Existence of Ring of Polynomial Forms in Transcendental over Integral Domain

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Theorem

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

Let $\left({D, +, \circ}\right)$ be an integral subdomain of $R$ whose zero is $0_D$.

Let $X \in R$ be transcendental over $D$


Then the ring of polynomials $D \left[{x}\right]$ in $X$ over $D$ exists.


Proof


Suppose that $D \left[{X}\right]$ exists.

Let $\displaystyle P \left({X}\right) = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be a typical element of $D \left[{x}\right]$.

Then $\displaystyle P \left({X}\right)$ corresponds to, and is completely described by, the sequence of coefficients $\left({a_0, a_1, \ldots, a_n, 0_D, 0_D, 0_D, \ldots}\right)$.


Consider the set $S$ of infinite sequences of elements of $D$ which are eventually $0_D$.

That is, whose elements are of the form $\left({b_0, b_1, \ldots, b_n, 0_D, 0_D, 0_D, \ldots}\right)$ where $b_0, \ldots, b_n \in D$.

Consider the polynomial ring over $S$ by defining the operations:

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

From Polynomial Ring of Sequences is Ring we have that $\left({S, +, \circ}\right)$ is a ring.



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