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The definition for an ideal to be principal is:

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

There is something that I am still confused about from the wording of the definition. Since $R$ is a commutative ring with a unit element, that implies that $R$ contains the identity element. If an ideal $I$ is a principal ideal, in non mathematical notational worded description, it means that all the element of the ideal $I$ can be generated by some single element $a\in R$. Nowhere does the definition state that $a$ has to be an element of $I$. Can we not have one of the following two scenarios $(1)$ where both $a$ and $r$ are in $R\setminus I$ or $(2)$ $a\in I$, $r\in R\setminus I$, but $ar \in I$.

I posted several times in regards to the concept of principal ideal because when I look over the proof of why the ideal of the set of polynomials with constant term being even, or similarly the constant terms being divisible by three, the usual proof is to show that why the ideals $I=\langle 2, x\rangle$ or $J=\langle3, x\rangle$ of the commutative ring $\mathbb{Z}[x]$, are not principal ideals. For either of these proofs, why can I not use the identity element in $\mathbb{Z}[x]$ as a generator to serve as the single element generator to obtain all the elements in either $I$ or $J$, even thought the identity element is neither in $I$ or $J$.

Thank you in advance.

mrtaurho
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Seth
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  • If $I$ is generated by some element $a\in R$ then $a$ lies in $I$ because $a=r\cdot a$ for $r=1$. – Alvaro Martinez Mar 30 '21 at 14:26
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    @AlvaroMartinez so the reason the identity element would not work in the case of $<2,x>$ because $1 \notin I$. Is that correct? – Seth Mar 30 '21 at 14:31
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    That's correct, the only ideal containing $1$ is $R$ itself. – Alvaro Martinez Mar 30 '21 at 14:34
  • @AlvaroMartinez is this similar to looking at which particular element of a group that can generate every element of a group to form cyclic group. So a non principal ideal is if the order of a group does not have prime order, then it requires more than one generator. – Seth Mar 30 '21 at 14:40
  • If $R$ is not unital, then it could be the case that $I$ is an ideal of $R$ generated by an element $a$ of $R \setminus I.$ For instance, take $R = 2 \Bbb Z$ and $I = 4 \Bbb Z = 2R.$ – Dylan C. Beck Mar 30 '21 at 14:43
  • @SethMai This situation is entirely analogous. Non-cyclic groups always require more than one generator as otherwise they would be cyclic by definition. – mrtaurho Mar 30 '21 at 14:43
  • @mrtaurho so if principal ideal is to rings what cyclic group is to groups, then can we determine whether a finite ideal is a principal ideal by looking at the order of the ideal, from the perspective of look at the ideal as a group. I am not sure if what I am asking is possible. – Seth Mar 30 '21 at 14:46
  • @SethMai I'd go only as far as saying that the concept is analogous. I'm not sure if what you describe is possible/effective. – mrtaurho Mar 30 '21 at 14:50
  • @mrtaurho I thought if the ideal is a finite ideal, then there must be some procedure similar to finite group whether the ideal is a principal ideal. I mean, the case for groups and to show whether a group is a cyclic group seems not to be difficult. I could be wrong since i have not seen cases where that is difficult. – Seth Mar 30 '21 at 14:53
  • @SethMai This might be possible in finite rings but I suspect it will stop working pretty quickly in infinite rings. But I don't know for sure. If you're dissatisfied with this rather vague statement maybe ask another question on this topic :) – mrtaurho Mar 30 '21 at 14:56

1 Answers1

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If the ring is unital (i.e. has a unit) then $a=1\cdot a\in I$. Hence the principal ideal $\langle a\rangle$ always contains its generator $a$ in such a setting. You have to be more careful if you ring is missing a unit, though. Regarding to your two cases: $(1)$ will never happen in a unital ring and $(2)$ always holds by definition.

Let's take a closer look at the ideal $I=\langle2,x\rangle<\mathbb Z[x]$. For $I$ to be principal we have to find some $f(x)\in\mathbb Z[x]$ such that $I\mathbf{=}\langle f(x)\rangle$. However, if you take $f(x)=1$ then $\langle f(x)\rangle=\mathbb Z[x]\supsetneq I$ missing the crucial equality. Slightly more general: any ideal containing a unit (a multiplicatively invertible element) will be the whole ring. See here for instance for the details.

J. W. Tanner
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mrtaurho
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  • I think the definition of principal ideal, when i first look at it, the thing I ask is, does the particular element that is chose have to be in the ideal in the first place. It just seems ambiguous. – Seth Mar 30 '21 at 14:43
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    @SethMai If you define a principal ideal as you did in your post there's is no ambiguity: the generator will always be in the ideal. A principal ideal generated by some element $a$ is the set of all multiplies of $a$, i.e. $r\cdot a$ for all $r\in R$; but among them is the trivial multiple $1\cdot a=a$ for $r=1$. We just don't spell it out that $a$ will be in $\langle a\rangle$ as this follows directly from the definition (in a unital ring). – mrtaurho Mar 30 '21 at 14:46
  • ah okay okay, thank you for the clarification. I sometimes still get confused when reading math definitions. Especially when there are multiple instances of quantifiers. I am not sure if mathematicians get tripped up over the wording of definitions. – Seth Mar 30 '21 at 14:51
  • @SethMai I think it's normal to struggle when encountering certain concepts for the first time. Glad to help! :) – mrtaurho Mar 30 '21 at 14:54
  • thank you so much for all your help and detail replies. I feel so back that I have to make three different post on the same concept. Feel like I am spamming the site. – Seth Mar 30 '21 at 14:59