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Be $F$ a field and let $I$ be any ideal in $F[X,X^{-1}]$. For any $f = \sum_{n\in \mathbb{Z}}a_nX^n \in F[X,X^{-1}]$, define deg$^-(f) := \min\{n \in \mathbb{Z} \mid a_n \neq 0\}$. Consider the set $\tilde{J} := \{X^{-\deg^-(f)}f \mid f \in I\}$ and let $J$ be the ideal generated by $\tilde{J}$ in $F[X]$. As $F[X]$ is a PID, we can let $j \in F[X]$ be such that $J = (j) = jF[X]$. I want to show that $jF[X,X^{-1}] \subseteq I$. For any $f \in jF[X,X^{-1}]$ there is a $g \in F[X,X^{-1}]$ such that $f = gj$. Now if $j$ were to be of the form $X^{-\deg^-(f')}f'$ for some $f' \in I$, we would be done. However, I can only see $j$ as a linear combination of elements of $I$ (with coefficients in $F[X]$), but that doesn't make it a multiple of any one such element.

user26857
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Jos van Nieuwman
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  • Since $j\in J$ we may write $j=\sum_{i=1}^nX^{-\deg^{-}(f_i)}f_ig_i$ with $f_i\in I$ and $g_i\in F[X]$. Now observe that $j\in I$. (I'm not sure why you consider $jg$ instead of $j$. As far as I know if the generator(s) of an ideal belong)s( to another ideal, then you get an inclusion.) – user26857 Mar 11 '19 at 18:29
  • If I am not mistaken, you want to show that $I=jF[X,X^{-1}]$. – user26857 Mar 11 '19 at 18:30
  • Correct, I've come to think that the person in this proof: https://math.stackexchange.com/questions/590219/how-to-prove-the-ring-of-laurent-polynomials-over-a-field-is-a-principal-ideal-d (see "Edit 2", the part beginning with "Let $f \in (j)$") does unnecessary things to conclude that $f \in I$. – Jos van Nieuwman Mar 11 '19 at 20:38

1 Answers1

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$F[X,X^{-1}]$ is the localisation $F[X]_X$, and hence is a P.I.D. Let $f_0$ a generator of the ideal $I$.

Claim: a generator of $J$ in $F[X]$ is $$j=X^{-\deg^{-}(f_0)}f_0. $$

Indeed, any Laurent polynomial $f$ in $I$ can be written as $\;f=f_0\,g,\enspace g\in F[X,X^{-1}]$. Note that $\;\deg^{-}(f)=\deg^{-}(f_0)+\deg^{-}(g) $, so that $$X^{-\deg^{-}(f)}f=X^{-\deg^{-}(f_0)}f_0\,X^{-\deg^{-}(g)}g(X^{-\deg^{-}(g)}g)j.$$

Jos van Nieuwman
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Bernard
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  • I apologise for my incompleteness. This whole exercise aims to prove that $F[X, X^{-1}]$ is a PID, so can most certainly not use that fact.. In fact, the question arose from reading this related post: https://math.stackexchange.com/questions/590219/how-to-prove-the-ring-of-laurent-polynomials-over-a-field-is-a-principal-ideal-d – Jos van Nieuwman Mar 10 '19 at 17:03
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    But this results from a far more general result: the localisation of a P.I.D. is itself a P.I.D. (that's why I wrote ‘hence’ in my answer). – Bernard Mar 10 '19 at 17:59
  • And without using PIDs $X$ is a unit so $(X^b f, \mathfrak{a}) = (f,\mathfrak{a})$ and $J = \overline{J}$ @JosvanNieuwman – reuns Mar 10 '19 at 19:47
  • Yes, absolutely. – Bernard Mar 10 '19 at 20:11
  • @Bernard Localisation is beyond the scope of my (first introduction to ring theory) course. – Jos van Nieuwman Mar 10 '19 at 21:47
  • @reuns Thanks, I totally forgot $X^{-1}$ was a unit here. By $\mathfrak{a}$, do you mean any element or subset of $F[X,X^{-1}]$?. By $\bar{J}$ do you mean $\tilde{J}$ or the set generated by $J$? – Jos van Nieuwman Mar 10 '19 at 21:47
  • The Fraktur font is the usual font for ideals, especially when there's a risk to confuse with elements of a ring. By $J$, I mean like you: the ideal generated by $\tilde J$. – Bernard Mar 10 '19 at 21:59
  • You speak of $\bar{J}$ though. Also, do you then mean that $(f, \mathfrak{a})$ is "the ideal generated by the element $f$ and the ideal $\mathfrak{a}$"? I'm not sure what that means... Is it "the ideal generated by the set $\mathfrak{a} \cup {f}$"? – Jos van Nieuwman Mar 10 '19 at 22:02
  • On second thoughts, this doesn't seem to get me that much closer. Are you sure this tells me that $j$ is expressible as a single multiple $X^{-\deg^-(f)}f$ for some $f \in I$? – Jos van Nieuwman Mar 10 '19 at 22:15
  • Previous question: $(f,\mathfrak a)$denotes the sum of the ideal generated by $f$ and the ideal $\mathfrak a$. Last question: don't you agree that $j$ and $f_0$ generate the same ideal in $F[X,X^{-1}]$? That's the main point of my suggestion – Bernard Mar 10 '19 at 22:36
  • @JosvanNieuwman: Are you supposed to know that $K[X,X^{-1}]$ is a noetherian ring? – Bernard Mar 10 '19 at 23:50
  • Let us continue this discussion in chat. – Jos van Nieuwman Mar 11 '19 at 00:07