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Math Help - Surjectivity of a natural ring homomorphism

  1. #1
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    Surjectivity of a natural ring homomorphism

    Hi,
    I am stuck on a question, I think I know the process required to solve it as we covered a similar example in class. I'm just a bit stuck...

    Let x=a+bi contained in the Gaussian integers, where a,b are coprime integers. I need to prove that the natural ring homomorphism
    f:Z --> Z[i]/(x) given by f(n) = n + (x) is surjective.

    So, if I have this right, I want to show that for all a,b in the natural numbers that a+ib+(x) =f(z) for some z contained in the natural numbers and I need to find i+(x) = f(z) = z+i
    i.e. that z-i is contained in (x). But what can z be? My problem I think is that I still don't fully understand what (x) is. All I know is that
    (x) = {(x)(a+ib) s.t a,b are integers} but I'm really confused!
    Any help/advice would be great! Thanks,
    Sooz
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    What we want to show is given u+iv we can find n so that n + (x) = u+iv + (x). This is equivalent to saying n - (u+iv) in (x) thus there exists s,t so that n - (u+iv) = (s+it)(a+bi) where a,b coprime (positive) integers. Thus, (n-u)+iv= (as - bt) + i(at+bs). Thus we want v = at+bs and n-u = as-bt. Since gcd(a,b)=1 it means there exists s,t so that v = at+bs. But then let n = u + (as - bt), which will make f(n) = u+iv + (x).

    In general this is what we do: Given z+(x) in Z[i]/(x) let s,t be integrers such that at+bs = Im(z) then let n = Re(z)+(as-bt).
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Since gcd(a,b)=1 it means there exists s,t so that v = at+bs. But then let n = u + (as - bt), which will make f(n) = u+iv + (x).

    In general this is what we do: Given z+(x) in Z[i]/(x) let s,t be integrers such that at+bs = Im(z) then let n = Re(z)+(as-bt).
    Hi,
    Thanks for the reply, that helps a lot, I was just wondering if I could clarify a few things.
    "But then let n = u + (as - bt), which will make f(n) = u+iv + (x)." How, by letting n = u + (as - bt) do we get f(n) = u + iv + (x)? I think I am looking at this wrong but if
    f(n) = n + (x) then f(u + (as - bt)) = u + (as - bt) + (x) and so (as - bt) = iv. But v = (at + bs)! I'm confusing myself...

    Also, if I want to show that the Kernel of f is (a^2 + b^2):
    Ker f = { z in Z : z in (x)}
    z=(x)(a + ib), can I say (x) = (u + iv)
    0 = (u +iv)(a +ib) = (au - bv) + i(av + bu) and we require av + bu = 0 and this occurs when a=u, b = -v which leaves us with
    0 = (a^2 + b^2). If this is how to do it, it is a very lucky guess on my part and I would still appreciate a better explaination. If (much more likely) this is complete rubbish, I would be very grateful for some guidance.
    Thanks,
    Sooz
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  4. #4
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    Quote Originally Posted by Sooz View Post
    Hi,
    Thanks for the reply, that helps a lot, I was just wondering if I could clarify a few things.
    "But then let n = u + (as - bt), which will make f(n) = u+iv + (x)." How, by letting n = u + (as - bt) do we get f(n) = u + iv + (x)? I think I am looking at this wrong but if
    f(n) = n + (x) then f(u + (as - bt)) = u + (as - bt) + (x) and so (as - bt) = iv. But v = (at + bs)! I'm confusing myself...
    What we want to do is to get: v = at+bs and n-u = as-bt. We are given a,b,u,v. Thus, it means we can find s,t so that v = at+bs since of relative primeness. Remember we are solving for n in the second equation is would mean that n = u+(as-bt) because we already determined the values of s,t from before. You can in fact check it, f(n) = n + (x), we claim that n + (x) = u+iv + (x), if and only if, n - (u+iv) in (x). This is true because n - (u+iv) = (a+bi)(s+it).
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  5. #5
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    Thanks! I understand! I am soooo grateful. Any chance you could comment on my question about the kernel?
    Thanks again,
    Sooz
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  6. #6
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    Quote Originally Posted by Sooz View Post
    Also, if I want to show that the Kernel of f is (a^2 + b^2):
    Ker f = { z in Z : z in (x)}
    z=(x)(a + ib), can I say (x) = (u + iv)
    0 = (u +iv)(a +ib) = (au - bv) + i(av + bu) and we require av + bu = 0 and this occurs when a=u, b = -v which leaves us with
    0 = (a^2 + b^2). If this is how to do it, it is a very lucky guess on my part and I would still appreciate a better explaination. If (much more likely) this is complete rubbish, I would be very grateful for some guidance.
    The kernel of f is the set { n in Z: f(n) in (x)} because (x) is the identity element for Z[i]/(x). Now f(n) = n + (x) if we want that n + (x) = (x) we require that n in (x) if and only if n = (c+id)(a+bi) for some integers c,d. Which means n = (ac - bd) + i(ad+bc), thus we need ad+bc = 0. But since gcd(a,b)=1 it is if ad+bc=0 then d=bt and c=-at for integer t. This means then n = ac - bd = -a^2t-b^2t = -t(a^2+b^2). We have shown that the kernel is the set of all multiples of a^2+b^2 (because t is an arbitrary integer).
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