# Atiyah-Macdonald, Exercise 2.17 (direct limit)

I have solved the following exercise, can you tell me if this is correct? Thanks.

2.17. Let $(M_i)_{i \in I}$ be a family of submodules of an $A$-module such that for each pair $i,j$ in $I$ there exists $k$ in $I$ such that $M_i + M_j \subset M_k$. Define $i \leq j$ to mean $M_i \subset M_j$ and let $\mu_{ij}: M_i \to M_j$ be the embedding of $M_i$ in $M_j$. Show that $$\varinjlim_n M_n = \bigcup M_n = \sum M_n$$

$\varinjlim_n M_n = \bigcup M_n$:

We want to show that the union together with inclusions $i_i : M_i \hookrightarrow \bigcup M_n$ satisfies the universal property of the direct limit of the direct system $M_i, \mu_{ij}$. To this end, let $Y$ be a module and $i_{i}^\prime$ be inclusions $i_{i}^\prime : M_i \hookrightarrow Y$ such that for $x_k \in M_k$ we have $i_i^\prime (\mu_{ki}(x_k)) = i_k^\prime(x_k)$. We want to show that there exists a unique homomorphism $\varphi: \bigcup M_n \to Y$ such that for all $i,j$: $\varphi \circ i_i = i_j^\prime \circ \mu_{ij}$. Define $\varphi: \bigcup M_i \to Y$ as follows: for $m$ in $\bigcup M_i$ there is $M_i$ such that $m \in M_i$. Set $\varphi(m) = i^\prime_j (\mu_{ij} (i_i^{-1}(m)))$. Now we have existence, since this (as a concatenation of homomorphisms) is clearly a homomorphism. (well-define since inclusions are injective). To verify uniqueness, let $\varphi^\prime$ be a second homomorphism making the diagram commute. Let $m \in M$ and for a random $k$, let $m_k = i_k^{-1}(m)$. Then $\varphi (m) = \varphi (i_k (m_k)) = i^\prime_j (\mu_{kj} (m_k)) = \varphi^\prime (i_k (m_k) ) = \varphi^\prime (m)$, hence showing uniqueness.

Similarly in the case $\sum M_n$.

Edit

Dear BenjaLim, or anyone else, I'm sorry but I don't understand why I cannot do $i_i^{-1}(m)$. Would someone explain it to me? Thank you.

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A priori, you have got your universal properties wrong. You need to fix that first. Redefine $i'_i : M_i \to Y$ as just any $A$ - module homomorphism from $M_i \to Y$ and then define $\varphi(m) = i'_i(m)$ and check that this is well defined. That is exactly what I did below. – user38268 Aug 3 '12 at 23:42
@BenjaLim Oh no! You are right! I am so stupid! Ok, I was tired when I wrote this, that's my excuse for making a stupid mistake like that : ) – Rudy the Reindeer Aug 4 '12 at 5:46

Edit: We first check that $\bigcup M_i$ is an $A$ - module: We will check the only non-trivial part here which is closure under addition. Suppose that we have $y,z \in \bigcup M_i$. Then there exist $j,k$ such that $y \in M_j$ and $z \in M_k$. Now by assumption there exists $l \in I$ such that $M_j + M_k \subseteq M_l$. Hence it is clear that $y+z \in M_l \subset \bigcup M_i$ showing that $\bigcup M_i$ is closed under addition.

I don't think your proof is quite right because you can't just say "concatenation of homomorphisms": how do you define $i^{-1}(m)$? I think your proof breaks down here because you cannot just do that. The way I would do it is as follows:

We want to prove that whenever we have maps $f_i : M_i \rightarrow N$ where $N$ is some $A$ - module and $f_i = f_j \circ \mu_{ij}$ whenever $i\leq j$, there is a unique $A$ - module homomorphism $L : \bigcup M_i \to N$. Take an element $x \in \bigcup M_i$. Then $x$ is in at least one $M_j$, so we can define $L(x)$ to be equal to $f_j(x)$. We need to check that this is well defined. Suppose $x$ is also in some other $M_k$. Then we know that we can choose an $l \in I$ such that $l \geq j,k$. By definition of $f_l$ we have that $f_j(x) = f_l\circ \mu_{jl}(x)$. But the right hand side is just $x$ viewed as an element of $M_l$ and then mapped under $f_l$. Similarly writing down $f_k(x) = f_l \circ \mu_{kl}(x)$, the right hand side is just $x$ viewed as an element of $M_l$ and then mapped under $f_l$. It follows that $f_j(x) = f_k(x)$ so that $L$ is well defined.

We now check that $L$ is an $A$ - module homomorphism; the only non-trivial part to check is that $L$ is additive. So suppose we have $x,y \in \bigcup M_i$. Then there is $j,k$ such that $x \in M_j$ and $y \in M_k$. Then we know that there exists an $l \geq j,k$ such that $x+y \in M_l$ and so $L(x+y) = f_l (x+y)$. Now $L(x) = f_j(x)$, $L(y) = f_k(y)$. Since

$$f_j(x) = f_l\circ \mu_{jl}(x), \hspace{4mm} f_k(y) = f_l\circ \mu_{kl}(y)$$

we can say that $f_j(x) + f_k(y) = f_l(x) +f_l(y)$. This is because $\mu_{jl}(x)$ and $\mu_{kl}(y)$ just views $x$ and $y$ respectively as elements of $M_l$. It follows that $L$ is additive. It now remains to check that $L$ is uniquely determined by the $f_i$. Suppose we have another map $\varphi : \bigcup M_i \longrightarrow N$ such that $\varphi \circ \tau_i = f_i$. Then it follows that

$$\varphi \circ \tau_i(x) = L \circ \tau_i(x)$$

given any $i \in I$ and $x \in M_i$. But then since the $\tau_i$ are inclusion maps this is effectively saying that $L(x) = \varphi(x)$ for all $x \in M_i$ for any $M_i$. Since $x$ is arbitrary we conclude that $L = \varphi$ showing uniqueness. This completes the proof that

$$\varinjlim M_i \cong \bigcup M_i.$$

Edit: I think you have your universal properties wrong: The universal property of the direct limit is this: If you have $N$ an $A$ - module and for each $i \in I$ let $\alpha_i : M_i \to N$ be an $A$ - module homomorphism such that $\alpha_i = \alpha_j \circ \mu_{ij}$ whenver $i \leq j$. The there exists a unique homomorphism $\alpha : M \to N$ such that $\alpha_i = \alpha \circ \mu_i$ for all $i \in I$.

Edit: Let me show you why it suffices to show that $\bigcup M_i$ satisfies the universal property of $\varinjlim M_i$ to prove that $\bigcup M_i \cong \varinjlim M_i$. Now we have inclusion maps $\tau_i : M_i \to \bigcup M_i$ such that clearly $\tau_i = \tau_j \circ \mu_{ij}$ for any $i \leq j$ because the $\mu_{ij}$ are just inclusion maps. So now we just need to show that $\bigcup M_i$ has the property that given compatible $A$ - module homomorphisms $f_i : M_i \to N$ for some $A$ - module $N$ such that $f_i = f_j \circ \mu_{ij}$ whenever $i \leq j$, there is a unique $A$ - linear map $L : \bigcup M_i \to N$ such that

$$f_i = L \circ \tau_i \hspace{3mm} \forall i\in I.$$

To see why we only need to verify this, suppose the result above about $\bigcup M_i$ holds. Then putting in place of $f_i$ the usual maps $\phi_i$ from $M_i$ to $\varinjlim M_i$ and $N = \varinjlim M_i$, we have a unique linear map $L : \bigcup M_i \longrightarrow \varinjlim M_i$ such that

$$\phi_i = L \circ \tau_i \hspace{3mm} \forall i \in I.$$

Recall how the maps $\phi_i$ are defined. We have $M_i \stackrel{\lambda_i}{\longrightarrow} \bigoplus_{i\in I} M_i \stackrel{ \mu}{\longrightarrow} \varinjlim M_i$ so

$$\phi_i \stackrel{\text{def}}{\equiv} \mu \circ \lambda_i$$

where $\mu$ is the canonical projection from the direct sum onto the direct limit, and the $\lambda_i$ the canonical injections from $M_i$ into the direct sum. Now because we have maps $\tau_i$ out of $M_i$ to $\bigcup M_i$ such that $\tau_i = \tau_j \circ \mu_{ij}$ whenever $i\leq j$, then by the universal property of the direct limit, there is a unique $A$ - module homomorphism $L': \varinjlim M_i \rightarrow \bigcup M_i$ such that

$$\tau_i = \phi_i \circ L' \hspace{3mm} \forall i \in I.$$

From here it is not hard to see that $L$ and $L'$ are mutual inverses so that $\bigcup M_i \cong \varinjlim M_i$.

$\hspace{6in} \square$

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Thanks mate, awesome answer! The inclusions are injective so I don't need to define their inverse: it's just the element. No? (Thanks for the edit.) – Rudy the Reindeer Aug 3 '12 at 12:36
@ClarkKent Reading your proof again, you have got your universal property wrong. A priori, you want maps $f_i : M_i \to Y$. – user38268 Aug 3 '12 at 12:47
Sorry that was a typo. I fixed it now. – Rudy the Reindeer Aug 3 '12 at 19:29
Can you tell me what you mean by "how do you define $i^{-1}(m)$"? – Rudy the Reindeer Aug 3 '12 at 19:36
@ClarkKent You have got your universal properties wrong. For example look at the universal property that I stated above. The maps $\alpha_i : M_i \to N$ are any maps, not necessarily inclusions. To paraphrase what you wrote above: " To this end, let $Y$ be a module and $i′_i$ be inclusions $i′_i:M_i \to Y"$. Why must your maps from $M_i$ to any $A$ - module $Y$ be inclusions? Already this does not make any sense. – user38268 Aug 3 '12 at 23:38