Inel comutativ.
Moderators: Bogdan Posa, Beniamin Bogosel, Marius Dragoi
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Inel comutativ.
Fie A un inel cu proprietatea ca \( x^{15}=x^4,(\forall)x\in A. \) Atunci A este comutativ.
-
Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
\( x^{15}=x^4 \Rightarrow 1+1=0 \)
Atunci avem: \( (x+1)^{2}=x^2+1 \) \( \Rightarrow (x+1)^4=x^4+1 \) iar \( (x+1)^{16}=x^{16}+1=x^{15}x+1=x^4x+1=x^5+1 \)
In relatia initiala , \( x^{15}=x^4 \) , trecem pe \( x \) in \( x+1 \) si obtinem: \( (x+1)^{15}=(x+1)^4 \Rightarrow (x+1)^{15}=x^4+1 \Rightarrow (x+1)^{16}=(x^4+1)(x+1) \)
\( \Rightarrow x^5+1=x^5+x^4+x+1 \Rightarrow x^4=x \)
Ridicand la puterea a 4-a ultima relatie obtinem: \( x^{16}=x^4 \Rightarrow x^{15}x=x \Rightarrow x^4x=x \Rightarrow x^2=x \Rightarrow \) A este un inel de tip Boole , deci este comutativ.
Atunci avem: \( (x+1)^{2}=x^2+1 \) \( \Rightarrow (x+1)^4=x^4+1 \) iar \( (x+1)^{16}=x^{16}+1=x^{15}x+1=x^4x+1=x^5+1 \)
In relatia initiala , \( x^{15}=x^4 \) , trecem pe \( x \) in \( x+1 \) si obtinem: \( (x+1)^{15}=(x+1)^4 \Rightarrow (x+1)^{15}=x^4+1 \Rightarrow (x+1)^{16}=(x^4+1)(x+1) \)
\( \Rightarrow x^5+1=x^5+x^4+x+1 \Rightarrow x^4=x \)
Ridicand la puterea a 4-a ultima relatie obtinem: \( x^{16}=x^4 \Rightarrow x^{15}x=x \Rightarrow x^4x=x \Rightarrow x^2=x \Rightarrow \) A este un inel de tip Boole , deci este comutativ.
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers