i want to show that $k[x,x^{-1}]$ (where $k$ is a field ) is a PID . i have some proof but seems too- straight forward to be correct take the natural homomorphism $\phi: K[x]\to S^{-1}K[x]$ where $S=(x^1,x^2...)$ for we know that for every ideal of $S^{-1}K(x)$ Lets say $I$ , $\phi^{-1}(I)$ is an ideal of $K[x]$ Which is PID so $\phi^{-1}(I)=(f)$ for some $f$ on $K[x]$ then is holds that the ideal $I$ is generated by $(f/1)$ ? if yes we've shown that $k[x,x^{-1}]$ is a PID?
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1Looks good to me. – George Giapitzakis Apr 08 '21 at 13:47
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Hi, welcome to Math.SE, I edited your post so that it uses MathJax. See here if you need some references : https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Roland Apr 08 '21 at 13:48
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6In general, any localization of a PID is itself a PID. – Geoffrey Trang Apr 08 '21 at 13:48
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@GeoffreyTrang yes ive read that online but i was not quite sure about the proof i gave above – Chris Kan Apr 08 '21 at 13:52
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It follows also from this duplicate. – Dietrich Burde Apr 08 '21 at 14:15