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why is that no homomorphism $\phi:\prod_{i\in\mathbb{N}}\mathbb{Z}\to\mathbb{Z}$ can send all $e_i$ to 1? In fact I saw a proof in MO using 2-adic integers...but I know very little about those topics. Could anyone prove it by other (possibly more elementary) means? I think I must show some contradiction about the value of $\phi(1,1,..,1,..)$ but can't formulate an argument.

Hanul Jeon
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John Ralph
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  • Fyi, a nice corollary of your fact is that one cannot extend the $\mathbb{Z}$-independent set ${ e_i : i \in \mathbb{N}}$ to a basis for $\prod_{i \in \mathbb{N}} \mathbb{Z}$. In fact, more is true. There is actually no basis whatsoever, but this is more difficult to prove. – Mike F Sep 27 '14 at 04:54
  • Could it be that you are the same as http://math.stackexchange.com/questions/947045 ? There I posted a link to http://mathoverflow.net/questions/10239 (which you already knew?) and there in Richard Stanley's answer you find an elementary argument. Why asking this here again? – Martin Brandenburg Sep 27 '14 at 14:49

1 Answers1

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Here is a recipe for creating contradictions from the assumption $\phi(e_i)=1$ for $i\ge1$.

Let us pick any sequence of integers $(a_i)_{i\ge1}$ with the following properties:

  1. $a_1\mid a_2\mid a_3\mid\cdots$ (i.e. each divides the next)
  2. $0<a_1+a_2+\cdots+a_n<a_{n+1}$
  3. $a_1+a_2+\cdots+a_n\to\infty$
  4. $a_{n+1}-(a_1+a_2+\cdots+a_n)\to\infty$

Then for every $n\ge1$ we have

$$\begin{array}{l} m:=\phi(a_1,a_2,a_3,\cdots) & =a_1\phi(e_1)+a_2\phi(e_2)+\cdots+a_n\phi(e_n)+a_{n+1}\phi({\rm etc}) \\ & \equiv a_1+a_2+a_3+\cdots+a_n \mod{a_{n+1}} \end{array} \tag{$\circ$}$$

We decomposed $m$ using finite additivity of $\phi$ and fact $(1)$. By $(\circ)$ and $(2)$, we know $m\ne0$ so the right side of the congruence must be the least positive residue (LPR) representing $m$ mod $a_{n+1}$. We know this modulus $a_{n+1}$ is bigger than $a_1+\cdots+a_n$ which diverges to $\infty$, so $a_{n+1}\to\infty$, and hence if the integer $m$ were positive its LPR mod $a_{n+1}$ would eventually be constant, but its LPR $\to\infty$, and so we must conclude $m$ is negative. But if $m$ is negative, its LPR mod $a_{n+1}$ must be $a_{n+1}$ minus a constant, at least eventually, and yet by $(4)$ we know $a_{n+1}$ minus the LPR diverges to $\infty$, another contradiction.

My original answer was with $a_k:=k!$. To check it satisfies $(2),(3),(4)$ notice

$$1!+2!+3!+\cdots+n!\le n!+n!+\cdots+n!=n\cdot n!=(n+1)!-n!.$$

anon
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  • This doesn't work; why can we not just set phi(1!,2!,3!,4!,...)=0? There's no contradiction in doing that - we just get, for instance, phi(1!,2!,3!,...,(n-1)!,0,(n+1)!,...)=-n!, but that's okay. The proof can't work if we only talk about the value of one point of phi, because the sequence 1!,2!,3!,... is non-zero on infinitely many points and not, therefore, a finite sum of the e_i; a homomorphism doesn't generally need countable additivity. – Milo Brandt Sep 27 '14 at 02:58
  • @Meelo $\phi(1,2,6,24,\cdots)=\phi(1,0,0,0,\cdots)+2\phi(0,1,3,12,\cdots)$ hence must be $1$ mod $2$, which $0$ is not. I am only using finite additivity in this argument, not countable additivity. – anon Sep 27 '14 at 03:00
  • Oh, I see; it might be good to note that, if s is the sequence of factorials, then s = (1!,2!,...,n!,0,0,...) + (n+1)!S for some other sequence S; otherwise, it's not immediately obvious why taking the residue is helpful. – Milo Brandt Sep 27 '14 at 03:03