Polynomial Forms over a Field is Principal Ideal Domain
Contents |
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.
$\blacksquare$
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.
$\blacksquare$
Proof of Corollary 1
Follows from Principal Ideal Domain is Unique Factorization Domain.
$\blacksquare$
Proof of Corollary 2
It follows from Principal Ideal of Irreducible Element that $\left({f}\right)$ is maximal for irreducible $f$.
Therefore by Maximal Ideal iff Quotient Ring is Field, $F \left[{X}\right] / \left({f}\right)$ is a field.
$\blacksquare$
Converse
A converse to this result is given by Polynomial Forms is PID Implies Coefficient Ring is Field.
Sources
- C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.25$
- C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.29$: Theorem $57$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous): $\S 65.2, \S 65.3$
