First Isomorphism Theorem/Rings

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

Let $\ker \left({\phi}\right)$ be the kernel of $\phi$.

Then:
 * $\operatorname {Im} \left({\phi}\right) \cong R / \ker \left({\phi}\right)$

where $\cong$ denotes ring isomorphism.

Proof
From Ring Homomorphism whose Kernel contains Ideal, let $J = \ker \left({\phi}\right)$.

This gives the ring homomorphism $\mu: R / \ker \left({\phi}\right) \to S$ as follows:


 * CommDiagFirstIsomTheorem.png

That is:
 * $\phi = \mu \circ \nu$

Then we have:
 * $\ker \left({\mu}\right) = \ker \left({\phi}\right) / \ker \left({\phi}\right)$

This is the null subring of $R / \ker \left({\phi}\right)$ by Quotient Ring Defined by Ring Itself is Null Ideal.

Then from Kernel is Trivial iff Monomorphism it follows that $\mu$ is a monomorphism.

From $\phi = \mu \circ \nu$, we have:
 * $\operatorname {Im} \left({\mu}\right) = \operatorname {Im} \left({\phi}\right)$

It follows that $\mu$ is an isomorphism.

Also known as
This result is also referred to as the First Fundamental Theorem on Ring Homomorphisms.

Also see

 * Isomorphism Theorems
 * Universal Property of Quotient Ring