I know that a polynomial ring over a field is a PID, does this property also hold for a polynomial ring over a skew field? Is there maybe something else that characterise the ideals in that ring ?
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See http://math.stackexchange.com/questions/476456/ideals-in-the-polynomial-ring-over-a-division-ring-are-free – user26857 Oct 27 '14 at 11:19
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Thank you, the link was usefull to me, the problem is solved now! – Николай Рыжков Oct 28 '14 at 18:32
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The argument for $F[x]$ being a PID for a field F carries over to the case with F a division ring.
Given a right ideal $T$, pick a monic elements minimal degree $g$ in $T$. Given a $y\in T$, you can still use a division algorithm (bring careful to multiply on the right) to find q,r such that $y=gq+r$ with the degree of r lower than g. Then you conclude r=0 and see that g generates $T$.
The left ideals are principal as well, and so are the ideals, being special cases.
rschwieb
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