Polynomial Forms over Field form Principal Ideal Domain/Proof 1

Proof
For any $d \in F \sqbrk X$, let $\ideal d$ denote the principal ideal of $F \sqbrk X$ generated by $d$.

Let $J$ be any ideal of $F \sqbrk X$. What we need to prove is that $J$ is a principal ideal.

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


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


 * If $J = F \sqbrk X$, then by Ideal of Unit is Whole Ring: Corollary $J = \ideal {1_F}$, and hence is a principal ideal.

Now suppose $J \ne \set {0_F}$ and $J \ne F \sqbrk X$.

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, $J = F \sqbrk X$.

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 Division Theorem for Polynomial Forms over Field, $f = q \circ d + r$ for some $q, r \in F \sqbrk X$ 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:
 * $f \in \ideal d$.

This reasoning shows that:
 * $J \subseteq \ideal d$

From property $(3)$ of the principal ideal $\ideal d$, we conclude that:
 * $\ideal d \subseteq J$

as $d \in J$.

Hence $J = \ideal d$.

These $2$ distinguished cases cover all of the possible ideals of $F \sqbrk X$.

Hence $F \sqbrk X$ is a principal ideal domain.