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Thread: Cosmology

  1. #901
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    One thing we can be sure of: the concept of left and right ideals is applicable only for non commutitive rings. In a commutative ring such as a modulus ring, left and right ideals are the same, since x times r and r times x are not different. So any time they start tlking about left and right ideals or splitting, you know they are talking about some non commutitive object.

    Everywhere I turn are statements I do not understand. All I can do is take them one at a time, putting them on the hold list until I can get to them. For instance, why and how there are exactly 21 different quadratic fields is still quite a mystery to me.

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    Since ideals at heart are instances of multiples, we always need to be able to see how any ideal is a multiple. Any time that becomes clear, we have understood the ideal. Even with Carmichael numbers the quest can be reduced to finding and understanding what this multiple is of. What is the generator and what does it leave in its tracks?

    In the case of Complex numbers there have to be two generators, as I see it, one for the integral x-axis and one for the imaginary y-axis. Even if they do not show it in many diagrams, the y-axis is really the i-axis, 1i, 2i, 3i etc. An example would be the lattice diagram in the following link:

    http://mathworld.wolfram.com/Ideal.html

  3. #903
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    Quote Originally Posted by desiresjab View Post
    In the case of Complex numbers there have to be two generators, as I see it, one for the integral x-axis and one for the imaginary y-axis...http://mathworld.wolfram.com/Ideal.html
    I am not too sure about my statement here. I mean, I still think there are two generators (are they called units, or not?), but I am not sure they are assigned to each axis. (1+i) is not even on an axis.

    * * * * *

    My oversight recently that not every Gaussian integer is a root of an equation, or something like that, was to due to a mix up with something I read. Even as I wrote it I knew it must be wrong, but I wrote it anyway because I thought I had read it on a long-time-without-sleep binge just prior to that. I must apologize for that embarrassment. I don't really know what I confused it with either.

    * * * * *

    On to the more interesting question of ways to factor 5 in the Complex numbers.

    We already know √4+i)(√4-i)=4-√4 i+√4 i+1=5.

    To write (2+i)(2-i)=4-2 i+2i i+1=5, is essentially a trivial change from √4 to 2, so is cheating and not a valid different factorization. But how about this?:

    (1+2i)(1-2i) = 1+-2i+2i-4(i)2 = 1+4 = 5

    That is definitely a different factorization of 5. Are there others? Maybe. I have not validated your claim yet. All it takes is the above to show lack of unique factorization. I am simply curious if there are more, or infinitely many, as you said, I believe.

    Hmmmm... I think it may be the case that these factorizations display a type of symmetry that is important later on, where exchanging a and b in the Gaussian integer does not change the result of an equation containing them. What we did in the factorizations above is exchange a and b.

    But somehow I feel I can make it work for 7 as well, which is supposed to be a Gaussian prime. Let's take a look.

    7 = √6+i)(√6-i)=6-√6 i+√6 i+1=7

    Doesn't exchanging a and b have to work?

    (1+√6i)(1-√6i)=1-√6i+√6i+6=7. Yes, it works.

    Now I really am confused, I thought 7 was a Gaussian prime, but I have found two different ways of factoring it that seem distinct. The method should work on any Gaussian integer, in fact. These two factorizations do not seem trivially different. What is going on???????
    Last edited by desiresjab; Yesterday at 05:05 PM.

  4. #904
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    Here is another factorization for 7.

    (2+√3i)(2-√3i)=4+3=7.

    Maybe those factors are not irreducible, so this factorization would not count. I do not know for sure. But there it is anyway, another factorization of 7. I could not find another one for 5.

  5. #905
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    Quote Originally Posted by desiresjab View Post
    On to the more interesting question of ways to factor 5 in the Complex numbers.

    We already know √4+i)(√4-i)=4-√4 i+√4 i+1=5.
    There are infinitely many ways to factor 5 in the complex numbers. Let c be a complex number. Since the complex numbers are a field, 5/c = d is a complex number. Multiply both sides by c and get 5 = cd.

    However, if one restricts attention to only the Gaussian integers https://en.wikipedia.org/wiki/Gaussian_integer, that is complex numbers like a + bi where a and b are in Z, then 5 should have a unique factorization into irreducibles (primes).

    Quote Originally Posted by desiresjab View Post
    To write (2+i)(2-i)=4-2 i+2i i+1=5, is essentially a trivial change from √4 to 2, so is cheating and not a valid different factorization. But how about this?:

    (1+2i)(1-2i) = 1+-2i+2i-4(i)2 = 1+4 = 5

    That is definitely a different factorization of 5. Are there others? Maybe. I have not validated your claim yet. All it takes is the above to show lack of unique factorization. I am simply curious if there are more, or infinitely many, as you said, I believe.
    However, you seem to have two different factorizations above. I will have to check this further. Also, I am not sure why the Gaussian integers are a principal ideal domain which would make it be a unique factorization domain. So I will look up some proof for that as well.

    Quote Originally Posted by desiresjab View Post
    Hmmmm... I think it may be the case that these factorizations display a type of symmetry that is important later on, where exchanging a and b in the Gaussian integer does not change the result of an equation containing them. What we did in the factorizations above is exchange a and b.

    But somehow I feel I can make it work for 7 as well, which is supposed to be a Gaussian prime. Let's take a look.

    7 = √6+i)(√6-i)=6-√6 i+√6 i+1=7

    Doesn't exchanging a and b have to work?

    (1+√6i)(1-√6i)=1-√6i+√6i+6=7. Yes, it works.

    Now I really am confused, I thought 7 was a Gaussian prime, but I have found two different ways of factoring it that seem distinct. The method should work on any Gaussian integer, in fact. These two factorizations do not seem trivially different. What is going on???????
    In the case of 7, note that √6 is not an integer, that is, an element of Z. Therefore √6+i and √6-i are not Gaussian integers, but complex numbers. In the field of complex numbers 7 is a unit and everything divides it, but not in the ring of Gaussian integers.

  6. #906
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    Quote Originally Posted by YesNo View Post
    There are infinitely many ways to factor 5 in the complex numbers. Let c be a complex number. Since the complex numbers are a field, 5/c = d is a complex number. Multiply both sides by c and get 5 = cd.

    However, if one restricts attention to only the Gaussian integers https://en.wikipedia.org/wiki/Gaussian_integer, that is complex numbers like a + bi where a and b are in Z, then 5 should have a unique factorization into irreducibles (primes).



    However, you seem to have two different factorizations above. I will have to check this further. Also, I am not sure why the Gaussian integers are a principal ideal domain which would make it be a unique factorization domain. So I will look up some proof for that as well.



    In the case of 7, note that √6 is not an integer, that is, an element of Z. Therefore √6+i and √6-i are not Gaussian integers, but complex numbers. In the field of complex numbers 7 is a unit and everything divides it, but not in the ring of Gaussian integers.
    Oh, yes, that is correct, they are complex numbers, not Gaussian integers.

  7. #907
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    On your next to last comment in your last post, I comment: All principal ideals are bsed on the multiple concept, if I have my reading straight this time. I have a sneaking suspicion that there are ideals based on other properties than simply "being a multiple of." I have a hunch Carmichael numbers might be non-principal ideals. But I seem to be about 50-50 on the hunches these days.

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    An ideal is also an additive subgroup. From that perspective, if there is more than one generator then one has to also consider not only the multiples of each of the generators, but also sums of anything generated from those two or more generators. One could generate the even numbers in the integers Z by using the principle ideal generated by 2 or the non-principal ideal generated by (2,4). However, those ideals are the same.

    Your earlier observation about factoring 5 in two different ways still has me puzzled.

    Edit: I think this resolves my earlier puzzlement:

    The two factorizations of 5 provided earlier are the same up to units in the Gaussian integers. So unique factorization, up to units, still holds.

    To see the significance of this, look at 10 in the integers. This factors as 10 = 2*5 but also as 10 = (-2)*(-5). Those are two different factorizations but not up to units since if I multiply (-2)*(-5) by 1 = (-1)*(-1) then (-2)*(-5)=2*5.

    In the case of Gaussian integers there are four units: 1, -1, i, -i. If I multiply 1+2i by 1 = (-i)(i), I get (-i)(i)(1+2i) = i(-i+2) = i(2-i). If I multiply 1-2i by 1 = (-i)(i), I get (-i)(i)(1-2i) = -i(i+2) = -1(2+i). So with 1 = (i)(-i), I get 5 = (2+i)(2-i) = 1(2+i)(2-i) = (i)(-i)(2+i)(2-i) = i(1-2i)(2-i) = i(2-i)(1-2i) = (1+2i)(1-2i).

    So the two factorizations are the same up to units.
    Last edited by YesNo; Today at 08:56 AM.

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