Integral closure

1. Oct 21, 2010

brian_m.

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

2. Oct 25, 2010

hochs

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 $$p$$ over $$\mathbb{Q}_p$$, $$p$$ 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 $$\mathbb{Q}(\sqrt{2})$$ is just $$\mathbb{Z}[\sqrt{2}]$$.

So for example, try to compute the ring of integers in $$\mathbb{Q}(17^{1/3})$$. 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.

3. Oct 26, 2010

disregardthat

$$y= p+q\sqrt{2}$$ is a general element of $$\mathbb{Q}(\sqrt{2})$$ 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).

4. Oct 26, 2010

hochs

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).

5. Feb 25, 2012

DonAntonio

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!

6. Feb 25, 2012

morphism

While I absolutely agree with you, was it necessary to resurrect this two-year-old thread just to say that?

7. Feb 25, 2012

DonAntonio

Oops, my bad! I'm still not used to this forum and I completely missed the posts' date.

8. Feb 25, 2012

hochs

9. Feb 25, 2012

hochs

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.

Last edited: Feb 25, 2012
10. Feb 25, 2012

morphism

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.

11. Feb 25, 2012

hochs

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.

12. Feb 25, 2012

hochs

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.

13. Feb 25, 2012

DonAntonio

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

14. Feb 25, 2012

hochs

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.

15. Feb 25, 2012

DonAntonio

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

Tonio

16. Feb 25, 2012

hochs

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.

Last edited: Feb 25, 2012
17. Feb 25, 2012

hochs

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?

18. Feb 25, 2012

DonAntonio

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

19. Feb 25, 2012

DonAntonio

Ad hominem? Well, that's another thing you don't know much about since the only one calling names and offending others has been you.

It never matters, of course, as you already quit the forum "long ago".
Good ridance.

20. Feb 25, 2012

hochs

Ok, clear case of a troll.

Time to print this and post it on the department walls here. By the way, we already had a lot of laugh ;)

Anyway, I'm out. waste of my time.