Second Isomorphism Theorem/Rings

Theorem
Let $R$ be a ring, and let:


 * $S$ be a subring of $R$.
 * $J$ be an ideal of $R$.

Then:


 * $(1): \quad S + J$ is a subring of $R$
 * $(2): \quad J$ is an ideal of $S + J$
 * $(3): \quad S \cap J$ is an ideal of $S$
 * $(4): \quad \displaystyle \frac S {S \cap J} \cong \frac {S + J} J$

where $\cong$ denotes group isomorphism.

This result is also referred to by some sources as the first isomorphism theorem.

Proof
The relations being defined can be illustrated by this commutative diagram:


 * CommDiagSecondIsomTheorem.png

$(1): \quad S + J$ is a subring of $R$

From Sum of All Ring Products is Additive Subgroup, $S + J$ is an additive subgroup of $R$.

Suppose $s, s' \in S, j, j' \in J$.

Then:
 * $\left({s + j}\right) \left({s' + j'}\right)$

so by the Subring Test $S + J$ is a subring of $R$.

$(2): \quad J$ is an ideal of $S + J$

Let $s + j \in S + J$ and let $j \in J$.

Then:

So $J$ is an ideal of $S + J$.

$(3): \quad S \cap J$ is an ideal of $S$

Let $\nu: R \to R / J$ be the natural epimorphism.

Let $\nu'$ be the restriction of $\nu$ to $S$.

Then $\nu': S \to R / J$ is a homomorphism.

The image of $\nu'$ is the set of all cosets $s + J$ for $s \in S$:
 * $\displaystyle \operatorname{Im} \left({\nu'}\right) = \frac {S + J} J$

Now, the kernel of $\nu'$ is the set of all elements of $S$ which are sent to $0_{S/J}$ by $\nu$.

That is, all the elements of $S$ which are also in $J$ itself, which is how the quotient ring behaves.

That is:
 * $\ker \left({\nu'}\right) = S \cap J$

and so from Kernel of Ring Homomorphism is Ideal, $S \cap J$ is an ideal of $S$.

$(4): \quad \displaystyle \frac S {S \cap J} \cong \frac {S + J} J$

This follows directly from the First Isomorphism Theorem.

Also known as
Because of the shape of its commutative diagram, this theorem is sometimes known as the parallelogram law.

Also see

 * Isomorphism Theorems