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...