7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $I$. $\endgroup$ – Jos van Nieuwman Mar 9 at 23:27
  • $\begingroup$ 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. $\endgroup$ – Jos van Nieuwman Mar 10 at 0:12
  • $\begingroup$ 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 $F[X,X^{-1}]$. $\endgroup$ – Jos van Nieuwman Mar 11 at 21:37
  • 1
    $\begingroup$ @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 $j\in I$. $\endgroup$ – user26857 Mar 11 at 21:50
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Alec Dec 3 '13 at 0:17
  • $\begingroup$ Second answer seems to me pretty clear. Tell me what exactly you don't understand. $\endgroup$ – user52045 Dec 3 '13 at 0:21
  • $\begingroup$ 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... $\endgroup$ – Alec Dec 3 '13 at 0:29
  • 1
    $\begingroup$ If you have ideal $I$ and you want to find ideal $J$ take generators of $I=<a_i/s_i>_i$ and then $J=<a_i>$. $\endgroup$ – user52045 Dec 3 '13 at 0:35
2
$\begingroup$

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 $I$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.