# Module homomorphisms $\prod \mathbb{Z} \rightarrow \mathbb{Z}$

Let $f\colon \prod \mathbb{Z} \rightarrow \mathbb{Z}$ be a $\mathbb{Z}$-module homomorphism. I want to show that $f(e_i) = 0$ for almost all $i$ where $e_i$ are the standard unit vectors. I assume $f(e_i) \neq 0$ for infinitely many $i$. We can assume without restriction that $f(e_i) > 0$ for all $i \geq 1$. Now is the following a valid argument and if not why? It is $f(\sum e_i) = \sum (f(e_i)) = \infty$, a contradiction. Or do I have to use a trickier argument?

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an homomorphism only commutes with finite sums. So $f(\sum e_i)$ exists, but it isn't related to $\sum f(e_i)$, and it doesn't imply that this sum should converge. I remember that this problem is quite tricky. – mercio Nov 25 '11 at 15:42
You should specify what your product is being taken over. – Dylan Moreland Nov 25 '11 at 17:06

Suppose $f : \prod \mathbb{Z} \to \mathbb{Z}$ is a group morphism. (I assume the product is indexed by $\mathbb{N}$.)

Let $(x)$ be a sequence of strictly positive numbers $(x_0, \ldots, x_n, \ldots)$ such that for all $n\ge 0$, $x_n > 2 |f(x_0, \ldots, x_{n-1}, 0, \ldots)|$ and for all $n \ge 1$, $x_n$ is a multiple of $x_{n-1}$ strictly larger than $x_{n-1}$.

Now, $(x_n)$ is increasing, so there exists $n_0$ such that $x_{n_0} > 2 |f(x)|$. Suppose $n\ge n_0$

$f(x) - f(x_0, \ldots, x_{n-1}, 0, \ldots) = f(0, \ldots, 0, x_n, \ldots) \equiv 0 \pmod {x_n}$ But $|f(x) - f(x_0, \ldots, x_{n-1}, 0, \ldots)| \le |f(x)| + |f(x_0, \ldots, x_{n-1}, 0, \ldots)| \lt x_n/2 + x_{n_0}/2 \le x_n$.

A number which is a multiple of something strictly bigger than itself has to be zero, so $f(x) = f(x_0, \ldots, x_{n-1}, 0, \ldots)$ for all $n \ge n_0$.

In particular, for all $n\ge n_0$, $f(x_n e_n) = f(x)-f(x) = 0$, and since $x_n > 0$, $f(e_n) = 0$

A similar trick can show that in fact, $f(x) = \sum_{k=0}^{n_0} x_k f(e_k)$ for all sequences $(x)$

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