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

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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:

\(\ds \map \phi {P_1 \times P_2}\) \(=\) \(\ds a_0 \times b_0\)
\(\ds \) \(=\) \(\ds \map \phi {P_1} \times \map \phi {P_2}\)

and:

\(\ds \map \phi {P_1 + P_2}\) \(=\) \(\ds a_0 + b_0\)
\(\ds \) \(=\) \(\ds \map \phi {P_1} + \map \phi {P_2}\)


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


The First Ring Isomorphism Theorem can then be used.


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Sources

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