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


 * If $$J = \left\{{0_F}\right\}$$, then $$J = \left({0_F}\right)$$, which is a principal ideal.


 * If $$J = F \left[{X}\right]$$, then $$J = \left({1_F}\right)$$, which is a principal ideal.


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

Thus $$J$$ contains a non-zero member.

By the well-ordering principle, we can introduce the lowest degree of a non-zero member of $$J$$ -- call it $$n$$.

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, $$J = F \left[{X}\right]$$.

So $$n \ge 1$$.


 * 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 $$< n$$.

Because $$J$$ is an ideal to which $$d$$ belongs, $$q \circ d \in J$$.

Since $$f \in J$$, $$f - q \circ d \in J$$, i.e. $$r \in J$$.

No non-zero member of $$J$$ has degree less than $$n$$, which leads us to the fact that $$r = 0_F$$.

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


 * Thus it follows that $$J \subseteq d$$.

But since $$J$$ is an ideal to which $$\left({d}\right)$$ belongs, $$\left({d}\right) = J$$.

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


 * Therefore $$J$$ is a principal ideal of $$F \left[{X}\right]$$, and as $$J$$ is arbitrary, every ideal of $$F \left[{X}\right]$$ is a principal ideal.

Therefore, by Ring of Polynomial Forms is Integral Domain, $$F \left[{X}\right]$$ is a principal ideal domain.

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

Proof of Corollary 2
Follows from the main result, by Principal Ideal of Irreducible Element and others.