08. Quotient Modules and the Isomorphism Theorems

Quotient Modules

Let $N$ be a submodule of $M$

Module $M / N$ is the quotient module of $M$ by $N$ If
For $r \in R$ and $m + N \in M / N$ then
$r . (m + N) = r . m + N$
Well-defined as if $m_{1} + N = m_{2} + N$ then $m_{1} - m_{2} \in N$ so $r . (m_{1} - m_{2}) \in N$ hence
$r . m_{1} + N = r . m_{2} + N$

Link to original

Module Homomorphism

Let $M_{1}, M_{2}$ be $R$ -modules then

$ϕ : M_{1} \to M_{2}$ is a $R$ -module homomorphism if

$ϕ (m_{1} + m_{2}) = ϕ (m_{1}) + ϕ (m_{2})$ for all $m_{1}, m_{2} \in M_{1}$

$ϕ (r . m) = r . ϕ (m)$ for all $r \in R$ , $m \in M_{1}$

So $ϕ$ respects addition and multiplication by rings elements with

$ker (ϕ) = {m \in M_{1} : ϕ (m) = 0}$ is a submodules of $M_{1}$
$im (ϕ) = {ϕ (m) : m \in M_{1}}$ is a submodules of $M_{2}$

Link to original

Submodule Correspondence lemma

Let $M$ be an $R$ -module
Let $N$ be a submodule of $M$

Let $q : M \to M / N$ be the quotient map

If $S$ is a submodule of $M$ then $q (S)$ is a submodule of $M / N$
If $T$ a submodule of $M / N$ then $q^{- 1} (T)$ is a submodule of $M$

Map $T \mapsto q^{- 1} (T)$ is an injective from submodules of $M / N$ to submodules of $M$ containing $N$
Thus $M / N$ correspond bijectively to submodules of $M$ which contain $N$

Proof

For $^{- 1} (T)$
If $m_{1}, m_{2} \in q^{- 1} (T)$ then
$q (m_{1}), q (m_{2}) \in T$
As $T$ is a submodule then
$q (m_{1}) + q (m_{2}) = q (m_{1} + m_{2}) \in T$
Hence $m_{1} + m_{2} \in q^{- 1} (T)$

If $r \in R$ then
$q (r . m_{1}) = r . q (m_{1}) \in T$
As $q (m_{1}) \in T$ and $T$ is a submodule so $r . m_{1} \in q^{- 1} (T)$
Hence $q^{- 1} (T)$ is a submodule of $M$

Similar process for $q (S)$

As $T$ is any subset of $M / N$ then $q (q^{- 1} (T)) = T$ as $q$ is surjective

Then $q^{- 1} (T)$ is a submodule in $M$ then map $S \mapsto q (S)$ is surjective in submodules in $M$ to submodules in $M / N$
Then $T \mapsto q^{- 1} (T)$ is an injective map

As $q (N) = {0} \subseteq T$ then for any submodule $T$ of $M / N$ then
$N \subseteq q^{- 1} (T)$
Hence image of map $T \mapsto q^{- 1} (T)$ consists of submodules of $M$ containing $N$

Suppose $S$ is an arbitrary submodule of $M$ and consider $q^{- 1} (q (S))$ then
$q^{- 1} q (S) = {m \in M : q (m) \in q (S)} = {m \in M : \exists s \in S such that m + N = s + N} = {m \in M : \exists s \in S such that m \in s + N} = S + N$
As $S$ contains $N$ then $S + N = S$ so $q^{- 1} (q (S)) = S$
Hence any submodule $S$ containing $N$ is preimage of submodule of $M / N$

Link to original

07 - Universal Property of Quotients

Universal Property of Quotients

Let $ϕ : M \to N$ be a homomorphism of $R$ -modules
Let $S$ be a submodule of $M$ with $S \subseteq ker (ϕ)$

Let $q : M \to M / S$ be quotient homomorphism then

Exists unique homomorphism
$\overline{ϕ} : M / S \to N$
such that
$ϕ = \overline{ϕ} \circ q$
With $ker (\overline{ϕ})$ is submodule $ker (ϕ) / S = {m + S : m \in ker (ϕ)}$

Proof

As $q$ surjective then
$\overline{ϕ} (q (m)) = ϕ (m)$
uniquely determines values of $\overline{ϕ}$ so $\overline{ϕ}$ is unique if it exists

If $m - m^{'} \in S$ then as $S \subseteq ker (ϕ)$ then
$0 = ϕ (m - m^{'}) = ϕ (m) - ϕ (m^{'})$
Hence $ϕ$ is constant on $S$ -cosets thus induces a map on
$\overline{ϕ} (m + S) = ϕ (m)$
As $\overline{ϕ}$ is a homomorphism then
By definition of module structure on quotient $M / S$ then
$ϕ = \overline{ϕ} \circ q$
As $\overline{ϕ} (m + S) = ϕ (m) = 0$ if and only if $m \in ker (ϕ)$
Hence
$m + S \in ker (ϕ) / S$

Link to original

Isomorphism Theorem for Modules corollary

First Isomorphism Theorem

Let $ϕ : M \to N$ is a homomorphism then

$ϕ$ induces an isomorphism
$\overline{ϕ} : M / ker (ϕ) \to im (ϕ)$

Second Isomorphism Theorem

Let $M$ be an $R$ -module
Let $N_{1}, N_{2}$ are submodules of $M$ then
$(N_{1} + N_{2}) / N_{2} ≅ N_{1} / (N_{1} \cap N_{2})$

Third Isomorphism Theorem

Let $N_{1} \subseteq N_{2}$ be submodules of $M$ then
$(M / N_{1}) / (N_{2} / N_{1}) ≅ M / N_{2}$

Proof

For First Isomorphism Theorem then
Apply Universal Property of Quotients to $S = ker (ϕ)$
As
$ker (\overline{ϕ}) = ker (ϕ) / ker (ϕ) = 0$
Hence $\overline{ϕ}$ is injective so
Induces isomorphism onto image from $\overline{ϕ} \circ q = ϕ$ which is $im (ϕ)$

For Second Isomorphism Theorem

Let $q : M \to M / N_{2}$ be the quotient map
$q$ restricts to homomorphism $p$ from $N_{1}$ to $M / N_{2}$ with image $(N_{1} + N_{2}) / N_{2}$

By First Isomorphism Theorem then need to check kernel of $p$ is $N_{1} \cap N 2$
If $n \in N_{1}$ and $p (n) = 0$ then $n + N_{2} = 0 + N_{2}$ hence $n \in N_{2}$ so $n \in N_{1} \cap N_{2}$

For Third Isomorphism Theorem

Let $q_{i} : M \to M / N_{1}$ for $i = 1, 2$
By Universal Property of Quotients for $q_{2}$ with $S = N_{1}$ then
Exists homomorphism $\overline{q}_{2} : M / N_{1} \to M / N_{2}$ induces map
$q_{2} : M \to M / N_{2}$
with kernel $ker (q_{2}) / N_{1} = N_{2} / N_{1}$ and $\overline{q}_{2} \circ q_{1} = q_{2}$
Hence $\overline{q}_{2}$ is surjective and result follows from first isomorphism theorem

Link to original

Oxford Maths Notes

Explorer

08. Quotient Modules and the Isomorphism Theorems

07 - Universal Property of Quotients

Graph View