# Integral closure

by brian_m.
Tags: closure, integral
 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
 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 $$p$$ 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.
 Sci Advisor P: 1,670 $$y= p+q\sqrt{2}$$ 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).
P: 39

## Integral closure

 Quote by Jarle $$y= p+q\sqrt{2}$$ 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).
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).
P: 606
 Quote by 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.

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!
HW Helper
P: 2,020
 Quote by 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!
While I absolutely agree with you, was it necessary to resurrect this two-year-old thread just to say that?
P: 606
 Quote by morphism 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.
P: 39
 Quote by 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).
P: 39
 Quote by hochs But for quadratic case, you can do this by **hand** and the ring of integers in $$\mathbb{Q}(\sqrt{2})$$ is just $$\mathbb{Z}[\sqrt{2}]$$.

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.
 HW Helper Sci Advisor 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.
P: 39
 Quote by 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.
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.
 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?
P: 606
 Quote by hochs 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
P: 39
 Quote by 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
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.
P: 606
 Quote by 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.

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

Tonio