Definition of Polynomial from Polynomial Ring over Sequence

Theorem
Let $\struct {R, +, \circ}$ be a ring with unity.

Let $\struct {P \sqbrk R, \oplus, \odot}$ be the polynomial ring over the set of all sequences in $R$:
 * $P \sqbrk R = \set {\sequence {r_0, r_1, r_2, \ldots} }$

where the operations $\oplus$ and $\odot$ on $P \sqbrk R$ be defined as:

Let $\struct {R \sqbrk X, +, \circ}$ be the ring of polynomials over $R$ in $X$.

Then $\struct {R \sqbrk X, +, \circ}$ and $\struct {P \sqbrk R, \oplus, \odot}$ are isomorphic.

Proof
Let $P \sqbrk R$ be the polynomial ring over $R$.

Consider the injection $\phi: R \to P \sqbrk R$ defined as:
 * $\forall r \in R: \map \phi r = \sequence {r, 0, 0, \ldots}$

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

So the set $\set {\sequence {r, 0, 0, \ldots}: r \in R}$ is a subring of $P \sqbrk R$ which is isomorphic to $R$.

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

Next we note that $P \sqbrk R$ contains the element $\sequence {0, 1, 0, \ldots}$ which we can call $x$.

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

and in general:
 * $x^n = \sequence {0, 1, 0, \ldots}^{n - 1} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, \ldots \paren n \ldots, 0, 1, 0, \ldots}$

for all $n \ge 1$.

Hence we see that:

So by construction, $R \sqbrk X$ is seen to be equivalent to $P \sqbrk R$.

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