K Is A Field If And Only If K[x] Is A Pid

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

Contents

• Proving it right away!
• Some remarks

Proving it right away!

$K$ is a field $⟺$ $K[x]$ is a PID

$⟹$ Let $K$ be a field. Let $I⊴K[x]$. Let $p(x)\in I$ be a polynomial with the smallest degree.

If $g(x)\in I⟹p(x)|g(x)$.

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.

$⟸$ 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 $⟨u,x⟩$ which is ideal of $K[x]$. Note that by PID, $\mathrm{\exists }f$ such that $⟨f⟩=⟨u,x⟩$. Now, clearly $f$ is constant by degree comparison. If $f$ is a unit then we are done. If not, then $\alpha (f)|\alpha (d)\mathrm{\forall }d\in ⟨f⟩$. 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 $⟨p⟩$ is a maximal ideal can be used when $D$ is $\mathbb{Q}[x]$ or any other PID.

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

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

And we are done!