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Thread: Z-module homomorphism

  1. #1
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    Z-module homomorphism

    Let $\displaystyle M=\{(a_i): a_i \in \mathbb{Z}, i\geq 1\}$ (M a $\displaystyle \mathbb{Z}$-module); let N be the Z-submodule of M generated by $\displaystyle \{e_n: n\geq 1\}$ (where $\displaystyle e_n = (0,0,\ldots,0, 1 \text{(nth place)}, 0, \ldots)$). Let $\displaystyle g:M\to\mathbb{Z}$ be a Z-module homomorphism such that g(N)={0}.
    Use elements in M of the form $\displaystyle (2^n a_n)$ and $\displaystyle (3^n b_n)$ ($\displaystyle a_n$, $\displaystyle b_n$ in Z) to show that g=0. Hint: If $\displaystyle x=(2^n a_n)$, use that, given $\displaystyle k\geq 1$, there exist y in N, z in M such that $\displaystyle x-y=2^k z$ to deduce that g(x)=0.
    What confuses me is if g(N)=0, since an element x of M can be written as $\displaystyle x=\sum_{n=1}^{\infty} a_n e_n$ why then does it not follow that, since $\displaystyle g(e_n)=0$ for each n, $\displaystyle g(\sum_{n=1}^{\infty} a_n e_n) = \sum_{n=1}^{\infty} a_n g(e_n) =0$ for any x in M? After that the purpose of the various k's is also a mystery to me.



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  2. #2
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    Quote Originally Posted by james123 View Post
    What confuses me is if g(N)=0, since an element x of M can be written as $\displaystyle x=\sum_{n=1}^{\infty} a_n e_n$


    No, it can't: there don't exist infinite sums here. N is the free $\displaystyle \mathbb{Z}-$module (i.e., the free abelian group) on $\displaystyle e_i$ , and every element there is a finite $\displaystyle \mathbb{Z}-$linear combination of the $\displaystyle e_i$'s.

    Tonio

    why then does it not follow that, since $\displaystyle g(e_n)=0$ for each n, $\displaystyle g(\sum_{n=1}^{\infty} a_n e_n) = \sum_{n=1}^{\infty} a_n g(e_n) =0$ for any x in M? After that the purpose of the various k's is also a mystery to me.



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  3. #3
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    Oh, finitely generated - of course! (slaps head). Now, if $\displaystyle x=(2^n a_n)$, given k>=1 I have $\displaystyle y = \sum_{i=1}^{k-1}2^i e_i \in N$, $\displaystyle z = (0 , \ldots , 0, a_{k}, a_{k+1}, \ldots) \in M$ so that

    $\displaystyle x - y = (2^n a_n) - \sum_{i=1}^{k-1}2^i e_i = (0 , \ldots , 0, 2^{k}a_{k}, 2^{k+1}a_{k+1}, \ldots) = 2^k(0 , \ldots , 0, a_{k}, a_{k+1}, \ldots) $
    $\displaystyle = 2^k z $

    as per the hint.

    As g(N)=0, taking g of both sides gives
    $\displaystyle g((2^n a_n)) = 2^k g(0, \ldots, 0, a_k, a_{k+1},\ldots)$
    for any k>=1.

    In particular, $\displaystyle 2^k g(0, \ldots, 0, a_k, a_{k+1},\ldots) = 2^{k+1} g(0, \ldots, 0, a_{k+1}, a_{k+2},\ldots)$

    so, upon subtracting:
    $\displaystyle 0 = -2 g(0, \ldots, 0, a_k, 0, \ldots) = -2a_k g(e_k)$.

    This would seem to imply x=(0,0,..) or g=0; in either case g(x)=0.

    But this can't be right - I could have deduced as much by not bothering with the powers of 2. What am I missing?

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    Quote Originally Posted by james123 View Post
    Oh, finitely generated - of course! (slaps head).

    No, not finitely generated! Infinitely generated, but every element in N is a finite linear combination of this infinite basis. (so slap it again ).

    From what book/site is this question? I kindda remember a very similar question, but cannot place it.


    Now, if $\displaystyle x=(2^n a_n)$, given k>=1 I have $\displaystyle y = \sum_{i=1}^{k-1}2^i e_i \in N$, $\displaystyle z = (0 , \ldots , 0, a_{k}, a_{k+1}, \ldots) \in M$ so that

    $\displaystyle x - y = (2^n a_n) - \sum_{i=1}^{k-1}2^i e_i = (0 , \ldots , 0, 2^{k}a_{k}, 2^{k+1}a_{k+1}, \ldots) = 2^k(0 , \ldots , 0, a_{k}, a_{k+1}, \ldots) $
    $\displaystyle = 2^k z $

    as per the hint.


    Hmmm...I think it'd rather be $\displaystyle y=\sum\limits_{i=1}^{k-1}2^ia_ie_i=(2a_1,2^2a_2,\dots,2^{k-1}a_{k-1},0,\dots)$ , so that:

    $\displaystyle x-y = (2a_1,2^2a_2,\dots)-(2a_1,\dots,2^{k-1}a_{k-1},0,\dots)=(0,\dots,0,2^ka_k,2^{k+1}a_{k+1},\dots )=$ $\displaystyle 2^k(0,\dots,0,a_k,2a_{k+1},2^2a_{k+1},\dots)$ , and thus:

    $\displaystyle g(x)=g(x)-g(y)=g\left(2^k(0,\dots,0,a_k,2^ka_{k+1},\dots)\ri ght)=2^kg(0,\dots,0,a_k,2a_{k+1},\dots)$

    It's easy to see that the above is true, mutandis mutandis, if instead powers of 2 we choose powers of 3. Now, as $\displaystyle gcd(2^k,3^n)=1\,\,\forall\,n\,,\,k\in\mathbb{N}$ ,

    then $\displaystyle \forall m\in\mathbb{Z}\,\,\,\exists\, r,s\in\mathbb{Z}\,\,\,s.t.\,\,\,2^kr+3^ks=m$

    Tonio

    As g(N)=0, taking g of both sides gives
    $\displaystyle g((2^n a_n)) = 2^k g(0, \ldots, 0, a_k, a_{k+1},\ldots)$
    for any k>=1.

    In particular, $\displaystyle 2^k g(0, \ldots, 0, a_k, a_{k+1},\ldots) = 2^{k+1} g(0, \ldots, 0, a_{k+1}, a_{k+2},\ldots)$

    so, upon subtracting:
    $\displaystyle 0 = -2 g(0, \ldots, 0, a_k, 0, \ldots) = -2a_k g(e_k)$.

    This would seem to imply x=(0,0,..) or g=0; in either case g(x)=0.

    But this can't be right - I could have deduced as much by not bothering with the powers of 2. What am I missing?

    Thanks again
    .
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