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In the following post: (Confusion between principal ideal and ideal) on clarification between the concepts of ideals and principal ideals, @Yury stated: "... if $I$ is a principal ideal then every element of $I$ is a multiple of $$ (for some fixed $\in I$) (2) Every ideal contains a principal ideal but not the other way around." I try to write out precisely what it means by not every ideal is a principal ideal in mathematical notation.

What I would like is to see if I can phrase what Yury stated in mathematical notations. I know this may sound a bit pedantic. I just would like to make sure I am crystal clear if I cross all the Ts and dot all the Is when it comes to formulating it in terms of the correct quantifiers.

Definition of an Ideal:

An $\textbf{ideal}$ of a ring $R$ is a subring $I$ of $R$ such that for all $x\in R$ and $y\in I$, both $xy \in I$ and $yx \in I$

Definition of Principal Ideal:

Let $R$ be a commutative ring with a unit element. An ideal $I$ of $R$ is $\textbf{principal}$ if there exists $d\in R$ such that $I=(d)= \{rd\mid r\in R\}.$ In this case, $d$ is said to $\textbf{generate} I$

Yury's statement part (1) would mean: if we given a commutative ring $R$ with unit element, for any ideal $I$ of $R$, there exists an $a\in R$ such that $(a) \subset I$, where $(a)=\{ar: \forall r \in R\}$ contains principal ideals trivially. For part (2) of his statement, in mathematical notation, it translate to: There exists an ideal $I$ of $R$ such that $I \not\subset (a)$, meaning there exists an ideal $I$ of $R$ and a $x \in I$ $x \neq ar$ for all $a$, $r \in R$
I am not certain if how I put Yury's statements in mathematical notations is accurate. IF someone can comment and point out any errors, it will be much appreciated. Thank you in advance

RobPratt
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Seth
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1 Answers1

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I'll stick to the commutative case for clarity. I'm not sure what @Yury means by "but not the other way around" but hopefully the following bullet points will clear up the confusion.

  • If $R$ is a commutative ring and $I$ is an ideal of $R$, then for every $a \in I$, we have that $(a) \subseteq I$. In particular since $0 \in I$, $I$ contains a principal ideal.
  • It is trivially true that every ideal is contained in a principal ideal, namely $I \subseteq R = (1)$.
  • There are some rings where every ideal is principal, these are called principal ideal rings.
  • There are also some rings with ideals which are not principal. The two classic examples of this are that the ideals $(2 , x) \subseteq \mathbb{Z}[x]$ and $(x , y) \subseteq k[x , y]$ (for $k$ any field) are not principal.
Nate Gallup
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  • I think I am having trouble sorting out all the quantifiers when I try to write out precisely what it means for an ideal $I$ of $R$ to not being principal. As you know, principal ideal describes the set of all multiples associated to a particular element of an ideal. When I look at your last point and their proofs and relates to a math description of an ideal not being principal. Does it mean that for some $x \in I$, $x \in ar$ for all $a$ and $r\in R$. Or is it for some $a \in R$ and for all $r \in R$. – Seth Mar 20 '21 at 03:40
  • the reason I posted the question originally, was due to having trouble understanding the logic behind the proofs of the examples from your last point. I tried to worked it out in my mind the precise mathematical formulation of what it means for an ideal not to be principal. I think the definition of an ideal of a ring and also that of for principal ideal, involves juggling and negating the correct quantifiers. Hence my difficulties. – Seth Mar 20 '21 at 03:53
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    An ideal $I$ in $R$ is not principal if there does not exist an element $a \in R$ such that $I = (a)$. – Nate Gallup Mar 20 '21 at 03:55
  • if there does not exist an element $a\in R$ and for any $r\in R$ such that $I=(a)$, doesn't that violate the statement that any ideal $I$ contains the trivial principal ideals. I mean, if I take any particular element $x \in I$, I can always generate it's multiples, $xr$ for all $r \in R$ – Seth Mar 20 '21 at 04:00
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    Yes for any $x \in I$, $(x) \subseteq I$, but if $I$ is not principal then this inclusion will always be strict. – Nate Gallup Mar 20 '21 at 04:42
  • the reason I am so confused with the quantifiers, is in the definition of principal ideal, when it is defined as ${\mid \in }$, does the truth statement mean the set of all $ar$ for some $ \in $ $r\in R$. ℎ ℎ ,ℎ ℎ $r$ variable is not clear to me. I know I am being super pedantic, but I would rather make sure I understand it now then getting tripped up on an exam. Thank you for taking the time. – Seth Mar 20 '21 at 06:14