Show that the direct sum of countably many $\mathbb{Z}$ copies is not finitely generated

Question

Let $C$ denote the collection of infinite sequences $(a_1, a_2, a_2, \ldots)$ of integers where all but finitely many $a_i$ are $0$.

I WTS that $C$ is not finitely generated as a $C$-module. ($C$, of course, is a ring under componentwise addition and multiplication).

Suppose not so that we have $F:= \{(a_{n,i}\}_{i=1}^{k}$ generating $C$. We know that there is some natural $N$ such that $a_{n,i} = 0$ for all $n\geq N$ and all $1\leq i \leq k$. Now it's seems clear to try to construct a sequence $(c_n)$ so that we get a nonzero component in a spot where any linear combination of the assumed basis will have a $0$. But I can't come up with one. And I'm not sure how to explicitly prove that no linear combination can equal this sequence anyway.

If I'm on the right rack, how would I finish this proof?

Answer 1

As you say, there is some natural $N$ such that $a_{n, i} = 0$ for all $n \geq N$ and all $1 \leq i \leq k$, in particular since $k$ is finite, there is a value of $N$ works for all $i$ (just take the maximum). So consider the position $N+1$, where $a_{N+1, i} = 0$ for all $i$ and hence any linear combination of the $a$'s has that spot equal to $0$. So just pick $c_{N+1} = 1$ and the others anything else, such as $0$.

Answer 2

How about $ c_k = 0 $ for all $ k \neq N $, and $ c_N = 1 $?