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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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