Definition of Polynomial from Polynomial Ring over Sequence

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Theorem

Let $\left({R, +, \circ}\right)$ be a ring with unity.


Let $\left({P \left[{R}\right], \oplus, \odot}\right)$ be the polynomial ring over the set of all sequences in $R$:

$P \left[{R}\right] = \left\{{\left \langle {r_0, r_1, r_2, \ldots}\right \rangle}\right\}$

where the operations $\oplus$ and $\odot$ on $P \left[{R}\right]$ be defined as:

\((1)\)   $:$   Ring Addition:       \(\displaystyle \left \langle {r_0, r_1, r_2, \ldots}\right \rangle \oplus \left \langle {s_0, s_1, s_2, \ldots}\right \rangle \)   \(\displaystyle = \)   \(\displaystyle \left \langle {r_0 + s_0, r_1 + s_1, r_2 + s_2, \ldots}\right \rangle \)             
\((2)\)   $:$   Ring Negative:       \(\displaystyle -\left \langle {r_0, r_1, r_2, \ldots}\right \rangle \)   \(\displaystyle = \)   \(\displaystyle \left \langle {-r_0, -r_1, -r_2, \ldots}\right \rangle \)             
\((3)\)   $:$   Ring Product:       \(\displaystyle \left \langle {r_0, r_1, r_2, \ldots}\right \rangle \odot \left \langle {s_0, s_1, s_2, \ldots}\right \rangle \)   \(\displaystyle = \)   \(\displaystyle \left \langle {t_0, t_1, t_2, \ldots}\right \rangle \)             where $\displaystyle t_i = \sum_{j \mathop + k \mathop = i} r_j \circ s_k$


Let $\left({R \left[{X}\right], +, \circ}\right)$ be the ring of polynomials over $R$ in $X$.


Then $\left({R \left[{X}\right], +, \circ}\right)$ and $\left({P \left[{R}\right], \oplus, \odot}\right)$ are isomorphic.


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:

$x^2 = \left \langle{0, 1, 0, \ldots}\right \rangle \odot \left \langle{0, 1, 0, \ldots}\right \rangle = \left \langle{0, 0, 1, 0, 0, \ldots}\right \rangle$
$x^3 = \left \langle{0, 0, 1, 0, 0, \ldots}\right \rangle \odot \left \langle{0, 1, 0, \ldots}\right \rangle = \left \langle{0, 0, 0, 1, 0, \ldots}\right \rangle$

... and in general:

$x^n = \left \langle{0, 1, 0, \ldots}\right \rangle^{n-1} \odot \left \langle{0, 1, 0, \ldots}\right \rangle = \left \langle{0, \ldots (n) \ldots, 0, 1, 0, \ldots}\right \rangle$

for all $n \ge 1$.

Hence we see that:

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

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

$\blacksquare$


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


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