06 - Isomorphism Theorems

Isomorphism Theorem corollary

First Isomorphism Theorem

Let $\varphi :R\to S$ is a homomorphism then

$\varphi$ induces isomorphism

$$\overline{\varphi }:R/\ker (\varphi )\to \mathrm{im}(\varphi )$$

Second Isomorphism Theorem

Let $R$ be a ring
Let $A$ be a subring of $R$
Let $I$ be an ideal of $R$ then

$$(A+I)/I\cong A/(A\cap I)$$

Third Isomorphism Theorem

Let $I_1\subseteq I_2$ be ideals in ring $R$ then

$$(R/I_1)/(I_2/I_1)\cong R/I_2$$

Proof

First Isomorphism Theorem
Apply Universal Property of Quotients to $I=\ker (\varphi )$ then

$$\ker (\overline{\varphi })=\ker (\varphi )/\ker (\varphi )=0$$

Hence $\overline{\varphi }$ is injective and hence induces isomorphism onto image from

$$\overline{\varphi }\circ q=\varphi =\mathrm{im}(\varphi )$$

Second Isomorphism Theorem
As $A$ is a subring and $I$ is an ideal then $A+I$ is a subring of $R$ containing $I$
Let $q:R\to R/I$ be quotient map

Then $q$ restricts to homomorphism $p$ from $A$ to $R/I$ with image $(A+I)/I$
By First Isomorphism Theorem then kernel of $p$ is $A\cap I$
(As if $a\in A$ with $p(a)=0$ then $a+I=0+I$ so $a\in I$ so $a\in A\cap I$)

Third Isomorphism Theorem
Let $q_i:R\to R/I_j$ for $j=1,2$
By Universal Property of Quotients for $q_2$ then exists homomorphism

$$\overline{q_2}:R/I_1\to R/I_2$$

induced by map $q_2:R\to R/I_2$ with kernel

$$\ker (q_2)/I_1=I_2/I_2$$

and $\overline{q_2}\circ q_1=q_2$

Hence $\overline{q_2}$ is surjective as $q_2$ is hence result follows by First Isomorphism Theorem

Isomorphism Theorem for Modules corollary

First Isomorphism Theorem

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

$\varphi$ induces an isomorphism

$$\overline{\varphi }:M/\ker (\varphi )\to \mathrm{im}(\varphi )$$

Second Isomorphism Theorem

Let $M$ be an $R$-module
Let $N_1,N_2$ are submodules of $M$ then

$$(N_1+N_2)/N_2\cong N_1/(N_1\cap N_2)$$

Third Isomorphism Theorem

Let $N_1\subseteq N_2$ be submodules of $M$ then

$$(M/N_1)/(N_2/N_1)\cong M/N_2$$

Proof

For First Isomorphism Theorem then
Apply Universal Property of Quotients to $S=\ker (\varphi )$
As

$$\ker (\overline{\varphi })=\ker (\varphi )/\ker (\varphi )=0$$

Hence $\overline{\varphi }$ is injective so
Induces isomorphism onto image from $\overline{\varphi }\circ q=\varphi$ which is $\mathrm{im}(\varphi )$

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 $N_1\cap N2$
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