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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.

user26857
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alecbz
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  • 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 09 '19 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 '19 at 00:12
  • 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 '19 at 21:37
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    @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 '19 at 21:50
  • https://math.stackexchange.com/questions/536624 – user26857 Apr 28 '21 at 07:59

2 Answers2

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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.

Community
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user52045
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  • 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. – alecbz Dec 03 '13 at 00:17
  • Second answer seems to me pretty clear. Tell me what exactly you don't understand. – user52045 Dec 03 '13 at 00:21
  • 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... – alecbz Dec 03 '13 at 00:29
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    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 03 '13 at 00:35
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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 $I$?

Josué Tonelli-Cueto
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