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Math Help - Prime/Maximal Ideals

  1. #1
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    Prime/Maximal Ideals

    I'm working on prime an maximal ideals. My partner and I are studying for our final exam and got conflicting answers.

    The question was to find all of the prime and maximal ideals of \mathbb Z_7. My answer was that because a finite integral domain is a field, the prime and maximal ideals coincide, but that there are no prime and maximal ideals for \mathbb Z_7.

    As for \mathbb Z_3 \times \mathbb Z_5 , what are the prime and maximal ideals, and more importantly, how in the world do we know that we have found them all?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by DanielThrice View Post
    I'm working on prime an maximal ideals. My partner and I are studying for our final exam and got conflicting answers.

    The question was to find all of the prime and maximal ideals of \mathbb Z_7. My answer was that because a finite integral domain is a field, the prime and maximal ideals coincide, but that there are no prime and maximal ideals for \mathbb Z_7.
    Right, since \mathbb{Z}_7 is a field all is easy.


    As for \mathbb Z_3 \times \mathbb Z_5 , what are the prime and maximal ideals, and more importantly, how in the world do we know that we have found them all?
    You may be overthinking this. You can clearly check for example that \mathbb{Z}_3\times\{0\} and \{0\}\times\mathbb{Z}_5 which are prime? Are they maximal? Well suppose that \mathbb{Z}_3\times\{0\}\subset I\subseteq \mathbb{Z}_3\times\mathbb{Z}_5 then check that \pi_2\left(I\right) is an ideal and since \mathbb{Z}_5 is a field.....


    etc.
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  3. #3
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    to sharpen it even further, suppose p,q are distinct primes. what can an ideal of Zp x Zq possibly be? remember Zp x Zq is isomorphic to Zpq, so an ideal has to be a subgroup of the additive group. but (Zpq,+) is cyclic, so any proper subgroup has to be generated by some element of Zpq that doesn't generate the whole group, that is, either n = kp, or n = rq. k <q, so kp is co-prime to q (and similarly r is co-prime to p), so the only non-trivial proper ideals of Zpq are (p) and (q), so the only non-trivial proper ideals of Zp x Zq are Zp x {0} and {0} x Zq.

    it's actually enlightening to see what happens in Z2 x Z3: it's easy to see that (1,1) is a generator, and we can make the assignment:

    (0,0) --> 0
    (1,1) --> 1
    (0,2) --> 2
    (1,0) --> 3
    (0,1) --> 4
    (1,2) --> 5, by considering multiples of (1,1). the inverse map sends k --> (k mod 2, k mod 3).

    play around with this ring a bit.
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  4. #4
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    What if I look at it in a brute force kinda way, is this an ok way of thinking about it?

    Every element in \mathbb Z_3\times \mathbb Z_5 generates a principal ideal:
    - (0,0) generates the zero ideal
    - (1,0) generates the same ideal as (2,0)
    - (0,1) generates the same ideal as (0,2), (0,3) and (0,4)
    - all the other elements are invertible and generate the entire ring.

    So we can look at three principal ideals: <(0,0)>, <(1,0)> and <(0,1)>. I don't think <(0,0)> is a prime (?) , but <(1,0)> and <(0,1)> are. Furthermore, can we say that these last two ideals are maximal?

    Can we say more generally that for the two fields F and K, {0} X K and F X {0} are the only prime and maximal ideals of F X K?
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