Ring Epimorphism from Integers to Integers Modulo m

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {\Z, +, \times}$ be the ring of integers.

Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.

Let $\phi: \struct {\Z, +, \times} \to \struct {\Z_m, +_m, \times_m}$ be the mapping defined as:

$\forall x \in \Z: \map \phi x = \eqclass x m$

where $\eqclass x m$ is the residue class modulo $m$.


Then $\phi$ is a ring epimorphism, but specifically not a ring monomorphism.


The image of $\phi$ is $\struct {\Z_m, +_m, \times_m}$.

The kernel of $\phi$ is $m \Z$, the set of integer multiples of $m$.


Proof

Let $a, b \in \Z$.

Then:

\(\ds \map \phi {a + b}\) \(=\) \(\ds \eqclass {a + b} m\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \eqclass a m +_m \eqclass b m\) Definition of Modulo Addition
\(\ds \) \(=\) \(\ds \map \phi a +_m \map \phi b\) Definition of $\phi$


\(\ds \map \phi {a \times b}\) \(=\) \(\ds \eqclass {a \times b} m\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \eqclass a m \times_m \eqclass b m\) Definition of Modulo Multiplication
\(\ds \) \(=\) \(\ds \map \phi a \times_m \map \phi b\) Definition of $\phi$

Hence $\phi$ is a ring homomorphism.


Now let $\eqclass a m \in \Z_m$.

By definition of residue class modulo $m$:

$\eqclass a m = \set {x \in \Z: \exists k \in \Z: z = a + k m}$

Setting $k = 0$:

$\map \phi a = \eqclass a m$

and so:

$\map {\phi^{-1} } {\eqclass a m} \ne \O$

Thus $\phi$ is a surjection.

Now setting $k = 1$, for example, we have that:

$\map \phi {a + m} = \eqclass a m$

and so:

$\map \phi a = \map \phi {a + m}$.

So $\phi$ is specifically not an injection.


It follows by definition that $\phi$ is a ring epimorphism, but specifically not a ring monomorphism.


Next we note that:

$\forall x \in \Z: \map \phi x \in \Z_m$

and so:

$\Img \phi = \Z_m$


Finally, we have that the kernel of $\phi$ is:

$\map \ker \phi = \set {x \in \Z: \map \phi x = \eqclass 0 m}$

Let $\map \phi x = \eqclass 0 m$

Then $x = 0 + k m$ for some $k \in \Z$.

That is, $x \in m \Z$ and so:

$\map \ker \phi \subseteq m \Z$

Now let $x \in m \Z$.

Then:

$\exists k \in \Z: x = 0 + k m$

and so by definition:

$\map \phi x = \eqclass 0 m$

So:

$m \Z \subseteq \map \ker \phi$

Hence:

$\map \ker \phi = m \Z$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Ring_Epimorphism_from_Integers_to_Integers_Modulo_m&oldid=544067"