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Let $k$ be a division ring. I want to show that every (right) ideal in $k[x]$ is free considered as a right $k[x]$-module.

That means if $I$ is an ideal in $k[x]$ we have to show that $I=f\cdot k[x]$ for an $f\in I$. My first intention was to use the divison algorithm, but as $k$ is just a division ring I guess it won't work?

Maybe somebody can help me with this...

Thanks!

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    $\begingroup$ Don't guess that it won't work, try it out and figure out if it does work or not. If it doesn't work you'll find out why it doesn't work along the way. $\endgroup$ – rschwieb Aug 26 '13 at 12:55
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    $\begingroup$ +1 to rschwieb's comment and answer for giving you the key piece of advice. You are right in that there are caveats. For further encouragement and bits about the caveats I recommend that you take a look at this answer by Arturo Magidin. $\endgroup$ – Jyrki Lahtonen Aug 26 '13 at 13:21
  • $\begingroup$ @JyrkiLahtonen Thanks for that interesting link. $\endgroup$ – Heffalump Aug 26 '13 at 13:40
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Hint: You're on the right track since you should be able to show that $k[x]$ is a principal right ideal domain. You can mimic the proof for fields that $F[x]$ is a PID, which you probably have seen by now.

Once you show all the right ideals of $k[x]$ are principal, consider such an ideal $aR$. If $aR\neq 0$, then of course $a\neq 0$, and there is a very suggestive mapping here from $R\to aR$ that you should investigate.

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  • $\begingroup$ Sorry for guessing and thank you for that hint! But there is no difference between showing that $k[x]$ ist a principal right ideal domain and showing that all ideals are free modules, right? $\endgroup$ – Heffalump Aug 26 '13 at 13:37
  • $\begingroup$ @Heffalump For commutative rings, it's true that if all right ideals are free, then you are looking at a PID. BUT for noncommutative rings, there exist rings whose right ideals are all free, but they are not principal right ideal domains. This wiki page talks a little about such rings, but don't worry about learning too much there :) $\endgroup$ – rschwieb Aug 26 '13 at 13:39
  • $\begingroup$ Or is the fact that for $R=k[x]$ and $a\neq 0$ the map $R\rightarrow aR$ is an isomorphism important for knowing that the module is free? In other words, are there principle ideals which aren't free considered as a module? for example in a ring which has zero-divisors? $\endgroup$ – Heffalump Aug 26 '13 at 13:49
  • $\begingroup$ I wrote my last comment after reading yours... $\endgroup$ – Heffalump Aug 26 '13 at 13:51
  • $\begingroup$ @Heffalump For any domain with nonzero $a$, $R\cong aR$, and indeed, all principal right ideals are free. Principal right ideal domains certainly have all right ideals free, then. I'm just saying that there is a difference between the two, since there are free ideal rings that aren't principal right ideal rings. $\endgroup$ – rschwieb Aug 26 '13 at 14:20

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