Quotient of Ring of Polynomials in Ring Element on Integral Domain by that Polynomial is that Domain

Theorem
Let $\struct {D, +, \times}$ be an integral domain.

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

Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.

Let $D \sqbrk X / \ideal X$ denote the quotient ring of $D \sqbrk X$ by the ideal of $D$ generated by $X$.

Then:
 * $D \sqbrk X / \ideal X \cong D$

Proof
Let $n \in \Z_{> 0}$ be arbitrary.

Let $P = a_n X^n + a_{n - 1} X^{n - 1} + \dotsb + a_1 X + a_0$ be a polynomial over $D$ in $X$.

Consider the mapping $\phi: D \sqbrk X \to D$ defined as:
 * $\forall P \in D \sqbrk X: \map \phi P = a_0$

Let:
 * $P_1 = a_m X^m + a_{m - 1} X^{m - 1} + \dotsb + a_1 X + a_0$


 * $P_2 = b_m X^m + b_{m - 1} X^{m - 1} + \dotsb + b_1 X + b_0$

We have that:

and:

Thus it is seen that $\phi$ is a homomorphism.

The First Isomorphism Theorem for Rings can then be used.