Artin defines an ideal $I$ as :

  1. $I$ is a subgroup of $R^+$
  2. If $a \in I$ and $r \in R$ , then $ra \in I$

And Principal Ideal is defined as

"In any ring, the set of multiples of a particular element $a$ , forms an ideal called a principal ideal generated by $a$"

My question is:

If the set of multiples of a particular element is called principal ideal then that automatically is one of the properties of an ideal (Prop 2), then is every ideal a principal ideal?

  • $\begingroup$ The key words are "a particular element $a$". $\endgroup$ Sep 9, 2012 at 5:22
  • $\begingroup$ @AlexBecker I just posted a comment to yuri's answer, can you chip in? Always found your explanations helpful $\endgroup$
    – Soham
    Sep 9, 2012 at 5:36
  • $\begingroup$ for example every field is a PID, because the only ideals of a field $F$ are $\{0\}$ and $F$. $\endgroup$
    – user118135
    Dec 29, 2013 at 12:24
  • 1
    $\begingroup$ Sounds like your ring is assumed to be commutative. Part of the confusion may be that a principal ideal is always an ideal. It is a special kind of an ideal $I$ with that extra property promising the existence of such an element $a$ that all the elements of $I$ are multiples of $a$. $\endgroup$ Dec 29, 2013 at 23:36
  • $\begingroup$ Every principal ideal is an ideal. Not every ideal is a principal ideal. (Contrast this with the fact that every differential equation is a stochastic differential equation but not every stochastic differential equation is a differential equation.) $\endgroup$ Dec 30, 2013 at 2:27

1 Answer 1


No. If $I$ is an ideal and $a\in I$ then every multiple of $a$ also belongs to $I$. But the converse is not true — there might be no one element $a\in I$ such that every element of $I$ is a multiple of $a$.

Consider for example the ring ${\mathbb Z}[x,y]$. Let $I$ be the set of polynomial $p(x,y)$ such that $p(0,0) = 0$. It is easy to verify that that $I$ is an ideal. However, there is no element $a\in I$ that divides every element in $I$. (In particular, there is no element $a\in I$ that divides both polynomials $x$ and $y$.)

  • $\begingroup$ Oh so you mean to say in a principal ideal every element is a multiple of $a$. So can I say every principal ideal also "contains" an ideal?(would have liked to use embedded but I didnt want to abuse the term) $\endgroup$
    – Soham
    Sep 9, 2012 at 5:30
  • 4
    $\begingroup$ (1) Yes, if $I$ is a principal ideal then every element of $I$ is a multiple of $a$ (for some fixed $a\in I$). (2) Every ideal contains a principal ideal but not the other way around. In the example from my post, $I$ is not contained in any non-trivial principal ideal. $\endgroup$
    – Yury
    Sep 9, 2012 at 5:35
  • $\begingroup$ No Kirk it was my inability to parse such an interesting concept, the "excitement" to understand this and my motor skills all jamming my thought process that I messed up what I wanted to say in the first place :) Thanks for your patience $\endgroup$
    – Soham
    Sep 9, 2012 at 5:39

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