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
)
, given k>=1 I have
,
 \in M)
so that
 - \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) )
as per the hint.
Hmmm...I think it'd rather be )
, so that: -(2a_1,\dots,2^{k-1}a_{k-1},0,\dots)=(0,\dots,0,2^ka_k,2^{k+1}a_{k+1},\dots )=)
)
, and thus: =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 =1\,\,\forall\,n\,,\,k\in\mathbb{N})
, then 
Tonio
As g(N)=0, taking g of both sides gives
) = 2^k g(0, \ldots, 0, a_k, a_{k+1},\ldots))
for any k>=1.
In particular,
 = 2^{k+1} g(0, \ldots, 0, a_{k+1}, a_{k+2},\ldots))
so, upon subtracting:
 = -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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About LinkBacks
(M a
-module); let N be the Z-submodule of M generated by
(where
). Let
be a Z-module homomorphism such that g(N)={0}.
and
(
,
in Z) to show that g=0. Hint: If
, there exist y in N, z in M such that
to deduce that g(x)=0.
why then does it not follow that, since
for each n,
for any x in M? After that the purpose of the various k's is also a mystery to me.
Originally Posted by james123 
