The Fields

[trfrm]

Salam sejahtera.

Excuse me ... .


A field $F$ is structure of algebra $(F,+,\cdot)$ which $(F,+)$ and $(F-\{0\},\cdot)$ is abelian group ... .

Until now, in abstract algebra, we have known three field : $\{0\}$, $\mathbb{R}$, and $\mathbb{C}$ (read respectively: point, real line, and complex plane) under usual additive opetration and usual multiplicative operation ... .

Topologically,

the point $\{0\}$ is seem like the $\mathbb{R}^0$ ... ,

the real line $\mathbb{R}$ is seem like the $\mathbb{R}^1$ ... , and

the complex plane $\mathbb{C}$ is seem like the $\mathbb{R}^2$ ... .



Are there the fields which like topologically $\mathbb{R}^n$ for $n>2$ ... ?


Thank you all ... .

Syukur kepada Allah.

[trfrm]

Horas.

@ TheObserver

A set of quarternion $(Q,+,\cdot)$ can form a ring ... , but not a field ... because the semigroup $(Q,\cdot)$ is not commutative ... . In simple example, $ij=-ji=k$, $jk=-kj=i$, $ki=-ik=j$ ... . To form a field, we have to add the requirement of commutation ... .

I'm sorry if I do mistakes ... . Thank you ... .

Allahu Akbar.

[trfrm]

Om Swastyastu.

You forgot one more condition: multiplication must be distributive over addition. The two operations in a field are not independent of each other: they are related by the distributive law.

I’m sorry ... . I had forgotten to add this requirement for building a field ... .
For $a,b,c\in{F}$, where $F$ is a field, then $a(b+c)=ab+ac$ ... .

When defining a field, mathematicians usually insist that $0\ne1$, that is, the additive identity and the multiplicative identity must be distinct. Hence $\{0\}$ is not normally considered a field since a field must contain at least two elements.

Why do the additive identity and the multiplicative identity must be distinct ... ? Is the distinction one of all conditions of a field ... ?

Isn’t $0+0=0$, $0\cdot0=0$, and $0\cdot(0+0)=0\cdot0+0\cdot0$ ... ?

http://www.thescienceforum.com/mathematics/32114-fields.html#post399075

Berkah Dalem Gusti.