Third Isomorphism Theorem/Rings

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $R$ be a ring.

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.


Proof

By Ideal Contained in Larger Ideal is Ideal of Larger Ideal, $J$ is indeed an ideal of $K$.

Hence $K / J$ is adequately defined.


In Ring Homomorphism whose Kernel contains Ideal‎, 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:

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


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 Ring Isomorphism Theorem applies.

$\blacksquare$


Also known as

The Third Ring Isomorphism Theorem is also referred to by some sources as the First Ring Isomorphism Theorem, and by others as the Second Ring Isomorphism Theorem.

There is little consistency in the literature.

It can also be called the Third Isomorphism Theorem for Rings.


Also see


Sources