Fourth Isomorphism Theorem

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

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

Also:
 * $$(1)$$: $$f_\phi$$ and its inverse both preserve subsets.


 * $$(2)$$: $$f_\phi$$ and its inverse both preserve ideals:
 * If $$J$$ is an ideal of $$R$$, then $$f_\phi \left({J}\right)$$ is an ideal of $$S$$
 * If $$J'$$ is an ideal of $$S$$, then $$f_\phi^{-1} \left({J'}\right)$$ is an ideal of $$R$$

Proof of Preservation of Subsets
From Subset Maps to Subset, we have:


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

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

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 Ring Epimorphism Inverse of Subring, we have that $$f_\phi^{-1} \left({U}\right)$$ is a subring of $$R$$ such that $$K \subseteq R$$ and so:
 * $$f_\phi^{-1} \left({U}\right) \in \mathbb K$$

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 $$f_\phi \left({V}\right)$$ is a subring of $$S$$ and so:
 * $$f_\phi \left({V}\right) \in \mathbb S$$

Proof that f is a Bijection
By Image of Preimage of Mapping, we have that:
 * $$U = f_\phi \left({f_\phi^{-1} \left({U}\right)}\right)$$

We also have from Preimage of Image of Mapping that:
 * $$V \subseteq f_\phi^{-1} \left({f_\phi \left({V}\right)}\right)$$

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

Thus $$\phi \left({r}\right) \in f_\phi \left({V}\right)$$ 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:
 * $$f_\phi^{-1} \left({f_\phi \left({V}\right)}\right) \subseteq V$$

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

So we have:
 * $$U = f_\phi \left({f_\phi^{-1} \left({U}\right)}\right)$$
 * $$V = f_\phi^{-1} \left({f_\phi \left({V}\right)}\right)$$

That is:
 * $$f_\phi \circ f_\phi^{-1} = I_\mathbb S$$
 * $$f_\phi^{-1} \circ f_\phi = I_\mathbb K$$

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

Hence by the definition of inverse of bijection, it follows that $$f_\phi: \mathbb K \to \mathbb S$$ is a bijection.

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

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

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

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

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

Hence ideals are preserved in both directions.