## Direct limit of modules & tensor product

Posted: February 16, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
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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 1$R$ 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).

## 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 $b = 0. \ \Box$

## 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. $\Box$

## Direct and inverse limits; basic examples

Posted: January 16, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
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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 $f_if=g_i. \ \Box$

## Direct and inverse limits; definition & uniqueness

Posted: January 12, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
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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.

## Direct and inverse systems; basic examples

Posted: January 12, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
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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 on$I$ 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. Clearly$f_{ii}$ 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 $f_{jk}f_{ij}=f_{ik}.$

## Direct and inverse systems; definition

Posted: January 12, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
Tags: ,

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.