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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 :-)

1 Answers 1

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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.2011-11-13
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    @Dylan: Yes, that needs clarification (but my previous comment was slightly off); I'm editing.2011-11-13
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    @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.2011-11-13
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    @Arturo: Masterful, as always :-)2011-11-13
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    @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.2012-04-18
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    @BenjaminLim: I mean that if $i_r=i_t$, then in fact those entries are in the **same** coordinate; so we can combine those two "tuples" into a single one that is still one of our generators: $$(\cdots,a_{i_r},\cdots, -f^{i_r}_s(a_{i_r}),\cdots)+(\cdots b_{i_r},\cdots,-f^{i_r}_s(b_{i_r}),\cdots) = (\cdots c_{i_r},\cdots,-f^{i_r}_s(c_{i_r}))\cdots)$$where $c_{i_r}=a_{i_r}+b_{i_r}$. This is still a generator. (2) **No.** $x$ is not necessarily $0$. $x$ is the element that *maps* to zero in the quotient, we are showing it maps to $0$ in the $s$ coordinate.2012-04-18
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    @ArturoMagidin 1) is clear to me now. For (2), I don't really get what you mean. We know that because $\delta_i$ is injective, $\delta_i(x) = \delta_i (f^{i_1}_s(x_{i_1,s}))$ means that $x = f^{i_1}_s(x_{i_1,s})$. Since $x_{i_1,s} = 0$, we have that $f^{i_1}_s(x_{i_1,s}) = 0$. How does this means that $x$ maps to zero in the $s$ coordinate, as you stated in your comment above?2012-04-18
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    @BenjaminLim: We have a sum of $n$-terms; they involve exactly $n+1$ different coordinates, say $i_1\lt i_2\lt\cdots\lt i_n\lt s$, with the first term involving $i_1$ and $s$, the second term involving $i_2$ and $s$, etc. Their sum has $x$ in the $i$th coordinate and $0$s elsewhere (for some $i$). Since none of the coordinates $i_1,\ldots,i_n$ can be cancelled in the sum (the only coordinate that could be nonzero in one of the terms and $0$ in the entire sum is the $s$th coordinate ) (cont)2012-04-18
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    @BenjaminLim: that means that at most one $a_{i_1},a_{i_2},\ldots,a_{i_n}$ can be nonzero (because none of those can be cancelled, but in the end we have at most one nonzero entry). Any element in which $a_{i_j}=0$ can be eliminated (it's just all $0$s), so we end up with $$(\cdots,x,\cdots) = (\cdots,a_{i_1},\cdots,f^{i_1}_s(a_{i_1}),\cdots)$$where $x$ on the LHS is in the $i$th coordinate (the element we started with). So either $i_1=i$, $x=a_{i_1}$, and $f^{i_1}_s(a_{i_1})) = f^{i}_s(x)=0$ (proving $x$ maps to $0$ in "finite time"); (cont)2012-04-18
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    @BenjaminLim; I assume that is what you meant by "almost at the end". If you meant "the last sentence" instead (which is *at* the end, rather than almost at the end), then that's the other possibility: $i=s$, $a_{i_1}=0$; but then we would have $x=f^{i_1}_s(a_{i_1}) = f^{i_1}_s(0)=0$, which would mean that our original element was *already* equal to $0$; if $x=0$, then it clearly "maps to $0$ in finite time": it *is* zero; the question is only interesting when $x\neq 0$ to begin with, so this case need not be considered.2012-04-18
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    @ArturoMagidin Ah Arturo I get it now so the case at the end means that *a priori* $x$ was already zero so we do not need to look at it. Thanks.2012-04-18