Ring of Polynomial Forms Always Exists

Theorem

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

Let $\left({D, +, \circ}\right)$ be an integral domain such that $D$ is a subring of $R$ whose zero is $0_D$ and whose unity is $1_D$.

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

Let $D \left[{X}\right]$ be the ring of polynomial forms in $X$ over $D$.

Then a ring of polynomial forms in $X$ over $D$ always exists.

Outline of Proof

• When $D \left[{X}\right]$ does exist, the $\displaystyle \sum_{k=0}^n {a_k \circ {X_1}^k}$ corresponds to the coefficient sequence $\left({a_0, a_1, a_2, \ldots, a_n, 0_D, 0_D, 0_D, \ldots}\right)$ and is completely described by it.

• On the other hand, to construct $D \left[{X}\right]$ we take the set $R$ of all infinite sequences of elements of $D$ that have the property of being zero from some point on, and define $+$ and $\circ$ on $R$ so as to mimic the way the coefficient sequences of polynomials behave in $+$ and $\circ$.

• We then prove:

1. The system $\left({R, +, \circ}\right)$ is now a ring;
2. The sequences $\left({a_0, 0_D, 0_D, 0_D, \ldots}\right)$ form a subring $\hat {D}$ of $R$ which is isomorphic to $D$;
3. The sequence $X = \left({0_D, 1_D, 0_D, 0_D, \ldots}\right)$ is transcendental over $\hat {D}$;
4. $\hat {D} \left[{X}\right]$ is the whole of $R$.

The conclusion is that we have constructed a ring $\hat {D} \left[{X}\right]$ such that $\hat {D} \cong D$.

• If we now agree to ignore the difference between the element $a_0 \in D$ and the corresponding $\left({a_0, 0_D, 0_D, 0_D, \ldots}\right) \in \hat D$, we can identify $\hat D$ with $D$.

Thus the ring we have created is $D \left[{X}\right]$ as required.