Existence of Ring of Polynomial Forms in Transcendental over Integral Domain

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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


This needs considerable tedious hard slog to complete it.
In particular: The following is an outline only
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Suppose that $D \sqbrk X$ exists.

Let $\ds \map P X = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be an arbitrary element of $D \sqbrk X$.

Then $\map P X$ 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:       \(\ds \sequence {r_0, r_1, r_2, \ldots} + \sequence {s_0, s_1, s_2, \ldots} \)   \(\ds = \)   \(\ds \sequence {r_0 + s_0, r_1 + s_1, r_2 + s_2, \ldots} \)      
\((2)\)   $:$   Ring Negative:       \(\ds -\sequence {r_0, r_1, r_2, \ldots} \)   \(\ds = \)   \(\ds \sequence {-r_0, -r_1, -r_2, \ldots} \)      
\((3)\)   $:$   Ring Product:       \(\ds \sequence {r_0, r_1, r_2, \ldots} \circ \sequence {s_0, s_1, s_2, \ldots} \)   \(\ds = \)   \(\ds \sequence {t_0, t_1, t_2, \ldots} \)      where $\ds t_i = \sum_{j \mathop + k \mathop = i} r_j s_k$

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


This needs considerable tedious hard slog to complete it.
In particular: To be proved: a) that the sequences $\tuple {a_0, 0_D, 0_d}$ form a subring $D'$ of $\struct {S, +, \circ}$ isomorphic to $D$, b) the sequence $\tuple {0_D, 1_D, 0_D, 0_D, \dotsc}$ is transcendental over $D'$, and c) that $D' \sqbrk X$ is the whole of $R$. Thus we have constructed $D' \simeq D$. If we now ignore the difference between $a_0 \in D$ and $\tuple {a_0, 0_D, 0_d} \in D'$ so that $D'$ is identified with $D$ the ring $D \sqbrk X$ has been constructed as required.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



Sources

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