First Isomorphism Theorem/Rings

From ProofWiki
Jump to navigation Jump to search

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:


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


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

The First Ring Isomorphism Theorem is also referred to as the First Fundamental Theorem on Ring Homomorphisms.

It can also be called the First Isomorphism Theorem for Rings.


Also see


Sources