# Third Isomorphism Theorem/Rings

< Third Isomorphism Theorem(Redirected from Third Isomorphism Theorem for Rings)

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## Theorem

Let $R$ be a ring, and let:

Then:

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

The validity of the material on this page is questionable.In particular: $K / J$ is not well-defined, since $J$ is not assumed to be an ideal of $K$You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Questionable}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

- $(2): \quad \dfrac {R / J} {K / J} \cong \dfrac R K$
- 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.

$\blacksquare$

## Also see

## Sources

- 1970: B. Hartley and T.O. Hawkes:
*Rings, Modules and Linear Algebra*... (previous) ... (next): $\S 2.2$: Homomorphisms: Theorem $2.11$