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Math Help - Urgently need help! If any1 can do this plzzz help! ><

  1. #1
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    Urgently need help! If any1 can do this plzzz help! ><

    Let Zp[i] denote the ring {a + bi | a,b E [Z]p, i^2=-1}.

    a) Show that if p is not prime, then Zp[i] is not an integral domain.
    b) Suppose p is prime. Show that every non-zero element of Zp[i] is a unit iff x^2 + y^2 does not equal 0modp for any pairs x,y E [Z]p.

    (Its a baby 'm' and 'p' next to the z's... couldn't find how to shrink letters! :S)
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by carmz View Post
    Let Zp[i] denote the ring {a + bi | a,b E [Z]p, i^2=-1}.

    a) Show that if p is not prime, then Zp[i] is not an integral domain.
    b) Suppose p is prime. Show that every non-zero element of Zp[i] is a unit iff x^2 + y^2 does not equal 0modp for any pairs x,y E [Z]p.

    (Its a baby 'm' and 'p' next to the z's... couldn't find how to shrink letters! :S)
    i really don't know if this is correct, however, this is my idea (at least for part a).

    if p is not prime, then at least \mathbb{Z}_p is not an integral domain since p=mn where for some m,n both not units (or not equal to 1) and hence \mathbb{Z}_p have zero divisors. Note that \mathbb{Z}_p \subset \mathbb{Z}_p[i] by taking b=0, thus \mathbb{Z}_p[i] cannot be an integral domain.
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  3. #3
    Oli
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    Am I right in thinking that the notation [Z]n for example means that if x=n, then x=0?
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  4. #4
    Oli
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    If so, for question 1:

    Let p=mn.
    Then m,n are in [Z]p.
    Hence (m+mi), (n+ni) are in Zp[i]
    (m+mi)(n+ni)
    =mn+mni+mni-mn
    =mni
    =0 (since mn=0)

    Hence there are zero divisors and it is not an integral domain.
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  5. #5
    Oli
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    For question 2.

    Suppose that every non-zero element is a unit (assumtion 1).
    Assume there are x,y in [Z]p such that x^2+y^2=0modp and get a contradiction:
    (x+yi)(x-yi)=x^2+y^2=0
    But these are in Zp[i] and so there exist z1,z2 such that z1(x+yi)=1 and z2(x-yi)=1 by assumption 1.
    Hence:
    1=1.1=z1(x+yi)z2(x-yi)=z1z2(x+yi)(x-yi)=z1z2(x^2+y^2)=z1z20=0
    So contradiction.
    Therefore every non-zero element a unit implies there are no s,y such that x^2+y^2=0modp
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  6. #6
    Oli
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    Now suppose x^2+y^2 does not equal zero for any x,y pairs.

    Show that if x,y are in [Z]p then:
    x/(x^2+y^2) is in [Z]p and that -y/(x^2+y^2) is in [Z]p
    (can't remember how you do this, but it is true and must be quite straightforward)
    Therefore:
    (x-yi)/(x^2+y^2) is in Zp[i]

    Then for all x+yi we can multiply by (x-yi)/(x^2+y^2) to get:
    (x+yi)(x-yi)/(x^2+y^2)=1
    So x+yi is a unit.

    So x^2+y^2 not equal to zero implies all x+yi are units.
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  7. #7
    Oli
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    "Show that if x,y are in [Z]p then:
    x/(x^2+y^2) is in [Z]p and that -y/(x^2+y^2) is in [Z]p
    (can't remember how you do this, but it is true and must be quite straightforward)"

    You say:
    x,y in Z[p]
    implies:
    x^2 and y^2 in Z[p]
    implies
    x^2+y^2 in Z[p]
    implies
    1/(x^2+y^2) in Z[p] (since Z[p] is an integral domain, the tricky step!)
    implies
    x/(x^2+y^2) is in [Z]p and that -y/(x^2+y^2) is in [Z]p
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