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 $X$.
Strictly speaking the definition in terms of Definition:Polynomial Form is needed here, with $X$ specifically being an Definition:Indeterminate (Polynomial Theory) or Definition:Transcendental over Integral Domain.
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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 $P \left[{R}\right]$.
$\blacksquare$
This article contains statements that are justified by handwavery, in particular:
The entirety of the above argument needs to be made properly rigorous.
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It can also be shown that this proof works for the general ring whether it be a ring with unity or not.
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): $\S 6.25$
