Archive for the ‘Direct and Inverse Limit’ Category

Theorem. Let R be a commutative ring, A an R-module and \{M_i, f_{ij} \} a direct system of R-modules. Then

\varinjlim (A \otimes_R M_i) \cong A \otimes_R \varinjlim M_i.

The above isomorphism is an isomorphism of R-modules. We will give the proof of this theorem at the end of this post. So let’s get prepared for the proof!

Notation 1R is a commutative ring, A is an R-module and \{M_i, f_{ij} \} is a direct system of R-modules over some partially ordered index set I. We will not assume that I is directed.

Notation 2. We proved in here that M=\varinjlim M_i exists. Recall that, as a part of the definition of direct limit, we also have canonical homomorphisms f_i : M_i \longrightarrow M satisfying the relations f_jf_{ij}=f_i, for all i \leq j. For modules we explicity defined f_i by f_i(x_i)=\rho_i(x_i) + N (see the theorem in here), but we will not be needing that.

Notation 3. Let A be an R-module. For all i,j \in I with i \leq j we will let M_i' = A \otimes_R M_i and f'_{ij}=\text{id}_A \otimes_R f_{ij}. By Example 5, \{M_i', f'_{ij} \} is a direct system of R-modules and so M'=\varinjlim (A \otimes_R M_i) exists. We also have canonical homomorphisms f_i' : M_i' \longrightarrow M' satisfying f_j'f_{ij}'=f_i', for all i \leq j.

Notation 4. For every i \in I we will let h_i = \text{id}_A \otimes f_i.

Remark 1.  Clearly h_i : M'_i \longrightarrow A \otimes_R M and h_jf_{ij}'=h_i, for all i \leq j, because

h_j f_{ij}' = (\text{id}_A \otimes f_j)(\text{id}_A \otimes f_{ij})=\text{id}_A \otimes f_jf_{ij}=\text{id}_A \otimes f_i = h_i.

 Lemma. Let X be an R-module. Suppose that for every i \in I there exists an R-module homomorphism g_i : M_i' \longrightarrow X such that g_jf_{ij}' = g_i, for all i \leq j. Let a \in A. Then

1) There exist R-module homomorphisms \nu_{i,a} : M_i \longrightarrow X such that \nu_{j,a} f_{ij}=\nu_{i,a}, for all i \leq j.

2) There exists a unique R-module homomorphism \nu_a : M \longrightarrow X such that \nu_a f_i = v_{i,a}, for all i \in I.

Proof. 1) \nu_{i,a} are defined very naturally: \nu_{i,a}(x_i)=g_i(a \otimes_R x_i), for all x_i \in M_i. Then

\nu_{j,a} f_{ij}(x_i)=g_j(a \otimes_R f_{ij}(x_i))=g_j f_{ij}'(a \otimes_R x_i)=g_i(a \otimes_R x_i)=\nu_{i,a}(x_i).

2) This part is obvious from the first part and the universal property of direct limit. (See Definition 1 in here) \Box

Proof of the Theorem. We will show that A \otimes_R M satisfies the conditions in the definition of M' = \varinjlim M_i' (see Definition 1 in here) and thus, by the uniqueness of direct limit, A \otimes_R M \cong M' and the theorem is proved. The first condition is satisfied by Remark 1. For the second condition (the universal property), suppose that X is an R-module and g_i : M_i' \longrightarrow X are R-module homomorphisms such that g_jf_{ij}'=g_i, whenever i \leq j. So the hypothesis in the above lemma is satisfied. For a \in A and i \in I let \nu_{i,a} and \nu_a be the maps in the lemma. Define the map \nu : A \times M \longrightarrow X by v(a,x)=v_a(x), for all a \in A and x \in M. See that \nu is R-bilinear and so it induces an R-module homomorphism f : A \otimes_R M \longrightarrow X defined by f(a \otimes_R x)=v_a(x). We also have

 fh_i(a \otimes_R x_i)=f(a \otimes_R f_i(x_i))=\nu_af_i(x_i)=\nu_{i,a}(x_i)=g_i(a \otimes_R x_i),

for all a \in A, \ i \in I and x_i \in M_i. Thus fh_i = g_i. So the only thing left is the uniqueness of f, which is obvious because, as we mentioned in the second part of the above lemma, \nu_a is uniquely defined for a given a \in A. \ \Box

Remark 2. If R is not commutative, M_i are left R-modules and A is a right R-module (resp. (R,R)-bimodule), then the isomorphism in the theorem is an isomorphism of abelian groups (resp. left R-modules).

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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

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

Definition. A partially ordered set (I, \leq ) is called a directed set if for every i,j \in I there exists some k \in I such that i \leq k and j \leq k.

Example 1. Let \{S_i \}_{i \in I} be a collection of subsets of a set, where I is a directed set and i \leq j iff S_i \subseteq S_j. Let \{S_i, f_{ij} \} be the direct system defined in Example 1 in this post. Then \varinjlim S_i=\bigcup S_i.

Proof. Let T:=\bigcup_{i \in I} S_i. For every i \in I we define f_i : S_i \longrightarrow T to be the inclusion map.

1) If s \in S_i and i \leq j, then f_j f_{ij}(s)=f_j(s)=s=f_i(s). Thus f_j f_{ij}=f_i.

2) Suppose that there exists a set X and the maps g_i: S_i \longrightarrow X such that g_jf_{ij}=g_i for all i \leq j. Let’s see how we have to define f. Suppose that f: T \longrightarrow X is a map with ff_i = g_i, for all i. Let s \in T. Then s \in S_i for some i and hence f(s)=ff_i(s)=g_i(s). Now, if s is also in S_j, then choosing k \in I with i \leq k and j \leq k we will have  g_i(s)=g_k f_{ik}(s)=g_k(s) and g_j(s)=g_kf_{jk}(s)=g_k(s). Thus g_i(s)=g_j(s). So we must define f(s)=g_i(s) whenever s \in S_i. We just proved that f is well-defined. Clearly ff_i=g_i for all i. \ \Box

Remark. Clearly the result in Example 1 holds if S_i were groups, rings, modules, etc.

Example 2. Let S be a multiplicatively closed subset of a commutative ring R with 1 \in S and 0 \notin S. Let \{R_i, f_{ij} \}, \ i,j \in S, be the direct system defined in Example 2 in this post. Then \varinjlim R_i =S^{-1}R.

Proof. Let T:=S^{-1}R and, for every i \in S, define f_i : R_i \longrightarrow T to be the inclusion map.

1) If i \leq j, then j=ri for some r \in R and hence \displaystyle \frac{r^na}{j^n}=\frac{a}{i^n} for all a \in R and n \geq 1. That means f_jf_{ij}=f_i.

2) Suppose that there exists a ring X and the ring homomorphisms g_i: R_i \longrightarrow X such that g_jf_{ij}=g_i for all i \leq j. Let f: T \longrightarrow X be a ring homomorphism such that ff_i = g_i, for all i. Let’s see how we have to define f.  Let t = a/i \in T. Then t \in R_i and thus f(t)=ff_i(t)=g_i(t). Also if t = a/i = b/j, then  choosing k=ij we’ll have i \leq k, \ j \leq k. Thus

g_i(t)=g_k f_{ik}(t)=g_k(ja/k)=g_k(a/i)=g_k(t)

and g_j(t)=g_k f_{jk}(t)=g_k(ib/k)=g_k(b/j)=g_k(t). So we must define f(a/i)=g_i(a/i). We just proved f is well-defined and clearly f is a ring homomorphism and ff_i=g_i for all i. \ \Box

Example 3. Let R be a commutative ring with 1 and let R[x] be the polynomial ring in the indeterminate x. Let L = \langle x \rangle, the ideal of R[x] generated by x. For every positive integer i let R_i = R[x]/L^i. Let \{R_i, f_{ij} \} be the inverse system in Example 3 in this post. Then \varprojlim R_i = R[[x]].

Proof. Let T:=R[[x]] and define the map f_i : T \longrightarrow R_i by f_i(h)=h + L^i for all i and h \in T.

1) If i \leq j and h \in T, then f_{ij}f_j(h)=f_{ij}(h + L^j)=h+ L^i=f_i(h). Thus f_{ij}f_j = f_i.

2) Suppose first that there exists a ring X and the ring homomorphisms g_i : X \longrightarrow R_i such that f_{ij}g_j=g_i for all i \leq j. Suppose also that there exists a ring homomorphism f: X \longrightarrow T such that f_if = g_i, for all i. Then for any positive integer i and all u \in X we will have g_i(u)=f_i f(u)=f(u) + L^i. So here is how we have to define f: for every positive integer i we have

g_i(u)=\sum_{m=0}^{i-1} a_{mi}(u)x^{i-1} + L^i,

where a_{mi}(u) \in R. Now we define

f(u)=\sum_{m=0}^{\infty} a_{m, m+1}(u) x^m.

Clearly f is well-defined and f(u+v)=f(u)+f(v) for all u,v \in X. So,  to prove that f is a ring homomorphism we must show that f(uv)=f(u)f(v) for all u,v \in X. To prove this, we only need to show that for every integere k \geq 0, the coefficient of x^k in both f(uv) and f(u)f(v) are equal. To do so, we begin with the fact that f_{ij}g_j=g_i for all i \leq j,which gives us

a_{i, i+1}(u)=a_{i, j+1}(u) \ \ \ \ \ \ \ \ \ \ (1)

for all i \leq j. Now, the coefficient of x^k in f(u)f(v) is

\sum_{i=0}^k a_{i,i+1}(u)a_{k-i, k-i+1}(v). \ \ \ \ \ \ \ \ \ \ (2)

On the other hand, the coefficient of x^k in f(uv) is a_{k,k+1}(uv). But, since g_{k+1}(uv)=g_{k+1}(u)g_{k+1}(v), we will have

a_{k,k+1}(uv)=\sum_{i=0}^k a_{i,k+1}(u)a_{k-i,k+1}(v). \ \ \ \ \ \ \ \ \ (3).

It is clear from (1), (2) and (3) that the coefficients of x^k in both f(uv) and f(u)f(v) are equal.

Finally, using (1), we have

f_if(u)=f(u) + L^i=\sum_{m=0}^{i-1}a_{m,m+1}(u)x^m + L^i=\sum_{m=0}^{i-1}a_{mi}(u)x^m + L^i = g_i(u)

and thus f_if=g_i. \ \Box

Definition 1. Let \{M_i, f_{ij} \} be a direct system in a category \mathcal{C}. Suppose there exists an object M and morphisms f_i : M_i \longrightarrow M, for every i, such that the following conditions are satisfied:

1) f_j f_{ij}=f_i, whenever i \leq j,

2) (The universal property) Suppose that there exist an object X and morphisms g_i : M_i \longrightarrow X, \ i \in I, such that g_j f_{ij}=g_i whenever i \leq j. Then there exists a unique morphism f: M \longrightarrow X such that ff_i = g_i, \ i \in I.

Then M is called the direct limit of our direct system and we will just write \varinjlim M_i = M.

Definition 2. Let \{M_i, f_{ij} \} be an inverse system in a category \mathcal{C}. Suppose that there exists an object M and morphisms f_i : M \longrightarrow M_i, for every i, such that the following conditions are satisfied:

1) f_{ij}f_j=f_i, whenever i \leq j,

2) If there exists some object X and morphisms g_i : X \longrightarrow M_i, for every i, such that f_{ij}g_j=g_i whenever i \leq j, then there must exist a unique morphism f : X \longrightarrow M such that f_if = g_i for all i.

Then M is called the inverse limit of our inverse system and we will just write \varprojlim M_i = M.

In our definitions I wrote “the” direct limit and “the” inverse limit. That is because “the” direct (resp. inverse) limit of a direct (resp. inverse) system, if it exists, is unique, up to isomorphism of course, as we are going to see:

Fact. The direct (resp. inverse) limit of a direct (resp. inverse) system is unique up to isomorphism if it exists.

Proof. I will prove the fact for direct limit only. The proof for inverse limit is similar. Suppose that both \{M, f_i \} and \{M', f'_i \} satisfy the conditions in Dedinition 1. Then, by condition 2), there exist (unique) morphisms f: M \longrightarrow M' and f': M' \longrightarrow M such that ff_i=f'_i and f'f'_i=f_i. Thus the morphism f'f : M \longrightarrow M satisfies (f'f)f_i=f_i. But we also have that the identity morphism \text{id}_M : M \longrightarrow M satisfies \text{id}_M f_i = f_i. Thus, by the uniqueness condition in 2), we must have f'f = \text{id}_M. Similarly ff'= \text{id}_{M'} and hence f is an isomorphism and f' is its inverse.

Remark. You might have found the relationships between the morphisms in the definition of direct (resp. inverse) systems and limits complicated but they are not! If you draw diagrams, then you will see that they just mean that those diagrams are commutative.

Example 1. In the category of sets let S be a set and \{S_i \}_{i \in I} a collection of subsets of S. We also have i \leq j iff S_i \subseteq S_j. The map f_{ij}: S_i \longrightarrow S_j is defined to be the inclusion map. Clearly \{S_i, f_{ij} \} is a direct system.

Example 2. Let R be a commutative ring and let I be a non-empty subset of R such that no element of I is nilpotent. The partial order \leq is defined onI by i \leq j iff i \mid j in R, i.e. j=ri for some r \in R. For every i \in I let R_i be the localization of R at \{1,i,i^2, \cdots \}. We need now to define the morphisms f_{ij}. If i \leq j, then j=ri for some r \in R and so we can define f_{ij}: R_i \longrightarrow R_j by f_{ij}(a/i^n)=(r^na)/j^n for all a \in R, \ n \geq 0. It is easy to see that f_{ij} is a ring homomorphism. Clearlyf_{ii} is the identity map over R_i because in this case we may choose r=1. Also, if i \leq j \leq k with j=ri and k=sj, then k=rsi and hence

f_{jk}f_{ij}(a/i^n)=f_{jk}((r^na)/j^n)=(s^nr^na)/k^n=((rs)^na)/k^n = f_{ik}(a/i^n).

Thus f_{jk}f_{ij}=f_{ik} and so \{R_i, f_{ij} \} is a direct system in the category of rings.

Example 3. Let R be a ring and let L be a two-sided ideal of R. Let I be the set of positive integers and put R_i = R/L^i for every i \in I. For every i \leq j the ring homomorphism f_{ij}: R_j\longrightarrow R_i is defined naturally by f_{ij}(r + L^j)=r+L^i. This ring homomorphisms are well-defined because if i \leq j, then L^j \subseteq L^i. Clearly f_{ii} is the identity map of R_i and if i \leq j \leq k, then

f_{ij}f_{jk}(r + L^k)=f_{ij}(r + L^j)=r+L^i = f_{ik}(r+L^k).

Thus \{R_i, f_{ij} \} is an inverse system in the category of rings.

Similarly, if A is an R-module and we put A_i = A/L^iA, then we will have the inverse system \{A_i, g_{ij} \} in the category of R-modules. Here g_{ij}: A_j \longrightarrow A_i is defined by g_{ij}(a + L^jA)=a+L^i A whenever i \leq j.

Example 4. Let G be a group and let \{N_i \}_{i \in I} be a family of normal subgroups of G which have finite index in G. We define the partial order \leq on I by i \leq j if and only if N_j \subseteq N_i. We also define, whenever i \leq j, the group homomorphism f_{ij}: G/N_j \longrightarrow G/N_i by f_{ij}(gN_j)=gN_i. It is easily seen that \{G/N_i, f_{ij} \} is an inverse system in the category of finite groups.

Example 5. Let R be a commutative ring, let A be an R-module and let \{M_i, f_{ij} \} be a direct system of R-modules over some partially ordered index set I. For every i \in I let M_i'=A \otimes_R M_i and for all i,j \in I with i \leq j define f'_{ij} : M_i' \longrightarrow M_j' by f'_{ij}=\text{id}_A \otimes_R f_{ij}. That means f'_{ij}(a \otimes_R x)=a \otimes_R f_{ij}(x), for all a \in A and x \in M_i.

Claim . \{M_i', f_{ij}' \} is a direct system of R-modules.

Proof.  So we need to prove two things to show that \{M_i', f'_{ij} \} is a direct system:

1) f'_{ii} = \text{id}_{M_i'}, for all i \in I. This is obvious because f_{ii} = \text{id}_{M_i} and hence, for every a \in A and x \in M_i we have f'_{ii}(a \otimes_R x)=a \otimes_R f_{ii}(x)=a \otimes_R x.

2) f'_{jk}f'_{ij}=f'_{ik}, whenever i \leq j \leq k. This also follows very easily from f_{jk}f_{ij}=f_{ik}.

Definition 1. Let I be a set partially ordered by \leq (we only need \leq to be reflexive and transitive) and let \mathcal{C} be a category. Let \{M_i \}_{i \in I} be a family of objects in \mathcal{C} and suppose that for every i, j \in I with i \leq j there exists a morphism f_{ij} : M_i \longrightarrow M_j such that

1) f_{ii}=\text{id}_{M_i}, for all i \in I,

2) f_{jk}f_{ij}=f_{ik} whenever i \leq j \leq k.

Then the objects M_i together with the morphisms f_{ij} is called a direct sysetem in \mathcal{C}. We will use the notation \{M_i, f_{ij} \} for this direct system.

If we change the direction of the arrows in the above definition, we’ll get the definition of a inverse system , i.e.

Definition 2. Let I be a set partially ordered by \leq (we only need \leq to be reflexive and transitive) and let \mathcal{C} be a category. Let \{M_i \}_{i \in I} be a family of objects in \mathcal{C} and suppose that for every i, j \in I with i \leq j there exists a morphism f_{ij} : M_j \longrightarrow M_i such that

1) f_{ii}=\text{id}_{M_i},

2) f_{ij}f_{jk}=f_{ik} whenever i \leq j \leq k.

Then the objects M_i together with the morphisms f_{ij} is called an inverse sysetem in \mathcal{C}. We will use the notation \{M_i, f_{ij} \} for this inverse system.