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Let $R=k$ be a field. Prove that every nonzero (prinicpal) ideal in $k[x]$ is generated by a unique monic polynomial.

I'm struggling to prove this result. Any help or suggestions is much appreciated.

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Hint: since nonzero multiples of a polynomial$~P$ necessarily have a degree at least $\deg(P)$, the monic polynomial generating your ideal$~I$ must have the lowest possible degree among elements of $I\setminus\{0\}$.

Show there is at least one monic polynomial among those lowest degree elements; then pick one, and call it $P$. Now for any $Q\in I$ you must show $P$ divides $Q$. To show this, do Euclidean division of $P$ by $Q$, and show that the remainder $R$ is necessarily zero. (This is where the choice of $P$ comes into play.)

Finally, it is clear that the multiples of $P$ with the same degree as$~P$ are just its scalar multiples. And among those, $P$ itself it the only monic one. This shows the uniqueness of the choice of$~P$.

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  • $\begingroup$ Do you have any hints on how to pick $P$? Thanks! $\endgroup$ – user428487 Sep 23 '17 at 2:44
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    $\begingroup$ @user428487 That depends on how the ideal is presented to you; you must for instance have your hands on some nonzero element of $I$ before you can really start, and that could already require some work (for instance if $I$ were defined as the ernel of some ring morphism). In case you do have a finite set of elements that is known to generate the ideal, you can apply Euclid's algorithm to compute their $\gcd$, which will be your$~P$. $\endgroup$ – Marc van Leeuwen Sep 23 '17 at 5:22

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