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Ring homomorphism from Z4 to Z8

  1. Nov 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Exhibit two examples of a ring homomorphism [itex]\phi[/itex] from Z4 to Z8, one that is one-to-one and another that is not. For each case, find ker([itex]\phi[/itex]) and describe Z4/ker([itex]\phi[/itex])

    2. The attempt at a solution
    Let [itex]\phi : \mathbb{Z}_4 \longrightarrow \mathbb{Z}_8[/itex] be the identity mapping such that a = a mod 4 for [itex]a \in \mathbb{Z}_4[/itex] Clearly, this is a ring homomorphism since for any a,b in Z4,

    [itex]\phi (a+b)=(a+b) mod 4=a mod 4+b mod 4=\phi(a)+\phi(b)[/itex]

    and

    [itex]\phi(ab)=ab mod 4=a mod 4 \cdot b mod 4 = \phi(a) \cdot \phi(b)[/itex]

    This homomorphism is injective (or one-to-one) as every element in Z4 is carried to exactly one element in Z8, but not surjective as the range Z4 a proper subset of the codomain Z8.

    The kernel of [itex]\phi[/itex] is defined as the elements of Z4 that are mapped by [itex]\phi[/itex] to the zero element in Z8, so ker [itex]\phi[/itex] is the trivial ring {0} and Z4/ker [itex]\phi[/itex] is the quotient ring formed by modding out the trivial ring.

    Let us define another homomorphism [itex]\varphi : \mathbb{Z}_4 \longrightarrow \mathbb{Z}_8[/itex] by [itex]\varphi(a)=(8a)[/itex] mod 8 for [itex]a \in \mathbb{Z}_4[/itex]. Now,

    [itex]\varphi(a+b)=(8a+8b) mod 8=8a mod 8+8b mod 8=\phi(a)+\phi(b)[/itex]

    Also,

    [itex]\varphi(ab)=(64ab) mod 8=0=0 \cdot 0=(8a mod 8)(8b mod 8)[/itex]

    Observe that all of the elements in Z4 map to the zero element in Z8 so the kernel of [itex]\varphi[/itex] is simply the whole of Z4 and Z4/ker [itex]\varphi[/itex] is simply the trivial ring {0}.

    My first homomorphism seems to be wrong... a mod 4 * b mod 4 can be, say 4, 6 or 9 while ab mod 4 cannot when a,b belong to the subset {2,3}. What should I do to rectify this? Also, I cannot think of a homomorphism between these two that isn't injective. Any suggestions? Thanks! :D
     
    Last edited: Nov 30, 2011
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  3. Nov 30, 2011 #2

    micromass

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    Firstly, what do you mean with ring homomorphism?? Specifically, must it preserve units?? Must [itex]\phi(1)=1[/itex]?? I guess not, but I'll ask anyway.

    Anyway, your first homomorphism is wrong. It is not a homomorphism. Indeed, one one hand we have

    [tex]\phi(2+2)= 2+2=4[/tex]

    but also

    [tex]\phi(2+2)=\phi(4)=\phi(0)=0[/tex]

    So our function does not satisfy [itex]\phi(a+b)=\phi(a)+\phi(b)[/itex].

    Now, as a hint to find a suitable homomorphism. What happens if I put [itex]\phi(1)=2[/itex]? Can I extent this to a homomorphism???
     
  4. Nov 30, 2011 #3
    Mmm, yes, we're dealing with ring homomorphisms without the [itex]\varphi(1)=1[/itex] condition. I thought of something similar to your hint, I'm guessing since a homomorphism should preserve [itex]\varphi(nx)=n(\varphi(x))[/itex], we have [itex]x \longmapsto ax[/itex] and by Lagrange's theorem, |a| divides both 4 and 8, so |a|=1,2 or 4 and a=0,2 or 4. So my [itex]x \longmapsto 0[/itex] homomorphism works and is one-to-one, but when I tried [itex]x \longmapsto 2x[/itex] and [itex]x \longmapsto 4x[/itex], I couldn't have [itex]\varphi(1 \cdot 1)=\varphi(1) \cdot \varphi(1)[/itex]. Could you give another nudge? Also, I'm entirely clueless as to how to have the homomorphism NOT be injective.
     
  5. Nov 30, 2011 #4
    Anyone? ):
     
  6. Nov 30, 2011 #5

    Hurkyl

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    Why is it one-to-one? :confused:

    Do you believe your work? If it's right, what can you conclude? Have you thought carefully enough about both your work and what it implies to be confident in that conclusion? (and if you're not confident, can you imagine ways to analyze the problem to gain confidence?)

    If you really don't want to deal with that, mouse over the below to get another comment
    The zero map is the only rng homomorphism. Are you sure the problem isn't asking about group homomorphisms, or module homomorphisms?



    P.S. I think the best convention is to use "ring" to refer to the version where a ring has to have a multiplicative identity, and "rng" for the version where it does not. ("rng" is "ring" without "i")

    Actually, that's not true. The best convention is, whenever there may be confusion, to explicitly state what convention you're using. :wink:
     
    Last edited: Nov 30, 2011
  7. Nov 30, 2011 #6
    OH WAIT, I am an idiot! Ahhh.

    As you two suggest, I think my professor is expecting a group homomorphism actually (without [itex]\varphi(1)=1[/itex] condition). She teaches off Gallian's text, which doesn't have the [itex]\varphi(1)=1[/itex] condition in the definition of a ring homomorphism... But I think I've exhausted the options as well, with [itex]\varphi(a)=0,2a,4a[/itex] being the only possibilities. I don't get why there could be two examples she's expecting... argh. And ohhh... rng and ring is an interesting convention!
     
  8. Dec 1, 2011 #7

    Hurkyl

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    There's another difference -- a group homomorphism doesn't have a [itex]\varphi(xy)=\varphi(x)\varphi(y)[/itex] condition.
     
  9. Dec 1, 2011 #8
    Ahh, that's right. I'll point out the problem in her assignment. Thanks very much to both of you!
     
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