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:

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

$(2): \quad f_\phi$ 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

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.

$\Box$

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

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

$\Box$

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

$f_\phi \left({V}\right) \in \mathbb S$

$\Box$

Proof that $f$ is a Bijection

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

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

We also have from Subset of Domain is Subset of Preimage of Image 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.

$\blacksquare$

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 Preimage of Ideal under Ring Epimorphism is 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.

$\blacksquare$

Sources

• B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra (1970): $\S 2.2$: Theorem $2.12$