Third Isomorphism Theorem/Rings

Theorem
Let $R$ be a ring, and let:


 * $J, K$ be ideals of $R$
 * $J$ be a subset of $K$.

Then:
 * $K / J$ is an ideal of $R / J$
 * where $K / J$ denotes the quotient ring of $K$ by $J$


 * $\dfrac {R / J} {K / J} \cong \dfrac R K$
 * where $\cong$ denotes ring isomorphism.

This result is also referred to by some sources as the first isomorphism theorem, and by others as the second isomorphism theorem.

Proof
In Ring Homomorphism whose Kernel contains Ideal‎, take $\phi: R \to R / K$ to be the natural epimorphism.

Then (from the same source) its kernel is $K$.

Thus we have that:
 * $\phi = \psi \circ \nu$

where $\psi : R / J \to R / K$ is a homomorphism.


 * CommDiagThirdIsomTheorem.png

As $\phi$ is an epimorphism then from Surjection if Composite is a Surjection we have that $\psi$ is a surjection.

So $\operatorname{Im} \left({\psi}\right) = \operatorname{Im} \left({\phi}\right) = R / K$ and the First Isomorphism Theorem applies.