The Isomorphism Theorems for Rings

Write $K = \ker\varphi$.

(1) $K$ is an ideal. Since $\varphi(0_R) = \varphi(0_R + 0_R) = \varphi(0_R)+\varphi(0_R)$, subtracting $\varphi(0_R)$ from both sides in the abelian group $(S,+)$ gives $\varphi(0_R) = 0_S$, so $0_R \in K$. If $a,b \in K$ then $\varphi(a-b) = \varphi(a) - \varphi(b) = 0_S - 0_S = 0_S$ (using that $\varphi$ is additive, hence a group homomorphism on $(R,+)$, so $\varphi(-b) = -\varphi(b)$); thus $a - b \in K$, so $K$ is an additive subgroup. For $r \in R$ and $a \in K$, $\varphi(ra) = \varphi(r)\varphi(a) = \varphi(r)\cdot 0_S = 0_S$, so $ra \in K$; similarly $\varphi(ar) = \varphi(a)\varphi(r) = 0_S$, so $ar \in K$. Hence $K$ absorbs multiplication by $R$ on both sides, so $K$ is a two-sided ideal.

(2) $\operatorname{im}\varphi$ is a subring. It is nonempty ($0_S \in \operatorname{im}\varphi$). If $\varphi(a), \varphi(b) \in \operatorname{im}\varphi$ then $\varphi(a) - \varphi(b) = \varphi(a-b) \in \operatorname{im}\varphi$ and $\varphi(a)\varphi(b) = \varphi(ab) \in \operatorname{im}\varphi$. So $\operatorname{im}\varphi$ is closed under subtraction and multiplication, hence a subring of $S$.

(3) Well-definedness of $\bar\varphi$. Suppose $a + K = a' + K$ in $R/K$, i.e. $a - a' \in K$. Then $\varphi(a) - \varphi(a') = \varphi(a-a') = 0_S$, so $\varphi(a) = \varphi(a')$. Hence $\bar\varphi(a+K) := \varphi(a)$ does not depend on the choice of coset representative, and $\bar\varphi$ is a well-defined function $R/K \to \operatorname{im}\varphi$.

$\bar\varphi$ is a ring homomorphism. For $a,b \in R$:

These use exactly the definitions of addition and multiplication in $R/K$ from Quotient Rings, which are well-defined because $K$ is an ideal by part (1).

$\bar\varphi$ is surjective. Any element of $\operatorname{im}\varphi$ has the form $\varphi(a)$ for some $a \in R$, and $\bar\varphi(a+K) = \varphi(a)$, so $\bar\varphi$ hits every element of $\operatorname{im}\varphi$.

$\bar\varphi$ is injective. Suppose $\bar\varphi(a+K) = \bar\varphi(b+K)$, i.e. $\varphi(a) = \varphi(b)$. Then $\varphi(a-b) = \varphi(a)-\varphi(b) = 0_S$, so $a - b \in K$, which means $a+K = b+K$. So $\bar\varphi$ is injective. Since $\bar\varphi$ is a bijective ring homomorphism, it is a ring isomorphism, and $R/K \cong \operatorname{im}\varphi$. $\blacksquare$

(1) $A+B$ contains $0 = 0+0$. If $a_1+b_1, a_2+b_2 \in A+B$ (with $a_i \in A, b_i \in B$), then

since $A$ is closed under subtraction and $B$ is closed under subtraction. For products,

Here $a_1a_2 \in A$ since $A$ is a subring; and $a_1 b_2, b_1 a_2, b_1 b_2 \in B$ since $B$ is an ideal (absorbing multiplication by any element of $R$, in particular by elements of $A \subseteq R$), and $B$ is closed under addition. So the whole sum lies in $A + B$ (write it as the sum $a_1a_2 \in A$ plus the ideal element $a_1b_2+b_1a_2+b_1b_2 \in B$). Hence $A+B$ is closed under subtraction and multiplication, so it is a subring of $R$.

(2) $B \subseteq A+B$ and $B$ is an additive subgroup, hence an additive subgroup of $A+B$. Since $B \trianglelefteq R$ and $A + B \subseteq R$, for any $x \in A+B \subseteq R$ we have $xB \subseteq B$ and $Bx \subseteq B$. So $B$ is an ideal of $A+B$.

(3) $A \cap B$ is an intersection of two additive subgroups of $R$, hence an additive subgroup of $A$ (it is a subset of $A$ closed under the group operations). For $a \in A$ and $x \in A \cap B$: $ax \in A$ (since $A$ is a subring and $a, x \in A$) and $ax \in B$ (since $x \in B$ and $B \trianglelefteq R$ absorbs multiplication by $a \in R$); so $ax \in A \cap B$. Similarly $xa \in A \cap B$. Hence $A \cap B$ is an ideal of $A$. $\blacksquare$

The subring/ideal claims are exactly the preceding Sum and intersection of a subring and an ideal lemma, so it remains to construct the isomorphism. Define $\varphi: A \to (A+B)/B$ by

This makes sense because $a \in A \subseteq A+B$, so $a+B$ is a genuine coset in $(A+B)/B$.

$\varphi$ is a ring homomorphism. For $a_1, a_2 \in A$,

using the definitions of addition and multiplication on $(A+B)/B$ (valid since $B \trianglelefteq A+B$ by the lemma).

$\varphi$ is surjective. An arbitrary element of $(A+B)/B$ has the form $(a+b)+B$ for some $a \in A, b \in B$. Since $b \in B$, $(a+b) - a = b \in B$, so $(a+b)+B = a+B = \varphi(a)$. Hence every coset in $(A+B)/B$ is hit.

Computing the kernel. We claim $\ker\varphi = A \cap B$. If $a \in A$ and $\varphi(a) = 0 + B$ (the zero coset), then $a \in B$ by definition of the coset equivalence ($a - 0 = a \in B$); combined with $a \in A$ this gives $a \in A\cap B$. Conversely if $a \in A \cap B$ then $a \in B$, so $a + B = 0 + B$, i.e. $\varphi(a) = 0$ in $(A+B)/B$. So $\ker\varphi = A \cap B$.

Conclusion. By the First Isomorphism Theorem applied to $\varphi: A \to (A+B)/B$, which is surjective with kernel $A \cap B$,

This is the desired isomorphism (stated in the other order by convention). $\blacksquare$

Step 1: $J/I$ makes sense and is an ideal of $R/I$. Since $I \subseteq J$, every coset $j + I$ with $j \in J$ is a well-defined element of $R/I$; set $J/I := \{j+I : j \in J\} \subseteq R/I$. This is an additive subgroup of $R/I$ (image of the additive subgroup $J$ under the additive group homomorphism $\pi: R \to R/I$, $\pi(r)=r+I$, restricted to $J$). For $r+I \in R/I$ and $j+I \in J/I$: $(r+I)(j+I) = rj + I$, and $rj \in J$ because $J \trianglelefteq R$; so $rj + I \in J/I$. Similarly $jr+I \in J/I$. Hence $J/I \trianglelefteq R/I$, and the quotient ring $(R/I)/(J/I)$ is defined.

Step 2: Construct a surjective homomorphism $R \to (R/I)/(J/I)$ with kernel $J$. Let $\pi_1: R \to R/I$ be the canonical projection $\pi_1(r) = r+I$, and let $\pi_2: R/I \to (R/I)/(J/I)$ be the canonical projection $\pi_2(x) = x + (J/I)$. Both are surjective ring homomorphisms (canonical projections onto quotient rings are always surjective ring homomorphisms, as recalled in the definition block). Let

As a composite of ring homomorphisms, $\psi$ is a ring homomorphism. As a composite of surjections, $\psi$ is surjective.

Step 3: Compute $\ker\psi = J$. We have $\psi(r) = 0$ (the zero coset of $(R/I)/(J/I)$, i.e. the coset $J/I$ itself) if and only if $\pi_1(r) = r+I \in J/I$, i.e. if and only if $r + I = j + I$ for some $j \in J$, i.e. if and only if $r - j \in I \subseteq J$ for some $j \in J$. Since $j \in J$ and $r - j \in J$ (as $I \subseteq J$), adding gives $r = (r-j) + j \in J$. Conversely if $r \in J$ then $r + I \in J/I$ trivially (take $j = r$), so $\psi(r) = 0$. Hence

Step 4: Apply the First Isomorphism Theorem. Since $\psi: R \to (R/I)/(J/I)$ is a surjective ring homomorphism with $\ker\psi = J$, the First Isomorphism Theorem gives

$\blacksquare$

Let $\mathcal{J}_I$ denote the set of ideals of $R$ containing $I$, and let $\mathcal{Q}$ denote the set of ideals of $R/I$. Define

$\Phi$ is well-defined, i.e. $\pi(J)$ is an ideal of $R/I$ for $J \in \mathcal{J}_I$. Since $I \subseteq J$, this was verified in Step 1 of the Third Isomorphism Theorem proof.

$\Psi$ is well-defined, i.e. $\pi^{-1}(\bar J)$ is an ideal of $R$ containing $I$, for any ideal $\bar J \trianglelefteq R/I$. Preimages of ideals under ring homomorphisms are always ideals: if $x, y \in \pi^{-1}(\bar J)$ then $\pi(x-y) = \pi(x)-\pi(y) \in \bar J$ (as $\bar J$ is an additive subgroup), so $x - y \in \pi^{-1}(\bar J)$; and for $r \in R$, $\pi(rx) = \pi(r)\pi(x) \in \bar J$ (as $\bar J$ absorbs multiplication by the arbitrary element $\pi(r) \in R/I$), so $rx \in \pi^{-1}(\bar J)$, and similarly $xr \in \pi^{-1}(\bar J)$. Also $\pi^{-1}(\bar J) \supseteq \pi^{-1}(\{0\}) = \ker\pi = I$ because $0 \in \bar J$. So $\Psi(\bar J) \in \mathcal{J}_I$.

$\Psi \circ \Phi = \mathrm{id}$. Let $J \in \mathcal{J}_I$. We must show $\pi^{-1}(\pi(J)) = J$. The inclusion $J \subseteq \pi^{-1}(\pi(J))$ is immediate (if $x \in J$ then $\pi(x) \in \pi(J)$, so $x \in \pi^{-1}(\pi(J))$). For the reverse inclusion, suppose $x \in \pi^{-1}(\pi(J))$, i.e. $\pi(x) = \pi(j)$ for some $j \in J$. Then $x - j \in \ker\pi = I \subseteq J$ (using the hypothesis $I \subseteq J$). Since $j \in J$ and $x - j \in J$, we get $x = (x-j)+j \in J$. Hence $\pi^{-1}(\pi(J)) \subseteq J$, giving equality. This step crucially uses $I \subseteq J$; without it, the argument fails (see the counterexample below).

$\Phi \circ \Psi = \mathrm{id}$. Let $\bar J \in \mathcal{Q}$. We must show $\pi(\pi^{-1}(\bar J)) = \bar J$. The inclusion $\pi(\pi^{-1}(\bar J)) \subseteq \bar J$ holds for any function's preimage. For the reverse inclusion, since $\pi$ is surjective, every element of $\bar J \subseteq R/I$ has the form $\pi(x)$ for some $x \in R$; then $x \in \pi^{-1}(\bar J)$ by definition, so $\pi(x) \in \pi(\pi^{-1}(\bar J))$. This uses surjectivity of $\pi$, which holds since $\pi$ is a canonical projection. Thus $\Phi$ and $\Psi$ are mutually inverse bijections between $\mathcal{J}_I$ and $\mathcal{Q}$.

Order preservation. If $J_1 \subseteq J_2$ in $\mathcal{J}_I$, then clearly $\pi(J_1) \subseteq \pi(J_2)$, i.e. $\Phi(J_1) \subseteq \Phi(J_2)$, since $\pi$ is a function. Conversely, if $\Phi(J_1) \subseteq \Phi(J_2)$, apply $\Psi$ (order-preserving by the same reasoning, since preimages of nested sets are nested) to get $J_1 = \Psi\Phi(J_1) \subseteq \Psi\Phi(J_2) = J_2$, using the bijectivity just proved. So $J_1 \subseteq J_2 \iff \Phi(J_1) \subseteq \Phi(J_2)$.

The quotient formula. For $J \in \mathcal{J}_I$, i.e. $I \subseteq J$, the isomorphism $(R/I)/(J/I) \cong R/J$ is exactly the Third Isomorphism Theorem, already proved above, and $J/I = \Phi(J)$ by definition.

The subring version. The identical argument works verbatim replacing "ideal" by "subring" throughout: preimages and images of subrings under a ring homomorphism are subrings (the same closure-under-subtraction-and-multiplication computations apply, using only that $\pi(x-y)=\pi(x)-\pi(y)$ and $\pi(xy)=\pi(x)\pi(y)$, without ever using the ideal-absorption property), and the same two inclusion arguments ($J \subseteq \pi^{-1}\pi(J)$ using $I \subseteq J$, and $\pi\pi^{-1}(\bar J) = \bar J$ using surjectivity) go through unchanged. $\blacksquare$