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Math Help - is ideal prime?

  1. #1
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    is ideal prime?

    Is ideal (1+3i, 3+11i, -1-11i) prime in \mathbb{Z}[i]?

    \mathbb{Z}[i] is Euclidean doman, so I have to check whether (1+3i, 3+11i, -1-11i) is an irreducible element?
    How do I do this?

    Thank you for your help.
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  2. #2
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    Here's what I did so far:

    I try to find GCD (1+3i, 3+11i, -1-11i)
    .

    Firstly, I find GCD (1+3i, 3+11i)


    N(1+3i)=10


    N(3+11i)=130

    3+11i=(1+3i)*4 -1-i

    1+3i=(-1-i)*(-2-i)+0

    GCD (1+3i, 3+11i)=-1-i

    Next,
    GCD (-1-i, -1-11i)
    -1-11i=(-1-i)*(6-5i)+0
    So
    GCD (-1-i, -1-11i)=-1-i

    Now I check is (-1-i) irreducible.

    N(-1-i)=1+1=2
    2=(a^2+b^2)(c^2+d^2)

    (a^2+b^2)\in\{+-1, +-2\}, (c^2+d^2)\in\{+-1, +-2\}

    So in any case, one of them is +-1, which means that (-1-i) is irreducible, so (1+3i, 3+11i, -1-11i) is generated by an irreducible element so it is prime.


    I'm pretty sure I made mistakes, could someone please correct me?
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  3. #3
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    Quote Originally Posted by georgel View Post
    Here's what I did so far:

    I try to find GCD (1+3i, 3+11i, -1-11i) .

    Firstly, I find GCD (1+3i, 3+11i)

    N(1+3i)=10

    N(3+11i)=130

    3+11i=(1+3i)*4 -1-i

    1+3i=(-1-i)*(-2-i)+0

    GCD (1+3i, 3+11i)=-1-i

    Next, GCD (-1-i, -1-11i)
    -1-11i=(-1-i)*(6-5i)+0 (*)
    So GCD (-1-i, -1-11i)=-1-i

    Now I check is (-1-i) irreducible.

    N(-1-i)=1+1=2
    2=(a^2+b^2)(c^2+d^2)

    (a^2+b^2)\in\{+-1, +-2\}, (c^2+d^2)\in\{+-1, +-2\}

    So in any case, one of them is +-1, which means that (-1-i) is irreducible, so (1+3i, 3+11i, -1-11i) is generated by an irreducible element so it is prime.


    I'm pretty sure I made mistakes, could someone please correct me?
    it's all correct except for (*): we actually have -1-11i=(-1-i)(6+5i). also it's easier to work with a+bi instead of -a-bi when you're working in an ideal.

    also note that if N(a+bi) is a prime number, then a+bi is always irreducible.
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  4. #4
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    Quote Originally Posted by georgel View Post
    Is ideal (1+3i, 3+11i, -1-11i) prime in \mathbb{Z}[i]?

    \mathbb{Z}[i] is Euclidean doman, so I have to check whether (1+3i, 3+11i, -1-11i) is an irreducible element?
    How do I do this?

    Thank you for your help.
    We can factor into irreducibles:
    1+3i = (1+i)(2+i)
    3+11i = -i(1+i)(1+2i)(2+3i)
    -1-11i = -i(1+i)(5-6i)

    Notice that 1+i is a common irreducible factor so (1+3i,3+11i,-1-11i) = (1+i).

    If you are unsure how I got those factorizations I can show you.
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    it's all correct except for (*): we actually have -1-11i=(-1-i)(6+5i). also it's easier to work with a+bi instead of -a-bi when you're working in an ideal.

    also note that if N(a+bi) is a prime number, then a+bi is always irreducible.
    Thank you! I'm pretty happy with the fact that that's all I messed up )

    Quote Originally Posted by ThePerfectHacker View Post
    We can factor into irreducibles:
    1+3i = (1+i)(2+i)
    3+11i = -i(1+i)(1+2i)(2+3i)
    -1-11i = -i(1+i)(5-6i)

    Notice that 1+i is a common irreducible factor so (1+3i,3+11i,-1-11i) = (1+i).

    If you are unsure how I got those factorizations I can show you.
    If it's not a problem, I'd be really grateful! We haven't done it in class, and I wasn't able to Google anything useful.
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  6. #6
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    Quote Originally Posted by georgel View Post
    If it's not a problem, I'd be really grateful! We haven't done it in class, and I wasn't able to Google anything useful.
    Let me find the factorization for 3+11i with the approach I used. You need to know some stuff about the arithmetic in the Gaussian integers. The first one is that if a+bi with a,b\not = 0 is irreducible then it means be the case that N(a+bi) = a^2+b^2 is a prime number in \mathbb{Z}. Since 3+11i is of the form a+bi where a,b\not = 0 it means all its factors cannot be real or pure imaginary i.e. they must have the same form a+bi where a,b\not = 0. Write 3+11i = \pi_1 \pi_2 ... \pi_k where \pi_j are irreducibles (and are not necessarily distinct) in \mathbb{Z}[i] which are neither purely real or imaginary. We see that N(3+11i) = N(\pi_1) ... N(\pi_k) and so 130 = N(\pi_1) ... N(\pi_k). Notice that 130 = 2\cdot \cdot 5\cdot 13. Also, each N(\pi_j) must be a prime number as explained. This forces k=3 and the \pi_1,\pi_2,\pi_3 can be relabeled so that N(\pi_1) = 2, N(\pi_2) = 5, N(\pi_3) = 13.

    Now we need to use some more arithmetic knowledge about \mathbb{Z}[i]. The only prime (up to associates) with N(\pi_1) = 2 is \pi_1 = 1+i. However, with N(\pi_2)=5 there are two primes (not associates) so that have their norm equal to 5. This is in fact true in general if N(\pi) = p where p is an odd prime in \mathbb{Z} then \pi, \bar \pi both have norm p and they are not associates. This does not happen with the even prime p=2 because \overline{1+i} = 1-i = -(-1+i) = -(i^2+i) = -i(1+i) and so the conjugate of 1+i is associate with 1+i. Conjugates turn out being non-associate if the norm is odd, not even. So we need to find a prime \pi_2 = a+bi so that a^2+b^2 = 5. The good news is that if a Gaussian integer has norm a prime it is automatically irreducible so we just need to find such a,b. It is clear that a+bi = 1+2i works, which tells us that other non-associate prime that has norm 5 is \overline{a+bi} = 1-2i. Same idea with N(\pi_3) = 13 it is easy to see that 2+3i works, which tells us the other non-associate prime is 2-3i.

    Therefore, one of these have to hold:
    3+11i = u(1+i)(1+2i)(2+3i)
    3+11i = u(1+i)(1-2i)(2+3i)
    3+11i = u(1+i)(1+2i)(2-3i)
    3+11i = u(1+i)(1-2i)(2-3i)
    Where u is a unit i.e. u\in \{1,-1,i,-i\} which corrects the primes to propely associated with 3+11i.

    If we perform the computations (for the first one) we get:
    3+11i = u(-11 + 3i), now we see that u=-i.
    Thus, 3+11i = -i(1+i)(1+2i)(2+3i).
    -----

    This computation is not the most pleasant thing to do but I think it is a lot faster than just dividing out the primes and trying to look for factors because this approach eliminates guesswork. It just reduces the problem into cases.

    You try doing 1+11i !
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  7. #7
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    I printed it and am now looking into it.
    Thank you so much for your effort!
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