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 $D \left[{X_1}\right], D \left[{X_2}\right]$ 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 $p, q \in \phi: D \left[{X_1}\right]$.

Suppose $\phi \left({p}\right) = \phi \left({q}\right)$.

Then the coefficients of $\phi \left({p}\right)$ and $\phi \left({q}\right)$ are equal, and $p_1 = q_1$.

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.

It remains to show that $\phi$ is a ring homomorphism.

Let $\displaystyle p = \sum_{k=0}^n {a_k \circ {X_1}^k}, q = \sum_{k=0}^n {b_k \circ {X_2}^k} \in D[X_1]$.

For convenience we set $a_k = 0$, $k > n$ and $b_k = 0$, $k > m$. We have:

Similarly for multiplication:

This completes the proof.

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.