Definition of Polynomial from Polynomial Ring over Sequence

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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:

\((1)\)   $:$   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 $\struct {R \sqbrk X, +, \circ}$ 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:

\(\displaystyle \sequence {r_0, r_1, \ldots, r_n, 0, \ldots \ldots}\) \(=\) \(\displaystyle \sequence {r_0, 0, 0, \ldots \ldots} \odot \sequence {1, 0, 0, \ldots}\)
\(\displaystyle \) \(\) \(\, \displaystyle \oplus \, \) \(\displaystyle \sequence {r_1, 0, 0, \ldots \ldots} \odot \sequence {0, 1, 0, \ldots}\)
\(\displaystyle \) \(\) \(\, \displaystyle \oplus \, \) \(\displaystyle \cdots\)
\(\displaystyle \) \(\) \(\, \displaystyle \oplus \, \) \(\displaystyle \sequence {r_n, 0, 0, \ldots \ldots} \odot \sequence {0, \ldots \paren n \ldots, 0, 1, 0, \ldots}\)
\(\displaystyle \) \(=\) \(\displaystyle r_0 \oplus r_1 \circ x \oplus r_2 \circ x^2 \oplus \ldots \oplus r_n \circ x^n\)

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

$\blacksquare$


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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