Why any field is a principal ideal domain?

According to the definition of P.I.D, first, a ring's ideal can be generated from a single element; second, this ring has no zero-divisor. This two conditions make a ring P.I.D.

But how to prove any field is P.I.D?


1 Answer 1


Let $F$ be a field and $I \subset F$ be a nontrivial ideal. Then if $a \in I$ is nonzero, we have that $1 = a^{-1} \cdot a \in I$, where $a^{-1}$ exists since $F$ is a field and $a \neq 0$. Since $1 \in I$, for every element $b \in F$, $b = b \cdot 1 \in I$, so we have that $I = F = \langle 1 \rangle$ if $I \neq \{0\}$.

In conclusion, the only ideals of a field $F$ are $\langle 0 \rangle = \{0\}$ and $\langle 1 \rangle = F$, which are both principal ideals.

  • $\begingroup$ Clear proof, so the key point which makes $F$ P.I.D, is that it has an element $1$, right? $\endgroup$
    – avocado
    Jun 9, 2013 at 1:36
  • 3
    $\begingroup$ The key point is that every nonzero element has a multiplicative inverse. The same proof works for division rings (rings where each element has a multiplicative inverse but multiplication is not necessarily commutative). Usually rings are taken to have a multiplicative identity $1$, but there are many examples of (unital) rings that are not PIDs (e.g. $\Bbb Z[x]$). $\endgroup$ Jun 9, 2013 at 1:42
  • 1
    $\begingroup$ But note that you can have PIDs that are not fields (e.g. $\Bbb Z$) and also you can have rings with no nontrivial proper ideals that are not fields (e.g. $M_{2 \times 2}(\Bbb R)$); the "key point" mentioned here is just for this specific proof. $\endgroup$ Jun 9, 2013 at 1:46
  • $\begingroup$ Good to know. If $p$ is not a prime, $interger \pmod p$ doesn't form a field, but does it form a P.I.D? $\endgroup$
    – avocado
    Jun 9, 2013 at 1:52
  • 1
    $\begingroup$ When $m$ is not a prime, then $\Bbb Z/m$ is not a PID because it isn't even an integral domain. Since $m$ is not prime, $m = qr$ for some integers $q, r \neq 1$, and hence $qr = 0 \pmod m$, implying that $q, r$ are zero divisors in $\Bbb Z/m$. $\endgroup$ Jun 9, 2013 at 1:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.