Showing that no homomorphism $ϕ:∏ _{i∈N} \Bbb Z→\Bbb Z$ can send all $e_i$ to 1 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. 
 A: 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:


*

*$a_1\mid a_2\mid a_3\mid\cdots$ (i.e. each divides the next)

*$0<a_1+a_2+\cdots+a_n<a_{n+1}$

*$a_1+a_2+\cdots+a_n\to\infty$

*$a_{n+1}-(a_1+a_2+\cdots+a_n)\to\infty$


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