# Fourth Isomorphism Theorem

## Contents

## Theorem

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

Let $K = \map \ker \phi$ 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 $\Img \phi$.

Let $\phi^\to: \powerset R \to \powerset S$ be the direct image mapping of $\phi$.

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:

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

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 $\Img \phi$.

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

- $\map {\paren {\phi^\to}^{-1} } U \in \mathbb K$

$\Box$

### 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 $\map {\phi^\to} V$ is a subring of $S$ and so:

- $\map {\phi^\to} V \in \mathbb S$

$\Box$

### Proof that $f$ is a Bijection

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

- $U = \map {\phi^\to} {\map {\paren {\phi^\to}^{-1} } U}$

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

- $V \subseteq \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$

Now let $r \in \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$.

Thus $\map \phi r \in \map {\phi^\to} V$ and so:

- $\exists v \in V: \map \phi r = \map \phi v$

So $\map \phi {r + \paren {-v} } = 0_S$ 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:

- $\map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V} \subseteq V$

Putting that together with $V \subseteq \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$ and we see:

- $V = \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$

So we have:

- $U = \map {\phi^\to} {\map {\paren {\phi^\to}^{-1} } U}$
- $V = \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$

That is:

- $\phi^\to \circ \paren {\phi^\to}^{-1} = I_\mathbb S$
- $\paren {\phi^\to}^{-1} \circ \phi^\to = I_\mathbb K$

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 $U = \map {\phi^\to} V$.

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

Similarly, let $U = \map {\phi^\to} V$ be an ideal of $S$.

Then by Preimage of Ideal under Ring Homomorphism is Ideal, $V = \map {\paren {\phi^\to}^{-1} } U$ 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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