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Integral closure

by brian_m.
Tags: closure, integral
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brian_m.
#1
Oct21-10, 11:37 AM
P: 6
Hello,

is anybody here, who can explain to me how to compute the integral closure of a ring (in another ring)?

Example: What is the integral closure of Z in Q(sqrt(2)) ?

Thank you!

Bye,
Brian
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hochs
#2
Oct25-10, 01:05 PM
P: 39
Generally, finding a ring of integers is rather annoying to do by hand. Basically you compute relative discriminants (or equivalently norms of differents) and localize (and complete at the primes ofc) to compute the discriminants separately according to whether it's tamely ramified, wildly ramified, unramified, or totally ramified.

The last step is generally easy, except when it's wildly ramified, for example if you have totally ramified of deg [tex]p[/tex] over [tex]\mathbb{Q}_p[/tex], [tex]p[/tex] prime. But I can leave this as an exercise to show that for galois case, totally ramified deg p gives valuation of the discriminant to be 2p-2 (there are two descriptions of discriminant, one in terms of norm of f'(a) for integral generator a, and another in terms of the roots of f(a)).

But for quadratic case, you can do this by hand and the ring of integers in [tex]\mathbb{Q}(\sqrt{2})[/tex] is just [tex]\mathbb{Z}[\sqrt{2}][/tex].

So for example, try to compute the ring of integers in [tex]\mathbb{Q}(17^{1/3})[/tex]. The discriminant of Z[17^(1/3)] is 3^3 * 17^2, and 3 is not totally ramified when you localize and complete at (3), so Z[17^(1/3)] can't be all of the ring of integers. Now you manually look for more elements, and find that (1 + 17^(1/3))^2 / 3 is integral over Z, adjoin it, and note that this time the discriminant is 3*17^2, so that's the ring of integers.
disregardthat
#3
Oct26-10, 01:25 PM
Sci Advisor
P: 1,742
[tex]y= p+q\sqrt{2}[/tex] is a general element of [tex]\mathbb{Q}(\sqrt{2})[/tex] where p and q are rational. First you will need to show that if y is a solution to a polynomial over Z, then it is a solution to a polynomial over Z of degree 2. Then find necessary and sufficient conditions for y such that y is the solution of such a polynomial (use the fact that p and q are rational).

hochs
#4
Oct26-10, 01:29 PM
P: 39
Integral closure

Quote Quote by Jarle View Post
[tex]y= p+q\sqrt{2}[/tex] is a general element of [tex]\mathbb{Q}(\sqrt{2})[/tex] where p and q are rational. First you will need to show that if y is a solution to a polynomial over Z, then it is a solution to a polynomial over Z of degree 2. Then find necessary and sufficient conditions for y such that y is the solution of such a polynomial (use the fact that p and q are rational).
This works for quadratic and some cubic extensions. For bigger extensions, this gets really nasty quite quickly, so you're better off using some more theory (differents, discriminants ,etc.)

In fact it's quite easy to see by simple calculations what is ring of integers in Q(sqrt(d)), but not so simple to find integral closure in Q(17^(1/3)) with simple calculations like this. (the example I gave above).
DonAntonio
#5
Feb25-12, 05:58 AM
P: 606
Quote Quote by hochs View Post
Generally, finding a ring of integers is rather annoying to do by hand. Basically you compute relative discriminants (or equivalently norms of differents) and localize (and complete at the primes ofc) to compute the discriminants separately according to whether it's tamely ramified, wildly ramified, unramified, or totally ramified.

The last step is generally easy, except when it's wildly ramified, for example if you have totally ramified of deg [tex]p[/tex] over [tex]\mathbb{Q}_p[/tex], [tex]p[/tex] prime. But I can leave this as an exercise to show that for galois case, totally ramified deg p gives valuation of the discriminant to be 2p-2 (there are two descriptions of discriminant, one in terms of norm of f'(a) for integral generator a, and another in terms of the roots of f(a)).

But for quadratic case, you can do this by hand and the ring of integers in [tex]\mathbb{Q}(\sqrt{2})[/tex] is just [tex]\mathbb{Z}[\sqrt{2}][/tex].

So for example, try to compute the ring of integers in [tex]\mathbb{Q}(17^{1/3})[/tex]. The discriminant of Z[17^(1/3)] is 3^3 * 17^2, and 3 is not totally ramified when you localize and complete at (3), so Z[17^(1/3)] can't be all of the ring of integers. Now you manually look for more elements, and find that (1 + 17^(1/3))^2 / 3 is integral over Z, adjoin it, and note that this time the discriminant is 3*17^2, so that's the ring of integers.


I think it's pretty ridiculous to talk about ramificated (primes), discriminants and stuff to someone asking so basic and beginner's a question as what's the integral closure of Z in Q(sqrt(2))....now common!
morphism
#6
Feb25-12, 07:57 AM
Sci Advisor
HW Helper
P: 2,020
Quote Quote by DonAntonio View Post
I think it's pretty ridiculous to talk about ramificated (primes), discriminants and stuff to someone asking so basic and beginner's a question as what's the integral closure of Z in Q(sqrt(2))....now common!
While I absolutely agree with you, was it necessary to resurrect this two-year-old thread just to say that?
DonAntonio
#7
Feb25-12, 08:18 AM
P: 606
Quote Quote by morphism View Post
While I absolutely agree with you, was it necessary to resurrect this two-year-old thread just to say that?

Oops, my bad! I'm still not used to this forum and I completely missed the posts' date.
hochs
#8
Feb25-12, 10:37 AM
P: 39
Quote Quote by hochs View Post
This works for quadratic and some cubic extensions. For bigger extensions, this gets really nasty quite quickly, so you're better off using some more theory (differents, discriminants ,etc.)

In fact it's quite easy to see by simple calculations what is ring of integers in Q(sqrt(d)), but not so simple to find integral closure in Q(17^(1/3)) with simple calculations like this. (the example I gave above).
Do people read?
hochs
#9
Feb25-12, 10:39 AM
P: 39
Quote Quote by hochs View Post

But for quadratic case, you can do this by **hand** and the ring of integers in [tex]\mathbb{Q}(\sqrt{2})[/tex] is just [tex]\mathbb{Z}[\sqrt{2}][/tex].
Do people read anymore?

And discriminants, ramified primes, etc.. are pretty basic concepts you learn in the first course in algebraic number theory. It's common for students to ask, "so, how do you in general find integral closure of number fields?" after having seen an example of direct (and very elementary exercise) calculation of finding integral closure of quadratic extensions.

I apologize so much for having given a student a little perspective on how discriminants can be used, and not so much about what's in every number theory textbook (which is to say, very basic calculation of integral closure in quadratic extensions).

So we should not answer that question until they have learned some more class field theory. Oh and while we're at that, we should not discuss these very basic ideas until we have discussed langlands.

Quite ridiculous.
morphism
#10
Feb25-12, 10:50 AM
Sci Advisor
HW Helper
P: 2,020
Come on, hochs, be reasonable. If the OP were familiar with a fraction of the ideas mentioned in your post, then they would most likely know where to look up an answer to their question, and would definitely not have asked how one would compute the ring of integers of Q(sqrt2). Note moreover that they didn't even use the phrase "ring of integers" - so perhaps they're not even familiar with any algebraic number theory at all.
hochs
#11
Feb25-12, 11:02 AM
P: 39
Quote Quote by morphism View Post
Come on, hochs, be reasonable. If the OP were familiar with a fraction of the ideas mentioned in your post, then they would most likely know where to look up an answer to their question, and would definitely not have asked how one would compute the ring of integers of Q(sqrt2). Note moreover that they didn't even use the phrase "ring of integers" - so perhaps they're not even familiar with any algebraic number theory at all.
I didn't realize it was such a sin to give a little more perspective on things. As I said earlier, it's clear that it's a matter of setting the standard then.

I was a CA (at other universities, "TA") and also taught class field theory several times in the past. Students always enjoyed my giving them more perspectives on how, in general, one may go about solving certain problems - what are the general approaches?. Instead of an expected 'standard' reply, which I expected someone else to give anyway (ex. "disregardthat").

I think I was pretty reasonable. I also get motivated to learn more things when I hear of new terms, and maybe the OP does the same, who knows? Do you know the OP so intimately well? I disagree with your pedagogy.
hochs
#12
Feb25-12, 11:05 AM
P: 39
Besides, the OP's question verbatim is:

is anybody here, who can explain to me how to compute the integral closure of a ring (in another ring)?

Example: What is the integral closure of Z in Q(sqrt(2)) ?

His main question was how to find the integral closure, not how to find integral closure of Z in Q(sqrt(2)).

From this I read that he was looking for a little more *general* answer.

Again I want to ask, do people read?
DonAntonio
#13
Feb25-12, 11:19 AM
P: 606
Quote Quote by hochs View Post
Do people read anymore?

And discriminants, ramified primes, etc.. are pretty basic concepts you learn in the first course in algebraic number theory. It's common for students to ask, "so, how do you in general find integral closure of number fields?" after having seen an example of direct (and very elementary exercise) calculation of finding integral closure of quadratic extensions.

I apologize so much for having given a student a little perspective on how discriminants can be used, and not so much about what's in every number theory textbook (which is to say, very basic calculation of integral closure in quadratic extensions).

So we should not answer that question until they have learned some more class field theory. Oh and while we're at that, we should not discuss these very basic ideas until we have discussed langlands.

Quite ridiculous.

Indeed quite. And we could easily answer that question without any ramification, disciminants and stuff at all as the calculations are pretty straightforward.

I think the OP's question more than revealed his/her basic level in this, and your answer could have just confused into thinking that all the stuff you mention is needed for those basic calculations in the case of rational quadratic extensions.

Tonio
hochs
#14
Feb25-12, 11:23 AM
P: 39
Quote Quote by DonAntonio View Post
Indeed quite. And we could easily answer that question without any ramification, disciminants and stuff at all as the calculations are pretty straightforward.

I think the OP's question more than revealed his/her basic level in this, and your answer could have just confused into thinking that all the stuff you mention is needed for those basic calculations in the case of rational quadratic extensions.

Tonio
Again, I disagree. I learned much better when there were things I wanted to look up.

It's just a matter of pedagogy, ok? And you're simply saying it's "ridiculous" without giving any real substance. I've already responded and countered your ridiculous argument. You're simply repeating something you already wrote without any further comment.

Anyway I got to go run a seminar now, so I'm going to leave this ridiculous discussion and maybe go laugh at this thread with colleagues.
DonAntonio
#15
Feb25-12, 12:18 PM
P: 606
Quote Quote by hochs View Post
Again, I disagree. I learned much better when there were things I wanted to look up.

It's just a matter of pedagogy, ok? And you're simply saying it's "ridiculous" without giving any real substance. I've already responded and countered your ridiculous argument. You're simply repeating something you already wrote without any further comment.

Anyway I got to go run a seminar now, so I'm going to leave this ridiculous discussion and maybe go laugh at this thread with colleagues.

Yeah, I bet you'll get lots of laughs at this thread with your colleagues...I know I will with mine.

Tonio
hochs
#16
Feb25-12, 01:01 PM
P: 39
Yea, I think our colleagues will laugh at how little substance you put into your arguments. I think it's clear at this point that you only know how to rant and don't really understand how debate works. You refuse to read my perfectly sound responses to your claims (which you only repeat like a cry-baby without giving any support).

I'll make it easy for you (who obviously lacks essential skills in communication) by giving you a big picture overview of what is happening, starting with the OP's question:

OP: Hi, I would like to know how people in general find integral closures.

hochs: ok, in the case of global fields, we make use of discriminants and differents. You may not know what these terms mean, but you can easily look them up in any basic algebraic number theory texts. For what it's worth, I'll take my time and write down what may be useful to you after you have read them. This will basically be a guide so that you know where to start.

(hochs spends time carefully writing up outlines and the relevant terms/jargons so that the OP has a guide on where to look).

disregardthat: Hey, for integral closure of Z in Q(sqrt(2)), you just do it by hand. (hochs of course mentioned it, but I'll stipulate it again. no problem!)

DonAntonio: OMG these stuff is too hard! I can't understand! (Riles at poor hochs).

hochs: re-iterates his points, and clarifies the OP's wording of the question to DonAntonio.

DonAntonio: OMG these stuff is too hard! I can't understand! (Riles at poor hochs).

DonAntonio: OMG these stuff is too hard! I can't understand! (Riles at poor hochs).

I'll add one more point here:

DonAntonio, when you're doing real math, you can't always expect to be spoon-fed the details of simple calculations. Your advisor (I'm assuming you're at most an undergrad, since you clearly don't seem to understand how mathematicians learn materials) will not give you all the details in the world. Often times you will hear lots of terms that may seem over your head, you might feel dejected at first, but you go home and look up/put in efforts into understanding the relevant concepts and how they're used. Welcome to Math.

-hochs, Ph.D, p-adic uniformization of shimura varieties.
hochs
#17
Feb25-12, 01:20 PM
P: 39
Look, I can't really careless for a forum of this caliber, so I quit this forum long ago - unfortunately the forum system decided to notify me by an email. I could have easily chosen to ignore, but I took the bite for your benefit. If you aren't willing to read, then I no longer find the need to stay on this forum/thread any longer. I have much better things to do with my career than trying to argue with a troll. I read your argument. I responded to it. Will you do the same or will you just continue to rant and allude to only ad hominem?
DonAntonio
#18
Feb25-12, 02:44 PM
P: 606
Quote Quote by hochs View Post
Yea, I think our colleagues will laugh at how little substance you put into your arguments. I think it's clear at this point that you only know how to rant and don't really understand how debate works. You refuse to read my perfectly sound responses to your claims (which you only repeat like a cry-baby without giving any support).

I'll make it easy for you (who obviously lacks essential skills in communication) by giving you a big picture overview of what is happening, starting with the OP's question:

OP: Hi, I would like to know how people in general find integral closures.

hochs: ok, in the case of global fields, we make use of discriminants and differents. You may not know what these terms mean, but you can easily look them up in any basic algebraic number theory texts. For what it's worth, I'll take my time and write down what may be useful to you after you have read them. This will basically be a guide so that you know where to start.

(hochs spends time carefully writing up outlines and the relevant terms/jargons so that the OP has a guide on where to look).

disregardthat: Hey, for integral closure of Z in Q(sqrt(2)), you just do it by hand. (hochs of course mentioned it, but I'll stipulate it again. no problem!)

DonAntonio: OMG these stuff is too hard! I can't understand! (Riles at poor hochs).

hochs: re-iterates his points, and clarifies the OP's wording of the question to DonAntonio.

DonAntonio: OMG these stuff is too hard! I can't understand! (Riles at poor hochs).

DonAntonio: OMG these stuff is too hard! I can't understand! (Riles at poor hochs).

I'll add one more point here:

DonAntonio, when you're doing real math, you can't always expect to be spoon-fed the details of simple calculations. Your advisor (I'm assuming you're at most an undergrad, since you clearly don't seem to understand how mathematicians learn materials) will not give you all the details in the world. Often times you will hear lots of terms that may seem over your head, you might feel dejected at first, but you go home and look up/put in efforts into understanding the relevant concepts and how they're used. Welcome to Math.

-hochs, Ph.D, p-adic uniformization of shimura varieties.


Hochs, thanx for adding to the Department's laughs.


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