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Thread: Subrings and subgroups of Z

  1. #1
    Super Member Bernhard's Avatar
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    Subrings and subgroups of Z

    In Dummit and Foote Ch. 10 Introduction to Rings on page 228 we read:

    "2Z is a subring of Z, as is nZ for any integer n. The ring Z/nZ is not a sub-ring (or a subgroup) of Z for any n $\displaystyle \geq $ 2."

    Can anyone help me explictly prove that the ring Z/nZ is not a sub-ring (or a subgroup) of Z for any n >= 2.

    Peter
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  2. #2
    Super Member TheChaz's Avatar
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    Re: Subrings and subgroups of Z

    Wouldn't this be since the (nontrivial) quotient group isn't even a subset of Z?
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    Super Member Bernhard's Avatar
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    Re: Subrings and subgroups of Z

    Hmmm ... yes, its a thought, definitely

    But then the example of D&F is a little peculiar!

    I was thinking that they were identifying $\displaystyle [a]_n $ with a, but of course if you do this you appear to have a subring (i think)

    I tried this with Z/3z and it appeared to be both a subgroup and a subring of Z - so maybe you are correct.

    Peter
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    Re: Subrings and subgroups of Z

    you need to show that there is no (injective or even non-zero) group homomorphism from $\displaystyle \mathbb{Z}/n\mathbb{Z}$ to $\displaystyle \mathbb{Z}.$ to see this just look at the image of $\displaystyle [n]=[0].$
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  5. #5
    Super Member Bernhard's Avatar
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    Re: Subrings and subgroups of Z

    Thanks.

    To clarify, are you essentially saying that because there are n elements in Z/nZ and an infinite number in Z there can be no injective mapping? Or is there more to it?

    [I guess I am finding it hard to figure out why "you need to show that there is no (injective or even non-zero) group homomorphism from Z/nZ to Z" (post by NonCommAlg - thanks) leads to the conclusion that "the ring Z/nZ is not a subring (or a subgroup) of Z"]

    I started looking at the idea of using a homomorphism to establish what we need and was bothered by the following:

    In Dummit and Foote section 7.3 we find:

    --------------------------------------------------------------------------------------------------------------------------------------
    "Proposition 5: Let R and S be rings and let $\displaystyle \phi $ : R $\displaystyle \rightarrow $ S be a homomorphism
    (1) The image of $\displaystyle \phi $ is a subring of S
    (2) the kernel of $\displaystyle \phi $ is a subring of S"
    --------------------------------------------------------------------------------------------------------------------------------------



    Now, consider the rings Z and Z/2Z

    The map $\displaystyle \phi $: Z $\displaystyle \rightarrow $ Z/2Z defined by $\displaystyle \phi $: (even integer) ---> $\displaystyle [0]_n $ and $\displaystyle \phi $: (odd integer) ---> $\displaystyle [1]_n $ is a homomorphism.

    Thus its image (which is Z/2Z) is a subring

    What have I got wrong? This does not square with D&Fs statement that "2Z is a subring of Z, as is nZ for any integer n. The ring Z/nZ is not a sub-ring (or a subgroup) of Z for any n $\displaystyle \geq $ 2."

    Can you please help clarify things.

    Peter
    Last edited by Bernhard; Sep 3rd 2011 at 11:05 PM.
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