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Fintely generated ideals 
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#1
Dec2306, 10:05 AM

P: 277

I would appreciate some help with developing a simple proof that the ideals in the ring of integers for a number field have the same rank as the ring of integers itself.
In other words, assuming from the start that all the ideals are finitely generated, all ideals require the same number of generators as the entire ring of integers itself. I find it easy to show that the ideals have the same rank or lower, but not that they have to have the same rank. Any help would be appreciated. Thanks. 


#2
Dec2306, 11:48 AM

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If I understand the question correctly, I think one way to see it is that O/I is a finite ring for any nontrivial ideal I of O.



#3
Dec2306, 12:40 PM

P: 277

That is what I really want to show. Maybe there is another way to do it? I know I can do it my module theory if I can show that and ideal I has the same rank as O. But I would be more than happy for a simpler way to show the end result.
I saw another proof, but I don't understand a few steps. One is that in any (nontrivial) ideal I in O we can always find a rational integer element of I, call it A. Then we get: [itex](A) \subseteq I [/itex] It is obvious that O/(A) is finite (that much I can prove), then they make the claim that we have the following relation which proves O/I finite: [itex]{O/(A) \over O/I} \cong I/(A) [/itex] But I have several confusions with this version. First, I'm not sure how we prove that we always have a rational integer in all proper ideals of O. Second, I think I understand the general relation above as one of the fundamental isomorphism theorems for rings. But I'm not sure how we go from O/(A) being finite and the above congruence to O/I is finite. So, help understanding this proof would be great. 


#4
Dec2306, 12:45 PM

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Fintely generated ideals
It's actually pretty easy to show that every nonzero ideal contains a rational integer: think about minimal polynomials.
That next step looks wrong; I think you meant [tex]\frac{ \mathcal{O} / \langle A \rangle }{ I / \langle A \rangle} \cong \mathcal{O} / I.[/tex] So... what is the relationship between O/A and (O/A) / (I/A)? Or, a roughly equivalent alternative... You know that [tex]\langle A \rangle \subseteq I \subseteq \mathcal{O}[/tex] so what does that tell you about O/A and O/I? (And O/0 and O/O) 


#5
Dec2306, 01:10 PM

P: 277

Your last question was something I also saw, they used this alone in one place I saw to say that O/I was finite since O/A was. I just don't understand why this is true. I do remember one of the correspondence theorems that there is a one to one correspondence between ideals between say A and O and ideals of O/A, and I is an ideal between A and O, but I don't see how this leads to finiteness either. I would really like some clarification on this if you could provide it. Aside from that (which I still very much would like) would the following be enough to prove O/I finite? (or rather that I has maximum rank in O, which then leads to finiteness): Assume we have the following integral basis for O: [itex]\omega_1 ... \omega_n[/itex] then since we know that I contains a rational integer A, then [itex]A \omega_1 ... A \omega_n[/itex] must all be in I as well, and this alone is certainly a submodule of maximum rank, so trivially I must also have maximum rank. Does that suffice? 


#6
Dec2306, 01:53 PM

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Your proof sounds right. But I think it's still worth understanding the other proofs.
Actually, the intuitive idea behind all of these things is this: if you mod out by (A), you have something left over. If you mod out by more than (A), you have even less. 


#7
Dec2306, 01:59 PM

P: 277

I would really like to understand the other proofs too, which is why I was asking for more info on them. And yes, to your very picky part, I can handle the part about the rank being less than or equal to, it was just the equal to exactly in this case that was bugging me.
I liked your last part, about the mods, that makes a lot of sense intuitively and is a nice way to think of it. Thanks. I miss that a lot ... a more general way of looking at these things. Unfortunately, I don't see how to apply that directly into a proof. So more tips on the other proofs would be very much appreciated. 


#8
Dec2306, 02:11 PM

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Oh bah, you caught the picky part before I decided it wasn't worth mentioning and edited it out.
I guess the most direct translation of the inutitive idea is that [itex]0 \subseteq A \subseteq B \subseteq C[/itex] induces surjective maps [itex]C \twoheadrightarrow C/A \twoheadrightarrow C/B \twoheadrightarrow 0[/itex]. The isomorphism theorems even tell you that the kernel of the middle map is B/A. (But that should be clear... what's the kernel of [itex]C \twoheadrightarrow C/B[/itex]?) I'm not really sure me explaining things will help you any further  the best thing at this point is probably to work it out as an exercise in using the isomorphism theorems (or even to work it out directly in terms of cosets). 


#9
Dec2306, 02:34 PM

P: 277

Though I am much more interested in understanding the other parts. Maybe we can start with this one. That if: [itex]<A> \subset I \subset O [/itex] Then this implies directly that if O/A is finite that O/I is as well. This was the proof given in one of my books, but they just said this was "trivially true", which i don't really agree with ... obviously. At least not yet. I understand conceptually how this relates to your comments about mods, so I understand abstractly why it should be true, but I'm still missing the practical connection to actually show it. 


#10
Dec2306, 02:35 PM

P: 277

I saw the second part of your response after I typed mine ... I'll look at it some more myself and see what I come up with.
By the way, it's that middle map that confuses me. Quotient to quotient with kernal another quotient somehow tends to lose me. 


#11
Dec2306, 02:40 PM

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Try working a concrete example. Say...
[tex]0 \subseteq (6) \subseteq (2) \subseteq \mathbb{Z} \twoheadrightarrow \mathbb{Z}_6 \twoheadrightarrow \mathbb{Z}_2 \twoheadrightarrow 0 [/tex] Or, how about [tex] 0 \subseteq \langle x \rangle \subseteq \langle x, y \rangle \subseteq F[x, y] \twoheadrightarrow F[y] \twoheadrightarrow F \twoheadrightarrow 0 [/tex] (where F is some ring. Maybe a field. Your call. The first surjection sends x>0 and y>y. The second sends y>0) I'm just trying to pick some examples where you can understand each component of the diagram in its own right. 


#12
Dec2306, 02:45 PM

P: 277

I think I just saw a possibly trivial way of looking at it. Let me know if this is solid reasoning.
If we have two elements [itex]r,s \in O[/itex] if they are in the same residue class mod A, in other words if A+r = A+s in O/A, then we must have [itex]rs \in A[/itex]. But of course since [itex]A \subset I[/itex] then [itex]rs \in I[/itex] as well. This means that any two elements in the same residue class in O/A must also be in the same residue class in O/I, from which is really is trivial that there can not be more residue classes in O/I then in O/A. That about right? 


#13
Dec2306, 02:48 PM

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Yep, that looks right! (Do you see the surjection O/A >> O/I now?)



#14
Dec2306, 02:52 PM

P: 277

I'll have to see if I can see that one as well. This is also the part where the kernel is supposed to be I/A right? This is still confusing me a bit. ... Hey, I just got the kernel part! If we are mapping O/A to O/I, then we are mapping A+r to I+r, and so the kernel is whenever r is in I, but that is obviously just I/A which is A+r where r is in I. So I see why I/A is the kernel. Cool. Getting there. 


#15
Dec2306, 03:00 PM

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#16
Dec2306, 03:02 PM

P: 277

Okay, then basic isomorphism theory gives that
[tex]\frac{ \mathcal{O} / \langle A \rangle }{ I / \langle A \rangle} \cong \mathcal{O} / I.[/tex] Which is what you had before. Assuming that O/I is the image of the map, or in other words that it is surjective as we have been discussing. But then isn't surjectivity itself enough to say that O/I is finite if O/A is? I feel like I'm missing one little step in showing surjectivity in either case ... 


#17
Dec2306, 03:05 PM

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#18
Dec2306, 03:12 PM

P: 277

A crack at surjectivity of the map from O/A to O/I:
An element of O/I is of the form I+r, but for every possible r here, we can map from A+r to that element. So all elements have a source in O/A. But the problem with this logic is it works the other way too as worded, and it can't be surjective in both directions or it would be an isomorphism and they are different sizes. So I must be making an oversimplification here that isn't valid. Maybe I need to look at residue classes instead? A+[r] to I+[r]? Since we showed there were at least as many residue classes O/I as in O/A all residue classes in O/I must have a source class in O/A, but it doesn't work the other way around because you are "splitting" the classes up again. Makes sense logically, but it doesn't feel very algebraic yet. There must be something trivial here I'm not seeing, right? 


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