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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?

Zhen Lin
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Soham
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  • The key words are "a particular element $a$". – Alex Becker Sep 09 '12 at 05:22
  • @AlexBecker I just posted a comment to yuri's answer, can you chip in? Always found your explanations helpful – Soham Sep 09 '12 at 05:36
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    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 $a$. – Jyrki Lahtonen Dec 29 '13 at 23:36
  • 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.) – Michael Hardy Dec 30 '13 at 02:27
  • for example every field is a PID, because the only ideals of a field $F$ are ${0}$ and $F$. –  Dec 29 '13 at 12:24

1 Answers1

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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 $y$.)

Yury
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  • 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) – Soham Sep 09 '12 at 05:30
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    (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. – Yury Sep 09 '12 at 05:35
  • 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 – Soham Sep 09 '12 at 05:39