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'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.