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:
\((1)\) | $:$ | 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.
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 25$. Polynomials