Third Isomorphism Theorem/Rings

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Let $R$ be a ring, and let:

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


$(1): \quad K / J$ 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.


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:

$\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.


Also see