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Hi TL
So I'm self studying number theory because my high school doesn't offer anything beyond calculus, and right now I'm learning about class groups of various number fields. I know how class group is defined, but actually computing some of them might do me good. Here's a particular example whose class group I've been trying to find but I'm a bit stuck.
Let Q be the field of rationals. What is the class group of the ring of integers of Q(17^(1/3))? Clearly the ring of integers (i.e. the integral closure of Z in the field Q(17^(1/3))) contains Z[17^(1/3)], but this thing has discriminant 3^3 * 17^2 over Z, and 3 splits into two distinct prime ideals, so this can't be all of the integral closure.
So the first step here is to find the ring of integers, and this clearly contains Z[17^(1/3)], but its discriminant is 3^3 * 17^2 as one can quickly calculate.
But it's too large, so it can't be all of the ring of integers.
Any help?
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well disguised brag blog
User was warned for this post
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Just to be clear, you're talking about the set of rational numbers extended to include the cube root of 17, right?
And this term class group is something I don't recognize. Probably because I'm not into Number Theory. Is this what you're talking about? http://planetmath.org/encyclopedia/InverseIdeal.html
Also, kudos for making it this far into the subject on your own before even getting out of high school. This stuff is what senior mathematics majors do.
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Finding the class group is fun, but it's usually the ring of integers that is pain in the ass to find (as is the case here!).
The discriminant you have is right, and at the prime 17 Q_17(17^(1/3)) is totally ramified of degree 3 over Q_17 (the 17-adic rationals), so Z[17^(1/3)] in fact generates the ring of integers over 17-adic rationals.
For the prime 3, it's as you said. We need to find some element of Q(17^(1/3)) that is integral over Z. You can sort of do this by writing down the elements of Q(17^(1/3)) as (a + b * 17^(1/3) + c * 17^(2/3)) and just doing some brute force minimal polynomial calculations. Well, I just did it and got (1 + 17^(1/3))^2 / 3 as something in Q(17^(1/3)) that is integral over Z, and not in Z[17^(1/3)].
So I presume Z[(1 + 17^(1/3))^2 / 3] is the ring of integers. The minimal polynomial of this (1 + 17^(1/3))^2 / 3 is p(x) = x^3 - x^2 - 11x - 12 over Q, and (I just asked WolframAlpha to compute the norm of p'((1 + 17^(1/3))^2 / 3) = discriminant for me just now) this has discriminant 3 * 17^2. We took care of the prime 17 (totally ramified there, no problem), so this has to be the ring of integers.
The question is of course how to find class group. Usually this is how you find it (and I just computed it to be trivial, i.e. the ring is a PID, but I'll let you work on this):
Find the Minkowski bound - this is the the bound where every fractional ideal of the class group can be multiplied by principal ideal so that its norm is less than the Minkowski Bound. If you calculate correctly, this should be 8 (this is what I got).
So you only need to check those prime ideals lying above 2, 3, 5, and 7. Since you know the ring of integers now, which is Z[(1 + 17^(1/3))^2 / 3] ~= Z[x]/(x^3 - x^2 - 11x - 12), you know exactly how 2,3,5, and 7 factorize into prime ideals over this ring. In this case you can actually check that those are all principals by finding right elements in the ideals with norms 2, 3, 5, and 7 respectively.
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On October 21 2010 12:25 Galois wrote:
Also, kudos for making it this far into the subject on your own before even getting out of high school. This stuff is what senior mathematics majors do.
Double kudos -- this is not even undergraduate level math at most universities, I'm just getting into this sort of number theory for the first time myself, and I'm in my second year of graduate school.
Edit: the above post is way more useful than what I could contribute.
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On October 21 2010 12:25 Galois wrote:Just to be clear, you're talking about the set of rational numbers extended to include the cube root of 17, right? And this term class group is something I don't recognize. Probably because I'm not into Number Theory. Is this what you're talking about? http://planetmath.org/encyclopedia/InverseIdeal.html
I'm pretty sure he wants to find class group, which is what you linked modulo the principal ideals. So class group (fractional ideals / principal ideals) measures how "far" the ring is away from being a PID.
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What a fun problem! I was gonna post a solution but I'm feeling a little tired, I'm gonna take a quick six year nap and when I wake up I'll get back to you
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wtf how do you self teach yourself such advanced math? did you buy a book or are you just learning stuff from the internet?
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On October 21 2010 14:04 madnessman wrote:
wtf how do you self teach yourself such advanced math? did you buy a book or are you just learning stuff from the internet?
I don't see why not, if you're sufficiently interested.
As for how, sure you can get books in number theory and peruse them on your own. There are also online communities where one can find resources from.
If you're interested in algebraic number theory, as this young fellow seems to be, there are many good sources nowadays, especially with reprints of Cassels & Frohlich by LMS being made available again (finally).
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^ yes, I'm actually using cassels & frohlich
Can you tell me more how knowing the ring of integers helps me find how primes split above?
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On October 22 2010 07:54 LastPrime wrote:
^ yes, I'm actually using cassels & frohlich
Can you tell me more how knowing the ring of integers helps me find how primes split above?
Ok, in most cases your ring of integers will be generated by one element, and in this case I proved above that it's generated by (1 + 17^(1/3))^2 / 3, which has minimal polynomial x^3 - x^2 - 11x - 12, so the ring of integers is isomorphic to Z[x]/(x^3 - x^2 - 11x - 12) = O. If you want to consider how a rational prime p factors in O, just mod out by the ideal generated by p and use chinese remainder theorem:
O/p ~= Z/p[x]/(x^3 - x^2 - 11x - 12), and the factors of p in O correspond precisely to the factors of x^3 - x^2 - 11x - 12 mod p.
For example, x^3 - x^2 - 11x - 12 = x(x+1)^2 (mod 3), so the prime ideal (3) factors in O into prime ideals (3) = (3, \alpha) * (3, (\alpha + 1))^2 where \alpha = (1 + 17^(1/3))^2 / 3.
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EPT
