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?

  • $\begingroup$ 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. $\endgroup$
    – user144221
    Commented Sep 23, 2016 at 0:42

1 Answer 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 $\deg g$.

  • 1
    $\begingroup$ How does $r = f - gq \Rightarrow r \in I$? $\endgroup$
    – Oliver G
    Commented Sep 22, 2016 at 18:50
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    $\begingroup$ 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 $f-gq\in I$. $\endgroup$
    – Kevin Long
    Commented Sep 22, 2016 at 18:51
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    $\begingroup$ Where did we use that F is a Field? $\endgroup$ Commented Oct 27, 2022 at 13:50
  • $\begingroup$ @GuilhermeCosta a bit late, but we assume that gx is monic in order to guarantee r unique, which we can do since F is a field. $\endgroup$
    – emesupap
    Commented Oct 29, 2023 at 17:10
  • $\begingroup$ I was hoping you could clear this up for me, doesn't assuming that the ring has a division algorithm pre-suppose that it is a principal ideal domain? My understanding is that in order to have a division algorithm, and ring must be a Euclidean Domain? By my lights, once you prove that it is a Euclidean domain, then you are a done, since all Euclidean domains are principal ideal domains. I think @user144221 mentioned this, but I wanted to be sure I understood him correctly. $\endgroup$ Commented Nov 23, 2023 at 21:43

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