Struggling to understand principal ideals in Ring Theory

Question

I studied group theory last year and the concept of $<g>$ or $<g,h>$ for elements $g,h$ of a group $G$ made complete sense as it was just the set of all the products of elements within $< >$ and their inverses just like how span works. Now when I get to Ring Theory (old edition with NO UNITAL ELEMENT requirement) the definition of $<r>$ and $<r,s>$ for $r,s$ in a ring $R$ just doesn’t make sense anymore as it’s no longer just the products (and addition) of the elements within $< >$ but instead involves elements outside of $< >$ as well.

I just can’t picture in my head what’s going on here and it’s making harder following my classes harder. For example one line in a proof we were doing was:

“Clearly $<a>^3 \subseteq RaR$“ ($R$ is a ring and $a$ is an element in $R$) but this isn’t at all clear for me.

No, it doesn't involve elements outside of $\langle\rangle$. The ideal in $R$ generated by $a$ is often denoted by $(a)$. What is $\langle a\rangle$ for you? — Dietrich Burde, Nov 16 '19 at 16:18
@DietrichBurde What do you mean? Yes it does of course, $2X\in<2>\subset \Bbb Z[X]$ but $X$ is not in the $<>$ here. — Arnaud Mortier, Nov 16 '19 at 16:21
Of course $X\in \langle 2\rangle =2\Bbb R[X]$ is in there. This means, it is an element of it. — Dietrich Burde, Nov 16 '19 at 16:23
It's in there, but it's not specifically inside the brackets. Well, maybe this is more clear with $\Bbb Z$, edited. — Arnaud Mortier, Nov 16 '19 at 16:23
I don't think that this is what the OP meant, because then for groups $\langle g\rangle $ also has not $g^2$ in there. — Dietrich Burde, Nov 16 '19 at 16:25
@DietrichBurde But you get $g^2$ just by multiplying $g$ by $g$, and both $g$ and $g$ are in there. — Arnaud Mortier, Nov 16 '19 at 16:26
Not with $g^{-1}$, though. That's not necessarily from $g\cdot g\cdots$. — Dietrich Burde, Nov 16 '19 at 16:27
@DietrichBurde Are you trolling here? Of course $g^{-1}$ comes from inverting $g$. Inverting and multiplying are your two natural things to do in a group. — Arnaud Mortier, Nov 16 '19 at 16:33
I’m confused it is defined to include the products of elements in $<>$ with those outside $<>$ — Partey5, Nov 16 '19 at 16:33
But $g^{-1}$ is not written in $(g)$ in the same way, as $X$ is not written in $(2)=2\Bbb Z[X]$. — Dietrich Burde, Nov 18 '19 at 08:47

Arnaud Mortier · Answer 1 · 2019-11-16T16:44:07.210

-1

Since you are specifying that your ring doesn't have a unit, you need to be careful when defining ideals: $\left<a\right>$ is not the set of all multiples of $a$, since otherwise $a$ would not necessarily be an element of $\left<a\right>$ as there is no unit.

$\left<a\right>$ is the smallest subset of $R$ that

contains $a$
is an additive subgroup
is stable by external multiplication with elements of $R$.

This last requirement, which is the one that bothers you, is actually quite natural. You can think of how an even number is still an even number when you multiply it by any number, odd or even (here we are looking at the set of even numbers, $\left<2\right>$, inside $\Bbb Z$).

See this question regarding why ideals play a more important role than subrings in ring theory.

edited Nov 16 '19 at 16:44

answered Nov 16 '19 at 16:33

Arnaud Mortier

27,276

What is $I$ ??. – Partey5 Nov 16 '19 at 16:35
@Anteater23 Sorry $R$. Edited. – Arnaud Mortier Nov 16 '19 at 16:36
So we’re defining it is a completely different way to group theory so that is the smallest ideal containing $a$? – Partey5 Nov 16 '19 at 16:38
1

@Anteater23 Absolutely. – Arnaud Mortier Nov 16 '19 at 16:38
Why do we want it defined this way though? What’s wrong with being the smallest SUBRING of R containing a? Note: by my book every ideal is a subring so ideal is a stronger condition. – Partey5 Nov 16 '19 at 16:41
How does the definition guarantee that $a$ lies in . I can’t see this. – Partey5 Nov 16 '19 at 16:44
@Anteater23 I've added a link that answers that. – Arnaud Mortier Nov 16 '19 at 16:44
How do we know a lies in ? – Partey5 Nov 16 '19 at 18:06
1

@Anteater23 By the very definition of $$, it's requirement #1. – Arnaud Mortier Nov 16 '19 at 18:11
Ok. So there’s no actual definition which describes the elements in ? – Partey5 Nov 16 '19 at 18:12
There is the other answer by rogerl which is valid when $R$ has a unit: in this case $\left<a\right>$ is the set of all multiples of $a$ by an element of $R$. Otherwise, you're right, there is no such description. – Arnaud Mortier Nov 16 '19 at 18:13
So is {a union ra union ar for all r in R} – Partey5 Nov 16 '19 at 18:15
@Anteater23 No, because $a+ra$ is not a multiple of $a$ either. – Arnaud Mortier Nov 16 '19 at 18:16
Yes but if a is in and ra is in then a+ra in is as it is additive subgroup no? – Partey5 Nov 16 '19 at 18:19
@Anteater23 Yes, which is why your description {a union ra union ar for all r in R} is wrong. It's more complicated and I don't believe there is a closed description of all elements. – Arnaud Mortier Nov 16 '19 at 18:22
So ={a+ra+ar} for all r in R :) ? – Partey5 Nov 16 '19 at 18:24
@Anteater23 Try to prove it, if it's true then the proof will be straightforward, and if not you will see why. – Arnaud Mortier Nov 16 '19 at 18:31

Struggling to understand principal ideals in Ring Theory

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