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Math Help - Gaussian Integers

  1. #1
    Super Member Deadstar's Avatar
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    Gaussian Integers

    In this part, you may assume any facts about the factorisation theory of Z[i], the ring of Gaussian Integers, and of Z provided that you state clearly the properties that you are using.
    Let p be an integer prime for which there is an element a in Z with a^2 + 1 = p. Write down a factorisation of p in Z[i] and show that it is a proper factorisation (that is, neither factor is a unit of Z[i]). Deduce that p is not a prime in the Gaussian integers. Show that the factors of p that you obtained are primes in Z[i].

    Thoughts...

    p = (a + i)(a - i)
    I'm not certain about this but i think that a prime cant be a unit? So in Z that would be \pm 1?
    If so then a \neq 0 hence since the only units in Z[i] are 1,-1,i ,-i. Neither factor is not a unit, i.e. proper.

    p is not prime since p doesn't divide either factor. This is because a \pm i is imaginary and p is real.

    Showing a \pm i are primes in Z[i]...
    a + bi is prime if a^2 + b^2 is an integer prime of the form 4k+1. So how do i show a^2 + 1 is of the form 4k+1?

    Actually a thought has just occurred... Since a^2 + 1 is prime, a^2 has to be even, i.e. a is even. So let a be of the form 2n. Then we have (2n)^2 + 1 = 4k + 1 so 4n^2 = 4k, hence k = n^2? So... ?

    One final note... a^2 + 1 could be even, but only if a=1, since 2 is the only even prime. But 1+i and 1-i are primes in Z[i] so that takes care of the a=odd case.
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    In this part, you may assume any facts about the factorisation theory of Z[i], the ring of Gaussian Integers, and of Z provided that you state clearly the properties that you are using.
    Let p be an integer prime for which there is an element a in Z with a^2 + 1 = p. Write down a factorisation of p in Z[i] and show that it is a proper factorisation (that is, neither factor is a unit of Z[i]). Deduce that p is not a prime in the Gaussian integers. Show that the factors of p that you obtained are primes in Z[i].

    Thoughts...

    p = (a + i)(a - i)
    I'm not certain about this but i think that a prime cant be a unit? So in Z that would be \pm 1?
    If so then a \neq 0 hence since the only units in Z[i] are 1,-1,i ,-i. Neither factor is not a unit, i.e. proper.

    p is not prime since p doesn't divide either factor. This is because a \pm i is imaginary and p is real.

    Showing a \pm i are primes in Z[i]...
    a + bi is prime if a^2 + b^2 is an integer prime of the form 4k+1. So how do i show a^2 + 1 is of the form 4k+1?

    Actually a thought has just occurred... Since a^2 + 1 is prime, a^2 has to be even, i.e. a is even. So let a be of the form 2n. Then we have (2n)^2 + 1 = 4k + 1 so 4n^2 = 4k, hence k = n^2? So... ?

    One final note... a^2 + 1 could be even, but only if a=1, since 2 is the only even prime. But 1+i and 1-i are primes in Z[i] so that takes care of the a=odd case.
    Define N: \mathbb{Z}[i]\to \mathbb{Z}[i] by N(\alpha) = \alpha \bar \alpha , so if \alpha=a+bi then N(\alpha) = a^2+b^2.
    It is easy to show that \alpha is a unit if and only if N(\alpha) = 1.
    To show that a\pm 1 is not a unit just notice that N(a\pm 1) = (a\pm 1)^2 > 1 (if |a|> 1).

    If N(\alpha) = p for a prime then \alpha must be prime. Because otherwise \alpha = \beta \gamma for non-units and so p=N(\alpha) = N(\beta)N(\gamma). Which is a problem because p is prime.
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