# 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:

$(1): \quad K / J$ is an ideal of $R / J$
where $K / J$ denotes the quotient ring of $K$ by $J$

$(2): \quad \dfrac {R / J} {K / J} \cong \dfrac R K$
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 Idealâ€Ž, take $\phi: R \to R / K$ to be the quotient 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.

This can be illustrated by means of the following commutative diagram:

$\begin{xy}\[email protected]+2mu@+1em{ R \ar[dr]_*{\phi} \ar[r]^*{\nu} & R / J \ar@{-->}[d]^*{\psi} \\ & R / K }\end{xy}$

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

So $\Img \psi = \Img \phi = R / K$ and the First Isomorphism Theorem applies.

$\blacksquare$