# How to prove the ring of Laurent polynomials over a field is a principal ideal domain?

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.

• I like the proof you included very much, except for the fact that I don't quite see why $j = x^{-\text{deg}^-(f')}f'$ for some $f' \in I$. i.e. I don't see why that $f'$ has to exists in $I$. – Jos van Nieuwman Mar 9 at 23:27
• I only see that $j$ is a linear combination of elements $f' \in I$, but that that doesn't mean that it's a multiple of any one such $f'$. If $I$ would be a principal ideal, then you could conclude this, but that is exactly what we're trying to prove. – Jos van Nieuwman Mar 10 at 0:12
• Come to think of it, it seems your key point in the segment: "Let $f \in (j)$. ... so $f$ itself is in $I$." is redundant. That is: you don't need $j$ to be of the form $X^{-\deg^-(f')}f'$ for $f' \in I$ to still have $j \in I$. It already follows from the fact that $j$ is a linear combination of $f' \in I$ with coefficients in $F[X,X^{-1}]$. – Jos van Nieuwman Mar 11 at 21:37
• @JosvanNieuwman The OP made a small mistake when assumed $j = x^{-\deg^-(f')}f'$ for some $f' \in I$, but $j$ is a linear combination of such things with coefficients in $F[X]$ and this immediately leads to $j\in I$. – user26857 Mar 11 at 21:50

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.

• From the linked question: "the ideal $I$ has the form $S^{−1}J$ with $J$ an ideal of $A$". I'm having trouble seeing why this is. – Alec Dec 3 '13 at 0:17
• Second answer seems to me pretty clear. Tell me what exactly you don't understand. – user52045 Dec 3 '13 at 0:21
• The thing I'm talking about is the very first sentence of the second answer. It's stated as if it's obvious but I'm not seeing it... – Alec Dec 3 '13 at 0:29
• If you have ideal $I$ and you want to find ideal $J$ take generators of $I=<a_i/s_i>_i$ and then $J=<a_i>$. – user52045 Dec 3 '13 at 0:35

1st HINT: Given an ideal $I$ of $F[x,x^{-1}]$, what can you say about $I\cap F[x]$? How is the information you obtain related to $I$?