Rings of Polynomials in Ring Elements are Isomorphic

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

Let $D$ be an integral subdomain of both $R_1$ and $R_2$.

Let $X_1, X_2 \in R$ be transcendental over $D$.

Let $D \sqbrk {X_1}, D \sqbrk {X_2}$ be the rings of polynomials in $X_1$ and $X_2$ over $D$.

Then $D \sqbrk {X_1}$ is isomorphic to $D \sqbrk {X_2}$.

Proof
First it is shown that the mapping $\phi: D \sqbrk {X_1} \to D \sqbrk {X_2}$ given by:


 * $\displaystyle \map \phi {\sum_{k \mathop = 0}^n a_k \circ X_1^k} = \sum_{k \mathop = 0}^n a_k \circ X_2^k$

is a bijection.

Let $p, q \in \phi: D \sqbrk {X_1}$.

Suppose $\map \phi p = \map \phi q$.

Then the coefficients of $\map \phi p$ and $\map \phi q$ are equal, and $p_1 = q_1$.

Thus, by definition, $\phi$ is an injection.

By the same argument, the mapping $\psi: D \sqbrk {X_2} \to D \sqbrk {X_1}$ defined as:


 * $\displaystyle \map \psi {\sum_{k \mathop = 0}^n a_k \circ X_2^k} = \sum_{k \mathop = 0}^n a_k \circ X_1^k$

is similarly an injection.

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

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

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

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

We have:

Similarly for multiplication:

This completes the proof.

Consequences
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.