Rings of Polynomials in Ring Elements are Isomorphic

Theorem
Let $R_1, R_2$ be commutative rings with unity.

Let $D$ be an integral domain such that $D$ is a subring of both $R_1$ and $R_2$.

Let $X_1 \in R_1, X_2 \in R_2$ each be transcendental over $D$.

Let $D \left[{X_1}\right], D \left[{X_2}\right]$ be the rings of polynomial forms in $X_1$ and $X_2$ over $D$.

Then $D \left[{X_1}\right]$ is isomorphic to $D \left[{X_2}\right]$.

Proof

 * First we need to show that the mapping $\phi: D \left[{X_1}\right] \to D \left[{X_2}\right]$ given by:


 * $\displaystyle \phi \left({\sum_{k=0}^n {a_k \circ {X_1}^k}}\right) = \sum_{k=0}^n {a_k \circ {X_2}^k}$

... is a bijection.

Let $p_1, q_1 \in \phi: D \left[{X_1}\right]$.

Suppose $\phi \left({p_1}\right) = \phi \left({q_1}\right)$.

Then as $X_2$ is transcendental over $D$, it follows from the definition that $p_1 = q_1$, and thus $\phi$ is an injection.

By the same argument, the mapping $\psi: D \left[{X_2}\right] \to D \left[{X_1}\right]$ defined as:


 * $\displaystyle \psi \left({\sum_{k=0}^n {a_k \circ {X_2}^k}}\right) = \sum_{k=0}^n {a_k \circ {X_1}^k}$

... is similarly injective.

Thus by Injection is Bijection iff Inverse is Injection, $\phi$ is a bijection.


 * Then we need to show that $\phi$ is a ring homomorphism.

Comment
Thus we see that the ring in which an integral domain is embedded is (to a certain extent) irrelevant -- it is only the integral domain itself which is important.