If $F$ is a field, then $F[x]$ is a principal ideal domain proof question

Solution 1:

This is bases on the Well-Ordering Principle: Any nonempty subset of $\mathbb{Z}$ that is bounded below has a smallest element.

In this context, we assume that the ideal $I$ is nonzero, and let $I^+=\{\deg f\mid f\in I\backslash \{0\}\}$. Then, $I^+$ is a nonempty subset of $\mathbb{Z}$ that is bounded below. Hence, $I^+$ has a minimal element which corresponds to a nonzero element of $g\in I$ of minimal degree.

Now, you clearly have $(g)\subset I$. For the converse, take $f\in I$ and use the division algorithm to write $$f=gq+r$$ where $\deg r<\deg g$. Since $r=f-gq$, it follows that $r\in I$. If $r\neq 0$, then $\deg r\in I^+$ and $\deg r<\deg g$. This contradicts the minimality of $\deg g$.