BranchesHistoryHelpLogin
Welcome guest
You're not logged in.
319 users online, thereof 0 logged in

Example: Examples of Ring Homomorphisms

Example 1

We look for all endomorphisms (i.e. self-homomorphisms) of the ring $(\mathbb Z,+,\cdot)$ of all integers with addition and multiplication. Let $f:\mathbb Z\to\mathbb Z$ be a ring homomorphism. This means that it has to fulfill the properties $$\begin{array}{rcl}
f(x + y)&=&f(x) + f(y),\\
f(x \cdot y)&=&f(x)\cdot f(y)\\
f(1)&=&1.
\end{array}$$
Then for the special case $n\ge 1$ we get $$f(n)=f(\underbrace{1+\ldots+1}_{n\text{ times}})=\underbrace{f(1)+\ldots+f(1)}_{n\text{ times}}=\underbrace{1+\ldots+1}_{n\text{ times}}=n.$$ This implies $$f(0)=f(0\cdot 1)=f(0)\cdot f(1)=0\cdot 1=0.$$ Thus $f(n)+f(-n)=0$, or $$f(-n)=-f(n)=-n.$$
It follows that $f:\mathbb Z\to\mathbb Z$ be a ring homomorphism if and only if $f$ is the identity function $f(n)=n$ for all $n\in\mathbb Z.$

Example 2

The complex conjugation $f:\mathbb C\to\mathbb C$ is a ring homomorphism, since $$\begin{array}{rcl}
(z_1+z_2)^*&=&z_1^*+z_2^*\\
(z_1\cdot z_2)^*&=&z_1^*\cdot z_2^*\\
(1)^*&=&1^*
\end{array}$$

Example 3

For any integer $m\ge 2,$ the function $f:\mathbb Z\to \mathbb Z_m$ sith $f(a)=a(m)$ of the commutative ring of integers $(\mathbb Z,+,\cdot)$ to the commutative ring of congruences $\mathbb Z_m$ is a ring homomorphism which is partially shown here. It remains to note that $$1\equiv 1(m)$$ for all $m\ge 2.$

| | | | created: 2020-06-28 11:11:31 | modified: 2020-06-28 15:45:35 | by: bookofproofs | references: [8311]

Edit or AddNotationAxiomatic Method

This work was contributed under CC BY-SA 4.0 by:

This work is a derivative of:

Bibliography (further reading)

[8311] Modler, F.; Kreh, M.: “Tutorium Analysis 1 und Lineare Algebra 1”, Springer Spektrum, 2018, 4. Auflage