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Proposition :

Every ideal in k[x] (polynomial ring) is a principal ideal

Proof :

Suppose that $I\subseteq K[x]$. Take $p(x)\in I$ such that $p(x)$ is monic and $deg(p(x))$ is minimal over all polinomials of positive degree.

Take any $f(x)\in I$. We can say that $f(x)=q(x)p(x)+r(x)$. Then, $deg(r(x))=0$, that is, $r(x)$ is zero or a constant polynomial.

In the first case, $f(x)\in (p(x))$, so $(p(x))=I$. It is worth mentioning that $(p(x))$ is the set of all multiples of $p(x)$.

In the second case, $r(x)=\alpha \neq 0\in K[x]$. How can I finish this case?

Sergio Durán Vega
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    One way to solve it is to instead choose p to be of minimal degree among all nonzero polynomials of I. Then $\alpha$ cannot be constant. If you want to continue with the current approach, think about what ideal is generated by $\alpha$. – CrackedBauxite May 19 '22 at 14:39
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    If both $f$ and $p$ are in $I$, is $f-qp$ in $I$? – Arturo Magidin May 19 '22 at 14:40

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