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Thread: Cosmology

  1. #1006
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    In a sense the task is large, but like a huge jigsaw puzzle. Once one gets a piece in place, it's there. Remembering is easier after one has forgotten it.

    My confusion with the earlier link was that it talked about Z[sqrt(-3)], but that is an example of a ring that is not a Dedekind domain. It shows that they exist. One needs to start with Q, the rationals. Then note that Z is the ring of integers, OQ, in Q. An algebraic integer in an algebraic number field is the root of a monic polynomial with integer coefficients. In the case of Q, the monic polynomial is x - n = 0 where n is an integer. We get the expected ring of integers, Z.

  2. #1007
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    What is true for 3 and -3 is true for all 4n+3 type primes, correct? That would make things a bit easier to organize.

  3. #1008
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    The -3 is a 4n+1 type. If you add 4 to -3 you get 1. The quadratic algebraic integers, those that are roots of monic polynomials of degree 2, separate into two groups depending on mod 4 (except for 2 which is handled separately).

    Two things come to mind:
    (1) How does one tell which are the integers in a quadratic number field, like Q(sqrt(-3)) or Q(sqrt(-5))?
    (2) Is the ring of integers, such as OQ(sqrt(-3)), a unique factorization domain?

    The first question is resolved by checking if the number is congruent to 1 or 3 mod 4. The second is resolved by using ideals, so unique factorization is possible in the Dedekind domains with ideals. I think the number of UFDs for algebraic numbers that are square roots of negative integers are finite, if one does not use ideals. I will have to check that.
    Last edited by YesNo; 03-13-2017 at 11:09 AM.

  4. #1009
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    Yes, only 3 is a 4n+3 type.

    I am confused by this notation: OQ(sqrt(-3)).

    Can you say it in English, please?

    * * * * *

    What I meant to say is that what is true of one 4n+3 number is true of another. Rather, I meant to ask if that is strictly true.

  5. #1010
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    The OQ(sqrt(-3)) is the notation for the ring of integers from the field of algebraic numbers Q(sqrt(-3)). I think the O notation comes from Dedekind. Sometimes that ring of integers can be represented by something like Z[sqrt(-3)], which is a ring of a + b*sqrt(-3) where a and b are in Z or where a and be are what we normally think of as integers, or "rational integers". Sometimes it is not, as in this particular case, which is why I think Z[sqrt(-3)] was used as an example. Because -3 is congruent to 1 mod 4, we also have integers that look like this (a + b*sqrt(-3))/2. So the notation using O gives the ring of integers from a field which could be different from just extending Z by the same algebraic number. In the case of Q, the rationals, OQ is just Z. So for the field Q we don't need new notation because there is no difference from what we would expect the ring of integers to be.

    One other notation that can be confusing: If one is extending a field the notion of "vector space" is used which is a special kind of "module" and one uses parentheses, such as, Q(sqrt(-3)). What makes it special is Q has all of its inverses. If one is extending a general ring (not necessarily a field) one uses the more general concept of a "module" and the notation changes to brackets such as Z[sqrt(-3)]. Here's some discussion of that difference: https://www.quora.com/What-is-the-di...le-over-a-ring
    Last edited by YesNo; 03-14-2017 at 09:44 AM.

  6. #1011
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    Mod 3 there is no object which when squared equals -1 (mod 3), for -1 is equivalent to 2 (mod 3).

    Mod 5 the case is different. We do not even have to adjoin √-1 to our ring, because -1 is already there, since (√-1)2 is equal to 4.

    This simple truth hung me up for quite a while.

    When something snags me I find it almost impossible to move foreward until I have resolved the problem. Finally I have resolved this one, partly at least. I think you are well beyond me now, I hope not out of yelling distance. A new plateau of understanding may now pour over me quickly.

    The article said its quotient field contained all numbers within itself. There is nothing I can square (mod 5) to get 3, however, so it still appears to matter what one is trying to adjoin.

  7. #1012
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    I think it is only mod 4 that is of interest here, not mod 3 or mod 5.

    For quadratic algebraic number fields, that is fields where the numbers are roots of quadratic equations like ax2+bx+c=0, the way to tell what algebraic numbers are integers is given by the rule that if we extend Q by the square root of a negative integer, then if that negative integer, -t, is congruent to 3 mod 4 then the algebraic integers are what we would expect them to be, numbers like a+b*sqrt(-t). If -t is congruent to 1 mod 4 then we have to also include a/2+b/2*sqrt(-t) as integers. This comes from using the quadratic formula to find x = (-b+-sqrt(b2-4ac)/2a. That 2 in the denominator does not cancel out in this case. The a = 1 because that is required for an algebraic number to be an algebraic integer. It has to be monic.

    Don't worry about being out of reach. If I can't explain it, then I don't really understand it well enough and I don't understand this myself all that well. Also, I might have some of this wrong.

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    I have been traveling again, and I will have to do even more later this week or early next. While I am away I have no computer to do research on, so I concentrate on organizing as many details as I can remember on these math subjects, or what I can put down on paper.

    Ideals and field extensions are parts of the classical theory of pure mathematics. I seriously doubt that ideals have found "many," applications outside of conducting more pure research. Also I would be only slightly more surprised if ideals had not already found "some," applications outside of pure mathematics.

    Almost nothing is known about fields beyond quadratic fields. It was only for some quadratic fields that ideals recovered unique factorization. In the future I expect much more to be known about higher order fields. Perhaps the complexity will eventually be unraveled by quantum computers. At such a time more applications would come into being. Some of them might be more reliable and secure encryption techniques, as happened with the congruence theory of Gauss, a math language which did not find its big application outside of pure mathematics for a full 200 years.

    Right now I feel like tackling more math. The harder it gets, the less I feel like diving in, so I had better take advantage of every time I feel like pushing my boundaries.

  9. #1014
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    There is probably a lot that is not known also because the questions haven't been asked, but I don't know what the limits are. I ran into this article on "monogenic" fields which have an example of a cubic field that is not monogenic: https://en.wikipedia.org/wiki/Monogenic_field

    So here is another technical term, "monogenic", and also, "power integral basis". These terms are more pieces in the jigsaw puzzle.

  10. #1015
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    Quote Originally Posted by YesNo View Post
    I think it is only mod 4 that is of interest here, not mod 3 or mod 5.

    For quadratic algebraic number fields, that is fields where the numbers are roots of quadratic equations like ax2+bx+c=0, the way to tell what algebraic numbers are integers is given by the rule that if we extend Q by the square root of a negative integer, then if that negative integer, -t, is congruent to 3 mod 4 then the algebraic integers are what we would expect them to be, numbers like a+b*sqrt(-t). If -t is congruent to 1 mod 4 then we have to also include a/2+b/2*sqrt(-t) as integers. This comes from using the quadratic formula to find x = (-b+-sqrt(b2-4ac)/2a. That 2 in the denominator does not cancel out in this case. The a = 1 because that is required for an algebraic number to be an algebraic integer. It has to be monic.

    Don't worry about being out of reach. If I can't explain it, then I don't really understand it well enough and I don't understand this myself all that well. Also, I might have some of this wrong.
    What I need are a few specific examples worked out with all the algebra. If I can see it just once I can figure out how the discriminant figures into this. The problem with any examples I have seen is that they suddenly introduce new variables to make the job harder. Often I cannot see which step to take. For instance, when they say it is obvious so and so is the minimum equation for so and so, I will not see why or how they got the answer.

    A few examples geared just for me would make everything clear, I am convinced, but that luxury usually does not exist in math.

    Of course I understand that if the discriminant is negative then we intorduce the complex numbers through a field extension. I cannot reproduce the algebra involved. I would have to see it once.
    Last edited by desiresjab; Yesterday at 06:39 PM.

  11. #1016
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    Here is a video describing the discriminant for a quadratic equation: http://www.virtualnerd.com/algebra-1...ant-definition You may already know this.

    If the square root of the discriminant is not an integer, then the root (which could be either a real or a non-real complex number) is not a rational number. That is, it is not in the field of rational numbers, Q. We could extend this field by creating an object that has all of Q plus one of these non-rational roots. That would also be a field, and its structure would be a two-dimensional vector space. One of the dimensions would have 1 as the base and the other would have this non-rational root, r. The base would look like this: (1,r). The field extension of Q is written like this Q(r). If a and b are arbitrary rational numbers, that is elements of Q, then a number in this new field would look like this: a + br.

    Now suppose we don't extend Q but instead extend Z, the ring of integers of Q. Z is not a field, but the extension would be similar and called a module. This is not always a Dedekind domain with ideals forming a unique factorization domain. However, we still could extend Z and see what we get. If we are extending a ring that is not a field the notation would be modified to Z[r], but the idea of the extension is the same. If a and b are integers, that is elements of Z, then a + br would be in Z[r].

    Since the ring of integers OQ(r) may be larger than Z[r], we need some way to tell when they are not the same. The discriminant does this. If the discriminate is congruent to 1 mod 4, then we have to include the root that has a 2 in the denominator which is what the earlier link referenced when it discussed Z[sqrt(-3)].

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