Direct limit of modules (2)

Posted: January 19, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
Tags: ,

In this post we will keep the notation in the previous part. We will assume that $I$ is a directed set and $\{M_i, f_{ij} \}$ is a direct systen of $R$-modules. In Remark 2 in there we proved that

$\varinjlim M_i=\{ \rho_i(x_i) + N : \ i \in I, x_i \in M_i \}.$

Our goal now is to find a necessary and sufficient condition for an element of $\varinjlim M_i$ to be $0$, i.e. we want to see when we will have $\rho_i(x_i) \in N$. Let’s start with a notation which will simlify our work.

Notation. For any $j,k \in I$ with $j \leq k$ and $y_j \in M_j$ we let $\alpha(j,k,y_j)=\rho_k f_{jk}(y_j)-\rho_j(y_j).$ Note that $\alpha(j,k,y_j)$ are just the generators of $N$ as an $R$-submodule of $M = \bigoplus M_i.$

Now two easy properties of $\alpha$:

Lemma 1. If $r \in R,$ then

1) $r\alpha(j,k,y_j)=\alpha(j,k,ry_j),$

2)  $\alpha(j,k,y_k) + \alpha(j,k,y'_k)=\alpha(j,k,y_k+y'_k),$

Proof. Trivial because $\rho_j, \rho_k$ and $f_{jk}$ are all $R$-module homomorphisms. $\Box.$

Lemma 2. If $j \leq k \leq t$ and $y_j \in M_j,$ then $\alpha(j,k,y_j)=\alpha(j,t,y_j)+ \alpha(k,t, - f_{jk}(y_j).$

Proof. This is also a trivial result of this fact that $f_{kt}f_{jk}=f_{jt}. \ \Box$

And the main result of this section:

Theorem. $\rho_i(x_i) \in N$ if and only if $f_{it}(x_i)=0$ for some $t \in I$ with $i \leq t.$

Proof. If $f_{it}(x_i)=0$ for some $t$ with $i \leq t,$ then $\rho_i(x_i)=\rho_i(x_i) - \rho_t f_{it}(x_i) \in N.$ Conversely, supppose that $\rho_i(x_i) \in N.$ Then by Lemma 1, part 1, we can write

$\rho_i(x_i) = \sum \alpha(j,k,y_j) \ \ \ \ \ \ \ (1).$

Let $t \in I$ be such that $i,j,k \leq t$ for all $j,k$ occuring in (1). Since  $\rho_t f_{it}(x_i)=\rho_t f_{it}(x_i) - \rho_i(x_i) + \rho_i(x_i),$ we will have

$\rho_t f_{it}(x_i) = \alpha(i,t,x_i) + \sum \alpha(j,k,y_j) \ \ \ \ \ \ \ \ \ (2),$

by (1). Applying Lemma 2 to (2) we will get

$\rho_t f_{it}(x_i) = \sum \alpha(\ell, t, z_{\ell}) \ \ \ \ \ \ \ \ \ (3),$

where the sum runs over $\ell.$ Note that by Lemma 1, part 2, we may assume that all the indices $\ell$ in (3) are distinct. We can rewrite (3) as

$\rho_t f_{it}(x_i) = \rho_t \left (\sum f_{\ell t}(z_{\ell}) \right) - \sum \rho_{\ell}(z_{\ell}) \ \ \ \ \ \ \ \ \ (4).$

But since every element of $\bigoplus M_i$ is uniquely written as $\sum \rho_j(x_j),$ in (4) we must have $\rho_{\ell}(z_{\ell})=0,$ and hence $z_{\ell}=0,$ for all $\ell \neq t.$ Thus (4) becomes

$\rho_t f_{it}(x_i) = \rho_t f_{tt}(z_t) - \rho_t(z_t) = 0,$

which gives us $f_{it}(x_i)=0. \ \Box$

Corollary. Let $\{M_i, f_{ij} \}$ be a direct system of rings. We proved in Remark 3 in this post that $M = \varinjlim M_i$ is also a ring. If each $M_i$ is a domain, then $M$ is also a domain.

Proof. Suppose that $a = \rho_i(x_i) + N, \ b = \rho_i(y_i) + N$ are two elements of $\varinjlim M_i$ and $ab = 0.$ That means $\rho_i(x_iy_i) \in N$ and hence, by the theorem, there exists some $t$ with $i \leq t$ such that

$0=f_{it}(x_iy_i)=f_{it}(x_i)f_{it}(y_i).$

Therefore, since $M_t$ is a domain, we have either $f_{it}(x_i)=0$ or $f_{it}(y_i)=0.$ Applying the theorem again, we get either $a=0$ or $b = 0. \ \Box$