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:

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

... 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:

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

... 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.