Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, can we conclude that R is a PID?

  • $\begingroup$ The first option doesn't work since in finite algebraic extension of $\,\Bbb Q\,$ , we know that the fractional ideals (of the Dedekind domain of integers of the extension) can always be generated by two elements, and thus the actual (integral) ideals too, but the ring can be far from principal. About the second option I can't tell right now, but what we know is that if every prime ideal is principal then the ring is principal. This is a non-trivial claim and if I remember correctly we need Zorn's Lemma to prove it (the set of all non-principal ideals and etc.) $\endgroup$
    – DonAntonio
    Commented Oct 22, 2012 at 4:57
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    $\begingroup$ Dear @DonAntonio: I don't think your example works, because the OP assumes that in his ring, every ideal generated by 2 elements is principal. What you are saying is that there are rings where every ideal is generated by at most two elements but where the rings are not principal. Certainly this does not violate what the OP is asking. $\endgroup$
    – Rankeya
    Commented Oct 22, 2012 at 5:00
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    $\begingroup$ Dear user, Note that your two conditions (every ideal gen. by $2$ elts. is principal, and every f.g. ideal is principal) are equivalent (by an induction on the number of generators). Regards, $\endgroup$
    – Matt E
    Commented Oct 22, 2012 at 5:20
  • $\begingroup$ Dear @Rankeya, you of course are right. I misread. Thanks. $\endgroup$
    – DonAntonio
    Commented Oct 22, 2012 at 10:48

3 Answers 3


A domain with every two-generated ideal principal $\rm\:(a,b) = (c)\:$ is called a Bezout domain. By induction this is equivalent to every finitely-generated ideal being principal. This does not imply that every ideal is principal (PID). Indeed, the ring of all algebraic integers is Bezout (Kaplansky, Commutative Rings, Theorem 102), but its ideal $\:(2^{1/2},\,2^{1/3},\ldots)\:$ is not principal.

A domain is a PID $\iff \rm\color{#0a0}{Bezout}$ and ACCP (= ascending chain condition on principal ideals). Direction $(\Rightarrow)$ is clear. For $(\Leftarrow)$ $ $ ACCP $\Rightarrow $ among $\rm(i)\subseteq I$ there is a maximal one $\rm(c),$ $\rm\color{#0a0}{hence}$ $\rm \,(c,i) = (c)\,\ \forall\,i\in I,\,$ so $\,\rm I=(c).\,$ [Or, said element-wise: ACCP implies divisibility is well-founded, so an ideal $\rm\,I\neq 0\,$ is generated by an element c minimal wrt to divisibility, since if $\rm\,\color{#c00}{c\nmid i}\in I\,$ $\rm\color{#0a0}{then}$ $\rm\:(c,i) = (d)\subset I\:$ and $\rm\:d\:|\:c\:$ properly (else $\rm\,\color{#c00}c\mid d\mid\color{#c00}i\,),$ contra minimality of $c\,$]. This is essentially an application of the Dedekind-Hasse Test for a PID.

If we are given a Bezout UFD $(\Rightarrow$ ACCP) then we can use the number of prime factors as a metric for the Dedekind-Hasse test. Then that the above $\rm\,c\in I\,$ is divisibility minimal translates to it being an element of $\,\rm I\,$ with least number of prime factors.


No, in general this is not true. For an example, in the ring of entire functions every finitely generated ideal is principal, but this ring is not a PID.

If, in addition, $R$ satisfies the ascending chain condition for principal ideals, then one can conclude that $R$ is a PID.


There is a theorem of Isaacs which says that if every prime ideal is principal then so is every ideal.

  • $\begingroup$ A simpler version suffices here, e.g. see this answer. $\endgroup$ Commented Apr 23, 2015 at 5:01
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    $\begingroup$ And what has this answer to do with the question? (Usually one reads the questions, not only the titles which are intended to be informative.) $\endgroup$
    – user26857
    Commented Jun 15, 2016 at 8:56

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