## 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. $\Box$