Recent content by hochs

  1. H

    Computing Integral Closure of Rings - Brian's Query

    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.
  2. H

    Computing Integral Closure of Rings - Brian's Query

    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...
  3. H

    Computing Integral Closure of Rings - Brian's Query

    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...
  4. H

    Computing Integral Closure of Rings - Brian's Query

    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...
  5. H

    Computing Integral Closure of Rings - Brian's Query

    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...
  6. H

    Computing Integral Closure of Rings - Brian's Query

    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...
  7. H

    Computing Integral Closure of Rings - Brian's Query

    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...
  8. H

    Solving Quadratic Field Problems with Quick Tips - PhysicsForums

    What book are you reading anyway? And you're trying to learn algebraic number theory without solid background in abstract algebra, particularly galois theory. Of course you run into troubles. If you're not doing these for homework and have no time-constraint, then I suggest you get some...
  9. H

    Congruence Class Relationships in \mathbb{Q}[\sqrt{3}]

    Alright I'm wasting way too much time on this triviality. Here's the complete ****ing proof, leave me alone now: if x = 1 (mod lambda) then x = 1 + a*(1 - w) + b*(1 - w)^2 for some integers a,b. Then x-1 = a*(1 - w) + b*(1 - w)^2 is divisible by 1-w clearly. x-w = 1 - w + a*(1 - w) + b*(1 -...
  10. H

    Congruence Class Relationships in \mathbb{Q}[\sqrt{3}]

    Just wow... are you a native English speaker? I told you, you need to show separately that x-1, x-w, and x-w^2 are all divisible by lambda. That's 3 statements to show. my god..
  11. H

    Congruence Class Relationships in \mathbb{Q}[\sqrt{3}]

    You just need to show that each of x-1, x-w, and x-w^2 are divisible by lambda to show that x^3 - 1 = (x-1)(x-w)(x-w^2) is divisible by lambda^3. The above sentence should've been super clear. If not, think more. So it suffices to show that each of x-1, x-w, and x-w^2 are divisible by...
  12. H

    Congruence Class Relationships in \mathbb{Q}[\sqrt{3}]

    I didn't. Just show that x-1, x-w, and x-w^2 are each divisible by 1-w. I can't give more hints than this. I feel like I'm doing your homework 100%.
  13. H

    Congruence Class Relationships in \mathbb{Q}[\sqrt{3}]

    if x = 1 + a*(1 - w) + b*(1 - w)^2 then x - 1 = a*(1 - w) + b*(1 - w)^2 = (1-w)*(a + b*(1 - w)) is clearly divisible by 1-w = lambda. It's exactly the same trivial algebraic verifications for the others.
  14. H

    Congruence Class Relationships in \mathbb{Q}[\sqrt{3}]

    For the first one, if x = 1 (mod lambda) then x = 1 + a*(1 - w) + b*(1 - w)^2. Then x-1, x-w, x-w^2 are all divisible by lambda, so clearly x^3 - 1 = (x-1)(x-w)(x-w^2) is divisible by lambda^3. Do the same for part B. Part C is trivial.
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