Polynomial Forms over Field form Principal Ideal Domain

Theorem
Let $\left({F, +, \circ}\right)$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $X$ be transcendental in $F$.

Let $F \left[{X}\right]$ be the ring of polynomial forms in $X$ over $F$.

Then $F \left[{X}\right]$ is a principal ideal domain.

Corollary 1
$F \left[{X}\right]$ is a unique factorization domain.

Corollary 2
If $f$ is an irreducible element of $F \left[{X}\right]$, then $F \left[{X}\right] / \left({f}\right)$ is a field.

Proof 1
Notation: for any $d \in F \left[{X}\right]$, let $\left({d}\right)$ be the principal ideal of $F \left[{X}\right]$ generated by $d$.

Let $J$ be any ideal of $F \left[{X}\right]$. What we need to prove is that $J$ is a principal ideal.

Let us first distinguish the following two cases for $J$:


 * If $J = \left\{{0_F}\right\}$, then by Zero Element Generates Null Ideal $J = \left({0_F}\right)$, and hence is a principal ideal.


 * If $J = F \left[{X}\right]$, then by Ideal of Unit is Whole Ring $J = \left({1_F}\right)$, and hence is a principal ideal.

Now suppose $J \ne \left\{{0_F}\right\}$ and $J \ne F \left[{X}\right]$.

Then $J$ necessarily contains a non-zero element.

By the well-ordering principle, we can introduce the lowest degree of a non-zero element of $J$. Denote this degree by $n$.

If $n = 0$, then $J$ contains a polynomial of degree $0$. This is a non-zero element of $F$.

As $F$ is a field, this is therefore a unit of $F$, and thus by Ideal of Unit is Whole Ring, $J = F \left[{X}\right]$.

Because the degree of a non-zero element is a natural number, we conclude that $n \ge 1$.

Now let $d$ be a polynomial of degree $n$ in $J$, and let $f \in J$.

By the Division Theorem for Polynomial Forms over a Field, $f = q \circ d + r$ for some $q, r \in F \left[{X}\right]$ where either $r = 0_F$ or $r$ is a polynomial of degree smaller than $n$.

Because $J$ is an ideal and $d \in J$, it follows that $q \circ d \in J$.

Since $f \in J$, we also conclude $r = f - q \circ d \in J$.

From the construction of $d$, it follows that we must have $r = 0_F$.

Therefore $f = q \circ d$, and thus $f \in \left({d}\right)$.

This reasoning shows that $J \subseteq \left({d}\right)$.

From property $(3)$ of the principal ideal $\left({d}\right)$, we conclude that $\left({d}\right) \subseteq J$ as $d \in J$.

Hence $J = \left({d}\right)$.

The distinguished cases cover all of the possible ideals of $F \left[{X}\right]$.

Hence $F \left[{X}\right]$ is a principal ideal domain.

Proof 2
We have that Polynomial Forms over Field is Euclidean Domain.

We also have that Euclidean Domain is Principal Ideal Domain.

Hence the result.

Proof of Corollary 1
Follows from Principal Ideal Domain is Unique Factorization Domain.

Proof of Corollary 2
It follows from Principal Ideal of Irreducible Element that $(f)$ is maximal for irreducible $f$.

Therefore by Maximal Ideal iff Quotient Ring is Field, $F[X]/(f)$ is a field.

Converse
A converse to this result is given by Polynomial Forms is PID Implies Coefficient Ring is Field.