Direct limit of modules (1)

Posted: January 17, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
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Theorem. Let $R$ be a ring and let $\{M_i, f_{ij} \}$ be a direct system of left $R$-modules over some partially ordered index set $I.$ Let $M= \bigoplus_{i \in I} M_i$ and for every $i \in I,$ let $\rho_i : M_i \longrightarrow M$ be the natural injection map, i.e. $\rho_i(x_i)=(x_j)_{j \in I},$ where $x_j=0$ for all $j \neq i.$ Define the $R$-submodule $N$ of $M$ by

$N = \langle \rho_jf_{ij}(x_i) - \rho_i(x_i): \ i \leq j, \ x_i \in M_i \rangle.$

Then $\varinjlim M_i = M/N.$

Proof. For every $i \in I$ define the $R$-module homomorphism $f_i : M_i \longrightarrow M/N$ by $f_i(x_i)=\rho_i(x_i) + N.$ Clearly if $i \leq j,$ then $\rho_j f_{ij}(x_i) - \rho_i(x_i) \in N$ and thus

$f_j f_{ij}(x_i) = \rho_j f_{ij}(x_i) + N = \rho_i(x_i) + N = f_i(x_i).$

So $f_j f_{ij} = f_i.$ Suppose now that there exists a left $R$-module $X$ and $R$-module homomorphisms $g_i : M_i \longrightarrow X$ such that $g_j f_{ij}=g_i$ whenever $i \leq j.$ We need to prove that there exists a unique $R$-module homomorphism $f: M/N \longrightarrow X$ such that $ff_i = g_i$ for all $i.$ Let’s see how this $f$ must be defined. Well, we have $g_i(x_i)=ff_i(x_i)=f(\rho_i(x_i) + N).$ Now, since every element of $M/N$ is in the form $\sum \rho_i(x_i) + N$ we have to define $f$ in this form: $f(\sum \rho_i(x_i) + N)=\sum g_i(x_i).$ The only thing left is to prove that $f$ is well-defined. To do so, we define the homomorphism $\hat{f} : M \longrightarrow X$ by $\hat{f}(\sum \rho_i(x_i)) = \sum g_i(x_i).$ So I only need to prove that $N \subseteq \ker \hat{f}.$ So suppose that $i \leq j$ and let $x_i \in M_i.$ Then

$\hat{f}(\rho_j f_{ij}(x_i) - \rho_i (x_i))=g_jf_{ij}(x_i) - g_i(x_i)=0,$

because $g_j f_{ij}=g_i$ for all $i \leq j. \ \Box$

Remark 1. The direct limit of a direct system of abelian groups always exists by the theorem because abelian groups are just $\mathbb{Z}$-modules.

We now show that if $I$ is a directed set, then $\varinjlim M_i$ will look much simpler.

Remark 2. If, in the theorem, $I$ is directed, then $\varinjlim M_i = \{\rho_i(x_i) + N : \ i \in I, \ x_i \in M_i \}.$

Proof. Let $a = \sum_{i \in J} \rho_i(x_i) + N$ be any element of $\varinjlim M_i.$ Since $I$ is directed and $J$ is finite, there exists some $k \in I$ such that $i \leq k$ for all $i \in J.$ Then for every $i \in J$ we’ll have

$\rho_i(x_i) +N = \rho_k f_{ik}(x_i) + \rho_i(x_i) - \rho_k f_{ik}(x_i) + N = \rho_k f_{ik}(x_i) + N.$

So if we let $y_k = \sum_{i \in J} f_{ik}(x_i) \in M_k,$ then $a = \rho_k (y_k) + N. \ \Box$

Remark 3. If, in the theorem, $I$ is directed and $\{M_i, f_{ij} \}$ is a direct system of rings, then $\varinjlim M_i$ is also a ring.

Proof. Since $M_i$ are all abelian groups, $\varinjlim M_i$ exists and it is an abelian group by the theorem. So we just need to define a multiplication on $\varinjlim M_i.$ Let $a,b \in \varinjlim M_i.$ By Remark 2 there exist $i,j \in I$ such that $a = \rho_i(x_i) + N, \ b = \rho_j(x_j) + N.$ Choose $k \in I$ with $i,j \leq k.$ Then by the definition of $N$

$a=\rho_i(x_i) + N = \rho_k f_{ik}(x_k) + N$

and thus $a= \rho_k (y_k) + N.$ Similarly $b = \rho_k(z_k) + N.$ Now we define

$ab = \rho_k(y_k z_k) + N.$

We need to show that the multiplication we’ve defined is well-defined. So suppose that $a = \rho_{\ell}(y_{\ell}) + N$ and $b = \rho_{\ell}(z_{\ell}) + N$ is another way of representing $a,b.$ Choose $n \in I$ such that $k, \ell \leq n.$ Then

$\rho_k(y_k z_k) + N = \rho_n f_{kn}(y_k z_k) + N = (\rho_nf_{kn}(y_k) + N)(\rho_n f_{kn}(z_k) + N). \ \ \ \ (1)$

But we also have

$\rho_n f_{kn}(y_k) + N = \rho_k(y_k) + N=a= \rho_{\ell}(y_{\ell}) + N = \rho_n f_{\ell n}(y_{\ell}) + N \ \ \ \ \ \ (2)$

and similarly

$\rho_n f_{kn}(z_k) = \rho_n f_{\ell n}(z_{\ell}) + N. \ \ \ \ \ (3)$

Plugging (2) and (3) into (1) will give us

$\rho_k(y_kz_k) + N =\rho_n f_{\ell n}(y_{\ell} z_{\ell}) + N = \rho_{\ell}(y_{\ell} z_{\ell}) + N,$

which proves that the multiplication we defined is well-defined. $\Box$