# Fourth Isomorphism Theorem

## Contents

## Theorem

Let $\phi: R \to S$ be a ring homomorphism.

Let $K = \ker \left({\phi}\right)$ be the kernel of $\phi$.

Let $\mathbb K$ be the set of all subrings of $R$ which contain $K$ as a subset.

Let $\mathbb S$ be the set of all subrings of $\operatorname{Im} \left({\phi}\right)$.

Let $\phi^\to: \mathcal P \left({R}\right) \to \mathcal P \left({S}\right)$ be the mapping induced on $\mathcal P \left({R}\right)$ by $\phi$.

Then its restriction $\phi^\to: \mathbb K \to \mathbb S$ is a bijection.

Also:

## Proof

### Proof of Preservation of Subsets

From Subset Maps to Subset, we have:

- $(a) \quad \forall X, X' \in \mathbb K: X \subseteq X' \implies \phi^\to \left({X}\right) \subseteq \phi^\to\left({X'}\right)$
- $(b) \quad \forall Y, Y' \in \mathbb S: Y \subseteq Y' \implies \left({\phi^\to}\right)^{-1} \left({Y}\right) \subseteq \left({\phi^\to}\right)^{-1} \left({Y'}\right)$

So $\phi^\to$ and its inverse both preserve subsets, and $(1)$ has been demonstrated to hold.

$\Box$

### Proof that Inverse Image is a Subring

Let $U \in \mathbb S$, that is, let $U$ be a subring of $\operatorname{Im} \left({\phi}\right)$.

From Preimage of Subring under Ring Homomorphism is Subring, we have that $\left({\phi^\to}\right)^{-1} \left({U}\right)$ is a subring of $R$ such that $K \subseteq R$ and so:

- $\left({\phi^\to}\right)^{-1} \left({U}\right) \in \mathbb K$

$\Box$

### Proof that Image is a Subring

Let $V \in \mathbb K$, that is, a subring of $R$ containing $K$.

From Ring Homomorphism Preserves Subrings, we have that $\phi^\to \left({V}\right)$ is a subring of $S$ and so:

- $\phi^\to \left({V}\right) \in \mathbb S$

$\Box$

### Proof that $f$ is a Bijection

By Subset of Codomain is Superset of Image of Preimage, we have that:

- $U = \phi^\to \left({\left({\phi^\to}\right)^{-1} \left({U}\right)}\right)$

We also have from Subset of Domain is Subset of Preimage of Image that:

- $V \subseteq \left({\phi^\to}\right)^{-1} \left({\phi^\to \left({V}\right)}\right)$

Now let $r \in \left({\phi^\to}\right)^{-1} \left({\phi^\to \left({V}\right)}\right)$.

Thus $\phi \left({r}\right) \in \phi^\to \left({V}\right)$ and so:

- $\exists v \in V: \phi \left({r}\right) = \phi \left({v}\right)$

So $ \phi \left({r + \left({-v}\right)}\right) = 0_S$ and so $r - v \in K$ by definition of kernel.

So:

- $\exists k \in K: r + k = v$

But by assumption, $K \subseteq V$ as $V \in \mathbb K$.

So $k \in V$ and so it follows that $r \in V$ as well, by the fact that $V$ is subring and so closed for $+$.

So:

- $\left({\phi^\to}\right)^{-1} \left({\phi^\to \left({V}\right)}\right) \subseteq V$

Putting that together with $V \subseteq \left({\phi^\to}\right)^{-1} \left({\phi^\to \left({V}\right)}\right)$ and we see:

- $V = \left({\phi^\to}\right)^{-1} \left({\phi^\to \left({V}\right)}\right)$

So we have:

- $U = \phi^\to \left({\left({\phi^\to}\right)^{-1} \left({U}\right)}\right)$
- $V = \left({\phi^\to}\right)^{-1} \left({\phi^\to \left({V}\right)}\right)$

That is:

- $\phi^\to \circ \left({\phi^\to}\right)^{-1} = I_\mathbb S$
- $\left({\phi^\to}\right)^{-1} \circ \phi^\to = I_\mathbb K$

where $I_\mathbb S$ and $I_\mathbb K$ are the identity mappings of $\mathbb S$ and $\mathbb K$ respectively.

By Composite of Bijection with Inverse is Identity Mapping, it follows that $\phi^\to: \mathbb K \to \mathbb S$ is a bijection.

$\blacksquare$

### Proof of Preservation of Ideals

Let $V \in \mathbb K$ be an ideal of $R$.

Let $U = \phi^\to \left({V}\right)$.

Then from Ring Epimorphism Preserves Ideals, $U$ is an ideal of $S$.

Similarly, let $U = \phi^\to \left({V}\right)$ be an ideal of $S$.

Then by Preimage of Ideal under Ring Homomorphism is Ideal, $V = \left({\phi^\to}\right)^{-1} \left({U}\right)$ is an ideal of $R$ such that $K \subseteq V$.

Hence ideals are preserved in both directions.

$\blacksquare$

## Also see

## Sources

- 1970: B. Hartley and T.O. Hawkes:
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