Existence of Ring of Polynomial Forms in Transcendental over Integral Domain
Theorem
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$ whose zero is $0_D$.
Let $X \in R$ be transcendental over $D$
Then the ring of polynomials $D \sqbrk X$ in $X$ over $D$ exists.
Proof
Suppose that $D \sqbrk X$ exists.
Let $\displaystyle \map P X = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be an arbitrary element of $D \sqbrk X$.
Then $\displaystyle \map P X$ corresponds to, and is completely described by, the ordered tuple of coefficients $\tuple {a_0, a_1, \dotsc, a_n, 0_D, 0_D, 0_D, \dotsc}$.
Consider the set $S$ of infinite sequences of elements of $D$ which are eventually $0_D$.
That is, whose elements are of the form $\tuple {b_0, b_1, \dotsc, b_n, 0_D, 0_D, 0_D, \dotsc}$ where $b_0, \ldots, b_n \in D$.
Consider the polynomial ring over $S$ by defining the operations:
\((1)\) | $:$ | Ring Addition: | \(\displaystyle \sequence {r_0, r_1, r_2, \ldots} + \sequence {s_0, s_1, s_2, \ldots} \) | \(\displaystyle = \) | \(\displaystyle \sequence {r_0 + s_0, r_1 + s_1, r_2 + s_2, \ldots} \) | |||
\((2)\) | $:$ | Ring Negative: | \(\displaystyle -sequence {r_0, r_1, r_2, \ldots} \) | \(\displaystyle = \) | \(\displaystyle \sequence {-r_0, -r_1, -r_2, \ldots} \) | |||
\((3)\) | $:$ | Ring Product: | \(\displaystyle \sequence {r_0, r_1, r_2, \ldots} \circ \sequence {s_0, s_1, s_2, \ldots} \) | \(\displaystyle = \) | \(\displaystyle \sequence {t_0, t_1, t_2, \ldots} \) | where $\displaystyle t_i = \sum_{j \mathop + k \mathop = i} r_j s_k$ |
From Polynomial Ring of Sequences is Ring we have that $\struct {S, +, \circ}$ is a ring.
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 64$. Polynomial rings over an integral domain: Footnote