Hi, today we will discuss something which a while ago when proving $\Bbb[Z]$ is UFD. I was confused why it was “obvious” that $\Bbb{Q}[x]$ is a UFD.

Contents

Proving it right away!

$K$ is a field $\iff$ $K[x]$ is a PID
$\implies$ Let $K$ be a field. Let $I\unlhd K[x]$. Let $p(x)\in I$ be a polynomial with the smallest degree.
If $g(x)\in I\implies p(x)|g(x)$.
If not then $g(x)=p(x)q(x)+r(x),\text{ deg }r(x)<\text{ deg }p(x)$. But that implies $r(x)\in I$. Not possible.
$\impliedby$ Given $K[x]$ is a PID. We need to show $K$ is a field. Say we want to show that $u$ is a unit. Consider $\langle u,x \rangle$ which is ideal of $K[x]$. Note that by PID, $\exists f$ such that $\langle f\rangle=\langle u,x\rangle$. Now, clearly $f$ is constant by degree comparison. If $f$ is a unit then we are done. If not, then $\alpha(f)|\alpha(d)\forall d\in \langle f \rangle$. But $\alpha(f)=1$.

Some remarks

  • I wanted to write a post on it to show the importance and the power this theorem holds. Essentially many of the lemmas I stated in the blog post ED implies PID implies UFD, such as, if $D$ is PID, then $p$ is is irreducible iff $\langle p\rangle $ is a maximal ideal can be used when $D$ is $\Bbb{Q}[x]$ or any other PID.

  • Also, we will deal with the field of fractions soon, so a quick result!

If $D$ is ID$\implies$ $Q(D)$ is a field $\implies Q(D)[x]$ is a PID $\implies Q(D)[x]$ is UFD.

And we are done!