\right \rangle$	$=$	$\displaystyle \left \langle{r_0, 0, 0, \ldots \ldots}\right \rangle \left \langle{1, 0, 0, \ldots}\right \rangle$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$+$	$\displaystyle \left \langle{r_1, 0, 0, \ldots \ldots}\right \rangle \left \langle{0, 1, 0, \ldots}\right \rangle$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$+$	$\displaystyle \cdots$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$+$	$\displaystyle \left \langle{r_n, 0, 0, \ldots \ldots}\right \rangle \left \langle{0, \ldots (n) \ldots, 0, 1, 0, \ldots}\right \rangle$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$+$	$\displaystyle r_0 + r_1 x + r_2 x^2 + \ldots + r_n x^n$	$\displaystyle $	$\displaystyle $	$\displaystyle $

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.

$\blacksquare$

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

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left \langle{r_0, r_1, \ldots, r_n, 0, \ldots \ldots}\right \rangle\)	\(=\)	\(\displaystyle \left \langle{r_0, 0, 0, \ldots \ldots}\right \rangle \left \langle{1, 0, 0, \ldots}\right \rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle \left \langle{r_1, 0, 0, \ldots \ldots}\right \rangle \left \langle{0, 1, 0, \ldots}\right \rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle \left \langle{r_n, 0, 0, \ldots \ldots}\right \rangle \left \langle{0, \ldots (n) \ldots, 0, 1, 0, \ldots}\right \rangle\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle r_0 + r_1 x + r_2 x^2 + \ldots + r_n x^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)