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I have been trying to learn Principal ideals for a while but I am seeing conflicting definitions for the same. Starting out with the definitions I have learned : -

(1) Ideal generated by subset of a ring -
If S is a subset of ring R and if U is an ideal of R such that
$\qquad$ (i) S $\subseteq$ U
$\qquad$ (ii) If V is any other ideal of R and S $\subseteq$ V $\implies$ U $\subseteq$ V,
then U is called an ideal of R generated by its subset S

(2) Principal Ideal - If an ideal U of ring R is generated by a single element i.e, S={a} where a $\in$ R then the ideal U is called a principal ideal of R generated by a.

What I understand from (2) is that U is a principal ideal of R if we have a particular element a of R such that a $\in$ U and for any other ideal of R (say) V, a$\in$ V $\implies$ U $\subseteq$ V. Is this understanding correct?

Using the definition of principal ideal in (2) we can prove that -

If R is a commutative ring with unity, then the set aR = {ar | r$\in$ R} is a principal ideal of the ring R generated by a

However while solving problems, such as in this answer, we have taken the definition of principal ideal as

In any ring, the set of multiples of a particular element a , forms an ideal called a principal ideal generated by a i.e, the set aR = { ar | r$\in$ R} is the principal ideal generated by a

Therefore my question is, Why should a principal ideal always be of the form aR? Is the set aR not a particular principal ideal of R and shouldn't there be other principal ideals which are not in the form aR?

Buddha
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    They are equivalent definitions. Say $P_a$ is the principal ideal of $a$ by your first definition. Then since $aR$ is a ideal containing $a$, we have $P_a\subset aR$. You can then show $aR\subset P_a$. – beeclu Oct 15 '24 at 18:52
  • First it should be noticed that, if $R$ is not assumed to be commutative, then the ideal generated by $a$ should be $I={x\in R:\text{exists }n\in\mathbb Z_+, s_1,\cdots, s_n, r_1,\cdots,r_n\in R, \text{subject to }x=\sum_{i=1}^n s_iar_i}$. This can be simplified to $Ra=aR$ if $R$ is commutative. Clearly $I$ above is an ideal and if an ideal contains $a$, it is forced to contain $I$ (You can ask for hint if the two assertions are not clear to you). Therefore $I$ is the ideal generated by $a$, by definition. – Asigan Oct 16 '24 at 15:48
  • By “A set $S$ is of the form $aR$” the author meant that exists $a\in R$ subject to $S=aR$. This expression is a bit informal (“exists… subject to” is more formal), but it should be clear from context. – Asigan Oct 16 '24 at 15:51

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