Atiyah Macdonald – Exercise 2.15 (direct limit)

Question

Atiyah-Macdonald book constructs the direct limit of a directed system $(M_i,\mu_{ij})$, (where $i\in I$, a directed set, and $i\leq j$) of $A$-modules as the quotient $C/D$, where $C=\bigoplus_{i\in I} M_i$, and $D$ is the submodule generated by all the elements of the forms $x_i-\mu_{ij}(x_i)$, where $x_i\in M_i$ for some $i$ and $i\leq j$. Let $\mu \colon C \longrightarrow M$ be the projection map, and let $\mu_i$ be the restriction of $\mu$ to $M_i$ (which is identified with its image in the direct sum).

Now Exercise 2.15 asks us to prove the following – Show that if $\mu_i(x_i)=0$, then there exists $j\geq i$ such that $\mu_{ij}(x_i)=0$.

I have tried the following. $\mu_i(x_i)=0$ implies that $x_i\in D$. That is, $x_i$ is a finite sum of the elements of the form $a_l(x_l-\mu_{lk}(x_l))$, where $a_l\in A$, $x_l\in M_l$ and $k\geq l$. So, $x_i=\sum_{i=1}^{n} a_{l_i}(x_{l_i}-\mu_{{l_i}{k_i}}(x_{l_i}))$. I don’t know how to proceed to find the index $j$ for which $\mu_{ij}(x_i)=0$. How do I proceed? Any help will be greatly appreciated!

Answer 1

People have pointed out that my hand-waving does not necessarily correspond to a proof. It was an absolute delight to revisit this proof. I have a new proof below, and it does not rely on the hand-wavy "minimality" hypothesis at all.

Let $x_i\in M_i$, and suppose $x_i\in D$. As you have noted, $x_i$ is a finite $A$-linear combination of elements of $C$ of the form $x_a-\mu_{ab}(x_a)$. Absorbing the coefficients from $A$ in the terms $x_a-\mu_{ab}(x_a)$, we get terms of the same form. So, let $i_1,i_2,\dots,i_k,j_1,j_2,\dots,j_k\in I$, $x_{(1)}\in M_{i_1}\dots,x_{(k)}\in M_{i_k}$ and suppose $i_r\le j_r$ for $r=1,2,\dots,k$ as well as $$x_i = (x_{(1)}-\mu_{i_1j_1}(x_{(1)}))+\dots+(x_{(k)}-\mu_{i_kj_k}(x_{(k)})).\qquad(1)$$ Since $\{i,j_1,j_2,\dots,j_k\}$ is a finite subset of $I$, which is directed, there exists $j_{\ast}$ such that $i\le j_{\ast}$ and $j_r\le j_{\ast}$ for $r=1,2,\dots,k$. Also, for $r=1,2,\dots,k$, we have $i_r\le j_r$, so $i_r\le j_{\ast}$.

For $a\in I$, let $\pi_{a}:C\to M_{a}$ be the homomorphism given by restricting the canonical projection $\prod_{b\in I}M_b\to M_{a}$ to $C$. On the one hand, we have $$\pi_a(x_i) = \begin{cases} x_i&\text{if $a=i$} \\ 0 &\text{if $a\ne i$}\end{cases}$$ Based on (1), we also find that $$\pi_a(x_i) = \sum_{i_b=a}x_{(b)} - \sum_{j_c=a}\mu_{i_cj_c}(x_{(c)}).$$ Then, $$\mu_{aj_{\ast}}(\pi_a(x_i)) = \sum_{i_b=a}\mu_{i_bj_{\ast}}(x_{(b)})-\sum_{j_c=a}\mu_{j_cj_{\ast}}(\mu_{i_cj_c}(x_{(c)}))$$ Summing over $a$, we have $$\sum_{a\in I}\mu_{aj_{\ast}}(\pi_a(x_i)) = \mu_{i_1j_{\ast}}(x_{(1)})-\mu_{j_1j_{\ast}}(\mu_{i_1j_1}(x_{(1)}))+\dots+\mu_{i_kj_{\ast}}(x_{(k)})-\mu_{j_kj_{\ast}}(\mu_{i_kj_k}(x_{(k)})).$$ The left side is $\mu_{ij_{\ast}}(\pi_i(x_i))$, and the right side is $0$. Therefore, $$\mu_{ij_{\ast}}(x_i) = 0.$$

Answer 2

Here is an alternative proof. The idea is to give another construction of the direct limit, a bit closer to the way it arises in algebraic geometry as germs of sections, and verify that construction satisfies the same universal property. This makes it clear where the hypothesis that the underlying poset is directed enters.

Let $N$ be the disjoint union of the underlying sets $$N=\coprod M_i,$$ and define an equivalence relation on $N$ by declaring $x \in M_i$ equivalent to $y \in M_j$ if there is some $k$ with $\mu_{jk}(y)=\mu_{ik}(x)$.

Let $M$ be the quotient of $N$ by this equivalence relation, write $\overline{x}$ for the equivalence class containing $x \in N$, and then define the structure of an $A$-module on $M$ by $$a \overline{x}=\overline{ax} \quad \text{and} \quad \overline{x}+\overline{y}=\overline{x+y}$$ for $a \in A$ and $x,y \in M_i$ for some $i$ (here we use that the index set is directed, which implies that if $x \in M_i$ and $y \in M_j$ then there are $k \geq i,j$ and $z,w \in M_k$ with $\overline{x}=\overline{z}$ and $\overline{y}=\overline{w}$).

One (I won’t do it here, but it is routine) now verifies that $M$ together with the canonical maps $\nu_i:M_i \to M$ is a direct limit of $(M_i,\mu_{ij})$. The $0$ element of $M$ is the image of $0 \in M_i$ for some (any) $i \in I$. So by definition of the equivalence relation on $N$, the kernel of $\nu_i$ is precisely the set of $x \in M_i$ such that $\mu_{ij}(x)=0$ for some $j \geq i$.

By uniqueness of direct limits, the same is true with the other construction.

For the sake of completion, here is an example showing that the hypothesis that the set is directed is necessary for the result to remain valid. Consider the poset represented by the diagram $$3 \leftarrow 1 \rightarrow 2,$$ with $1 <2,3$ and no order relation between $2$ and $3$. Take all three groups $M_1=M_2=M_3=\mathbf{Z}$, with maps the identity $\mathrm{1}:M_1 \to M_2$ and the zero map $0:M_1 \to M_3$. In the direct limit, every element of $M_1$ goes to zero. Hence so does every element of $M_2$.

Answer 3

Here's another proof. The result is immediate from the following claim:

Claim: Let $E \subseteq C$ be the subset consisting of all finite sums $\sum_{j \in J} x_j$ (where $x_j \in M_j$) that satisfy the following property: there exists a "threshold" $k \in I$ such that 1) $j \leq k$ for all $j \in J$ and 2) $\sum_{j \in J} \mu_{jl}(x_j) = 0$ for all $l \geq k$. Then $D = E$.

Technical detail. In the above statement, we allow terms $x_j$ to be equal to $0$. In particular, the subset $J \subseteq I$ is not uniquely determined by the element $\sum_{j \in J} x_j \in C$. However, the above property is evidently still well-defined (though the exact "threshold" $k$ may change).

We first verify that $E$ is a submodule of $C$. Let $\sum_{j \in J} x_j$ and $\sum_{j \in J'} y_j$ be elements of $E$ with respective "thresholds" $k, k' \in I$. By defining some $x_j$ and $y_j$ to be $0$, we may assume without loss of generality that $J = J'$ (increasing $k$ and $k'$ if necessary). Then for all $l \geq \max\{k, k'\}$, we have $$ \sum_{j \in J} \mu_{jl}(x_j + y_j) = \sum_{j \in J} \mu_{jl}(x_j) + \sum_{j \in J} \mu_{jl}(y_j) = 0, $$ so $\sum_{j \in J} x_j + \sum_{j \in J'} y_j \in E$. Next, let $a \in A$. For all $l \geq k$, we have $$ \sum_{j \in J} \mu_{jl}(ax_j) = a\sum_{j \in J} \mu_{jl}(x_j) = 0, $$ so $a(\sum_{j \in J} x_j) \in E$. We have shown that $E$ is closed under addition and scalar multiplication, so $E$ is a submodule of $C$.

Now, observe that $E$ contains the generators of $D$, so $D \subseteq E$. Conversely, every sum $\sum_{j \in J} x_j \in E$ can be written $\sum_{j \in J} (x_j - \mu_{jk}(x_j)) \in D$, so $E \subseteq D$. This proves the claim.