[Math] How to prove the ring of Laurent polynomials over a field is a principal ideal domain

abstract-algebraring-theory

Disclaimer: this is a homework question. I'm looking for direction, not an answer.

Given a field $F$, show that $F[x,x^{-1}]$ is a principal ideal domain.

I'm unsure how to proceed. Would it be better to prove this directly? (ie, let $I$ be an ideal, show that $I = (f)$ for some $f \in F[x,x^{-1}]$). The proof that polynomials are a PID would involve (I imagine) division by remainder and use of degree, both of which don't seem to have obvious parallels for Laurent polynomials. Should I try and devise some parallels and mimic the proof for polynomials? Or is this overkill? (or just wrong?)

edit 1: I guess the approach I mentioned above amounts to showing that Laurent polynomials are a euclidean domain (we know that euclidean domain => principle ideal domain, so this would be sufficient) Are they a euclidean domain though? (it seems like this would have been the question if they were, instead of asking if they were a PID).

edit 2: I've spat this out, think most parts of it are correct, though it seems kind of ugly/cumbersome (but that might just be me trying to spell things out more than is needed):

Given a Laurent polynomial $f \in F[x,x^{-1}]$, define its "negative
degree" $\deg^-(f)$ to be the largest power of $x^{-1}$ that appears
in $f$.

Let $I$ be an ideal of $F[x,x^{-1}]$. Note that $\{x^{-\deg^-(f)}f \mid f \in I\} \subseteq F[x]$. Let $J$ be the
ideal in $F[x]$ generated by this set. $F[x]$ is a principal ideal
domain, so $J$ is a principal ideal and we have $J = (j)$ for some $j \in F[x]$.

We claim $I = (j)$ (now meaning an ideal of $F[x,x^{-1}]$).

Let $f \in I$. Then $x^{-\deg^-(f)}f \in J$, meaning $f = x^{\deg^-(f)}g$, where $g = x^{-\deg^-(f)}f$ is in $J$. Because $g \in
J = (j)$, there exists $g' \in F[x]$ such that $g = g'j$, and thus $f = (x^{\deg^-(f)}g')j$ is a multiple of $j$, so $f \in (j)$.

Let $f \in (j)$. Then $f = gj$ for some $g \in F[x,x^{-1}]$. But note
that $j = x^{-\deg^-(f')}f'$ for some $f' \in I$, so $f = gx^{-\deg^-(f')}f'$, so $f$ is a multiple of an element of an ideal
$I$, so $f$ itself is in $I$.

This shows $I = (j)$. So an arbitrary ideal of $F[x,x^{-1}]$ is
principal, so $F[x,x^{-1}]$ is a principal ideal domain.

I kind of feel like I still don't "get" the proof (I more-or-less see how each part works with the others but I'm having trouble seeing the bigger picture), though this may be due to a poor handle on ideals in general.

Best Answer

Polynomial ring over a field $k$ is a PID. Notice that $S^{-1}k[x]=k[x,x^{-1}]$, where $S=\{x^i:i\in \mathbb N\}$. Now use the fact that localization of a PID is a PID.

Related Question