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

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

By a previous theorem, we know that $F[x]$ is an integral domain. Now, let $I$ be an ideal in $F[x]$. If $I = \{0\}$, then $I = \langle 0 \rangle$. If $I \neq \{0\}$, then among all the elements of $I$, let $g(x)$ be one of minimum degree. We will show that $I = \langle g(x) \rangle$. Since $g(x) \in I$, we have $\langle g(x) \rangle \subseteq I$. Now let $f(x) \in I$. Then, by the division algorithm, we may write $f(x) = g(x)q(x) + r(x)$, where $r(x) =0$ or deg $r(x) \lt$ deg $g(x)$. Since $r(x) = f(x) - g(x)q(x) \in I$, the minimality of deg $g(x)$ implies that the latter condition cannot hold. So, $r(x) = 0$ and, therefore, $f(x) \in \langle g(x) \rangle$. This shows that $I \subseteq g(x)$.

My question is: What is a polynomial of minimum degree in this context? And how does this minimality imply that deg $r(x) \lt$ deg $g(x)$ cannot hold?

• A side remark: what you use is the Euclidean algorithm, and the same reasoning shows that any Euclidean domain is a PID. For instance, you would use the same argument to show that Z is a PID.
– user144221
Commented Sep 23, 2016 at 0:42

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 $\deg g$.
• How does $r = f - gq \Rightarrow r \in I$? Commented Sep 22, 2016 at 18:50
• Ideals are closed under addition, and any multiple of something in an ideal is in the ideal (by definition). $f\in I$ and $g\in I$, so $gq\in I$, which means that $f-gq\in I$. Commented Sep 22, 2016 at 18:51