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When defining a principal ideal, e.g $I$

$I=aR=(ar:r \in R)$

(an ideal generated by a single element of the ring)

do we require the ring to have an identity element and if so, in what manner does this affect the structure of the principal ideal? More precisely, is the requirement to have an identity element necessary to allow the element that generates the principal ideal, to belong to it in a trivial fashion
($a \in I$)?

PS Hopefully this question is not a matter of semantics, involving the ages-old ring vs rng definition i.e. whether we consider the identity element as part of every ring or not.

EDIT:
Ok, I think I made some progress: If we do not assume that the ring $R$ has an identity element, but do assume that it is a commutative ring and every ideal of the ring is principal-that is, $R$ is a Principal Ideal Domain, we can show that $R$ indeed has an identity element as a result of this assumption.

Indeed, let $R$ be a ring where every ideal $I⊆R$ is a principal ideal. Then, since $R$ is an ideal of $R$, and since every ideal of $R$ is a principal ideal,there must exist an element $a\in R $ such that $R=<a>=aR$
Since $a\in R$, there must be an element $r \in R$, such that $a=ar$.
Let $b \in R$. It follows $b=ac$ for some $c \in R$.
We have $b=ac=(ar)c=(ac)r$.
So $r$ acts as an identity element for a random $b \in R$
We define $r$ to be the identity element of $R$.

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  • $\begingroup$ It is not clear what you are asking. The definition of a principal ideal does not require the ring to have $1$. I think you are asking whether $a \in aR$ for every $a \in R$ implies that $R$ has $1$. Your use of the phrase "in a trivial fashion" and your PS make it unclear what your question actually is. And your recent edit seems to be about principal ideal rings (or rngs?) Please clarify. $\endgroup$ – Rob Arthan Jan 17 '16 at 22:05
  • $\begingroup$ I was asking precisely if the definition of a principal ideal requires the ring to have $1$. In the text I was using it stated in the principal ideal definition that the ring does have $1$, but did not explain in any manner why it was necessary. Prior to the principal ideal definition it was ambiguous on whether a ring has $1$ or not. $\endgroup$ – MathematicianByMistake Jan 17 '16 at 22:10
  • $\begingroup$ The edit assumes that we have an rng and goes on to show-hopefully correctly-that if the rng in question has the property of each of its ideals being principal, then it is a ring. I am not sure if this answers the question though.. $\endgroup$ – MathematicianByMistake Jan 17 '16 at 22:13
  • $\begingroup$ "Are principal ideal rngs rings?" is a good question. P $\endgroup$ – Rob Arthan Jan 17 '16 at 22:19
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    $\begingroup$ The lovely creature did more than just distract me - it typed "P" and then ENTER for me $\ddot{\frown}$. I look forward to your update. $\endgroup$ – Rob Arthan Jan 17 '16 at 22:49
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do we require the ring to have an identity element

You may define a principal ideal of a commutative ring (or a principal right ideal of a ring) that way, but it is not standard. The problem is that you have no guarantee $a$ is in $aR$ if that is your definition. This is what having an identity does for you, and it is supposed to be an expectation of and ideal "generated by an element."

Take for example the rng $2\Bbb Z$ and the principal ideal generated by 2 via your definition: $2\notin 2(2\Bbb Z)=4\Bbb Z$.

but do assume that it is a commutative ring and every ideal of the ring is principal-that is, is a Principal Ideal Domain,

A ring whose ideals are all principal is called just a principal ideal ring.

in what manner does this affect the structure of the principal ideal?

I'm not sure what you want here. You are defining is structure explicitly. For rings not necessarily having identity, the normal definition of the principal right ideal is $a\Bbb Z +aR$, and the normal definition of the two sided principal ideal is the set of elements of the form $na+ \sum r_ias_i $ where n is an integer and the sum is finite, and $r_i,s_i$ are ring elements.

In the standard sense, $R=2\Bbb Z$ is a principal ideal domain without identity, but in your sense it is not (the entire ring can't be of the form $aR$) Requiring every ideal to be of that form is a very strong condition.

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  • $\begingroup$ I think you might want to be more explicit about the fact that OP's definition is not the standard meaning of "principal ideal" in a rng and state the standard definition more prominently, since it is not clear to me that OP is aware of this. $\endgroup$ – Eric Wofsey Jan 18 '16 at 4:50
  • $\begingroup$ My main question has been answered-regarding the need to have $a$ in $<a>$ and including the $1$ as a means to achieve it. Also, thanks to @EricWofsey because I was not aware of the various definitions of the principal ideal and the differences in meaning they entail. $\endgroup$ – MathematicianByMistake Jan 18 '16 at 12:54
  • $\begingroup$ Sir, i know comments are not right place to ask questions. But my question is related to above answer. Sir can you please help me in it. Here is link. math.stackexchange.com/q/3638315/168676 $\endgroup$ – Akash Patalwanshi Apr 23 '20 at 3:09

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