# First Isomorphism Theorem/Rings

## Theorem

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

Let $\map \ker \phi$ be the kernel of $\phi$.

Then:

$\Img \phi \cong R / \map \ker \phi$

where $\cong$ denotes ring isomorphism.

## Proof

From Ring Homomorphism whose Kernel contains Ideal, let $J = \map \ker \phi$.

This gives the ring homomorphism $\mu: R / \map \ker \phi \to S$ as follows:

$\begin {xy} \[email protected] + 2mu@ + 1em { R \ar[rr]^*{\nu} \ar[rdrd]_*{\phi} & & R / \map \ker \phi \ar[dd]_*{\mu} \\ \\ & & S } \end {xy}$

That is:

$\phi = \mu \circ \nu$

Then we have:

$\map \ker \mu = \map \ker \phi / \map \ker \phi$

This is the null subring of $R / \map \ker \phi$ by Quotient Ring Defined by Ring Itself is Null Ring.

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

From $\phi = \mu \circ \nu$, we have:

$\Img \mu = \Img \phi$

It follows that $\mu$ is an isomorphism.

$\blacksquare$

## Also known as

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