Zero image of an element in the direct limit of modules

Let $\left(M_i,f_j^i\right)_{i,j \in I, i \le j}$ be a directed family of modules over some ring. Assume there is an index $k \in I$ such that there exists $x_k \in M_k$ whose image is zero in $\varinjlim M_i$. Why there must exist $j \in I$ such that $j \ge k$ and $f_j^k(x_k)=0$? In particular, how do we see this from the interpretation of $\varinjlim M_i$ as $\left(\bigoplus M_i\right) / N$ where $N$ is the submodule of $\bigoplus M_i$ generated by elements of the form $x_{ij}$ with $i \le j$, whose $i^{th}$ component is $x \in M_i$, their $j^{th}$ component is $-f^i_j(x)$ and the rest components are zero?

A hint would be appreciated :-)

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If I understand you correctly, you want to show that if an element maps to $0$ at the limit, then it must map to $0$ "in finite time".

One construction of the direct limit is to take the disjoint union of the $M_i$, say as ordered pairs $(x,i)$ with $i\in I$ and $x\in M_i$), modulo the equivalence relation $$(x,i)\sim (y,j)\Longleftrightarrow\text{there exists }k\in I, i,j\leq k\text{ such that }f^i_k(x) = f^j_k(y).$$ Then define the operations by $$[(x,i)] + [(y,j)] = [(f^i_k(x)+f^j_k(y),k)]$$ where $k\in I$ is any index such that $i,j\leq k$; and $$a[(x,i)] = [(ax,i)]$$ for any $a\in R$.

In this contruction, the maps $f_i\colon M_i\to\lim\limits_{\rightarrow}M_j$ is given by $f_i(x) = [(x,i)]$. If $f_i(x)=[(0,i)]$, then by definition of the equivalence relation there exists $k\in I$, $i\leq k$, such that $f^i_k(x) = f^i_k(0) = 0$. Thus, $x$ maps to $0$ in "finite time".

To interpret it in your definition, an element that has $x$ in the $i$th coordinate and $0$s elsewhere lies in $N$. Then there is a finite number of generators of $N$ whose sum equals this element. Hence, there is a finite set of pairs of indices, $i_1\lt j_1$, $i_2\lt j_2,\ldots, i_k\lt j_k$, and elements $x_{i_1,j_i}\in M_{i_1},\ldots,x_{i_k,j_k}\in M_{i_k}$ such that $$\delta_i(x) = \sum_{r=1}^k \bigl(\delta_{i_r}(x_{i_r,j_r}) - \delta_{j_r}(f^{i_r}_{j_r}(x_{i_r,j_r})\bigr)$$ where $\delta_n$ represents the embedding into the $n$th coordinate.

Clarified/Corected. Now let $s$ be strictly greater than all the coordinates that occur. We can rewrite each $$\delta_{i_r}(x_{i_r,j_r}) - \delta_{j_r}(f^{i_r}_{j_r}(x_{i_r,j_r}))$$ as $$\delta_{i_r}(x_{i_r,j_r}) - \delta_s(f^{i_r}_s(x_{i_r,j_r})) - \delta_{j_r}(f^{i_r}_{j_r}(x_{i_r,j_r})) + \delta_s(f^{j_r}_s(f^{i_r}_{j_r}(x_{i_r,j_r}))),$$ since $$\delta_s(f^{j_r}_s(f^{i_r}_{j_r}(x_{i_r,j_r}))) = \delta_s(f^{i_r}_s(x_{i_r,j_r}))),$$ so that we may assume that $j_1=j_2=\cdots = j_r = s$.

Then, combining any pairs $(i_r,s)$ and $(i_t,s)$ with $i_r=i_t$, we may further assume that $i_1,\ldots,i_k,s$ are pairwise distinct.

But if all the $i_1,\ldots,i_k,s$ are pairwise distinct, and the sum is equal to the single-coordinate term $\delta_i(x)$, then there can be at most one pair $(i_1,s)$, one of the indices is $i$, and the entry in the other index is $0$. If $i=i_1$, then $x_{i_1,s} = x$ and $f^{i_1}_s(x)=0$ and we are done. If $i=s$, then $x_{i_1,s}=0$, so $x=f^{i_1}_s(x_{i_1,s}) = 0$, and we are done again.

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Oh, nevermind. I got confused about which way the arrows go. Everything makes sense! Sorry to bother you. –  Dylan Moreland Nov 13 '11 at 0:13
@Dylan: Yes, that needs clarification (but my previous comment was slightly off); I'm editing. –  Arturo Magidin Nov 13 '11 at 0:18
@Dylan: No, actually you were right; we have no warrant for assuming the indices are comparable first-component to first-component. Instead, what we can do is just push everything "high enough" and then combine; it's actually closer to a direct translation of the argument with the other construction. Take a look. –  Arturo Magidin Nov 13 '11 at 0:53
@Arturo: Masterful, as always :-) –  Manos Nov 13 '11 at 16:12
@ArturoMagidin I have been reading your answer above, and two things I don't understand. 1) What do you mean by "combining pairs $(i_r,s)$ and $(i_t,s)$ with $i_r = i_t$ we may assume that $i_1,\ldots, i_k,s$ are pairwise distinct"? 2) Do you mean almost right at the end that since $x = f^{i_1}_s(x_{i_1,s}) =0$, then $f^{i_1}_s(x) = 0$? Thanks, Ben. –  user38268 Apr 18 '12 at 3:51