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Thread: is ideal prime?

  1. #1
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    is ideal prime?

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

    $\displaystyle \mathbb{Z}[i]$ is Euclidean doman, so I have to check whether $\displaystyle (1+3i, 3+11i, -1-11i)$ 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$\displaystyle (1+3i, 3+11i, -1-11i)$
    .

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


    $\displaystyle N(1+3i)=10$


    $\displaystyle N(3+11i)=130$

    $\displaystyle 3+11i=(1+3i)*4 -1-i$

    $\displaystyle 1+3i=(-1-i)*(-2-i)+0$

    GCD$\displaystyle (1+3i, 3+11i)=-1-i$

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

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

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

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

    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$\displaystyle (1+3i, 3+11i, -1-11i)$ .

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

    $\displaystyle N(1+3i)=10$

    $\displaystyle N(3+11i)=130$

    $\displaystyle 3+11i=(1+3i)*4 -1-i$

    $\displaystyle 1+3i=(-1-i)*(-2-i)+0$

    GCD$\displaystyle (1+3i, 3+11i)=-1-i$

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

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

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

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

    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 $\displaystyle -1-11i=(-1-i)(6+5i).$ also it's easier to work with $\displaystyle a+bi$ instead of $\displaystyle -a-bi$ when you're working in an ideal.

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

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

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

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

    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 $\displaystyle -1-11i=(-1-i)(6+5i).$ also it's easier to work with $\displaystyle a+bi$ instead of $\displaystyle -a-bi$ when you're working in an ideal.

    also note that if $\displaystyle N(a+bi)$ is a prime number, then $\displaystyle a+bi$ 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:
    $\displaystyle 1+3i = (1+i)(2+i)$
    $\displaystyle 3+11i = -i(1+i)(1+2i)(2+3i)$
    $\displaystyle -1-11i = -i(1+i)(5-6i)$

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

    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 $\displaystyle 3+11i$ with the approach I used. You need to know some stuff about the arithmetic in the Gaussian integers. The first one is that if $\displaystyle a+bi$ with $\displaystyle a,b\not = 0$ is irreducible then it means be the case that $\displaystyle N(a+bi) = a^2+b^2$ is a prime number in $\displaystyle \mathbb{Z}$. Since $\displaystyle 3+11i$ is of the form $\displaystyle a+bi$ where $\displaystyle a,b\not = 0$ it means all its factors cannot be real or pure imaginary i.e. they must have the same form $\displaystyle a+bi$ where $\displaystyle a,b\not = 0$. Write $\displaystyle 3+11i = \pi_1 \pi_2 ... \pi_k$ where $\displaystyle \pi_j$ are irreducibles (and are not necessarily distinct) in $\displaystyle \mathbb{Z}[i]$ which are neither purely real or imaginary. We see that $\displaystyle N(3+11i) = N(\pi_1) ... N(\pi_k)$ and so $\displaystyle 130 = N(\pi_1) ... N(\pi_k)$. Notice that $\displaystyle 130 = 2\cdot \cdot 5\cdot 13$. Also, each $\displaystyle N(\pi_j)$ must be a prime number as explained. This forces $\displaystyle k=3$ and the $\displaystyle \pi_1,\pi_2,\pi_3$ can be relabeled so that $\displaystyle N(\pi_1) = 2, N(\pi_2) = 5, N(\pi_3) = 13$.

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

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

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

    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 $\displaystyle 1+11i$ !
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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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