Direct limit of modules (2)

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

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