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Problema din gazeta cu calcule intr-un inel

Posted: Tue Mar 18, 2008 7:40 am
by Beniamin Bogosel
Fie \( (A,+,\cdot) \) un inel cu cel putin doua elemente astfel incat: \( xy=1 \Rightarrow\ yx=1,\ B=\{x \in A |x^{2}+1=0\} \neq \emptyset \) si \( xa=ax, \) \( \forall x \in B, \) \( \forall a \in A \). Sa se arate ca daca \( a,b \in A \) si \( a^{3}+b^{3}=0 \), atunci \( ab=1+b^{2}a^{2} \) daca si numai daca \( ba=1+a^{2}b^{2} \).

D. M. Batinetu - Giurgiu, GM 2/1994

Posted: Fri Jan 16, 2009 7:37 pm
by Marius Mainea
Fie \( \alpha\in A \) astfel incat \( \alpha^2=-1 \) .

Presupunand ca \( ab=1+b^2a^2 \) avem

\( (b+\alpha a^2)(a+\alpha b^2)=ba-a^2b^2+\alpha(a^3+b^3)=1+\alpha\cdot 0=1 \)

asadar

\( (a+\alpha b^2)(b+\alpha a^2)=1 \) sau

\( ab-b^2a^2+\alpha(b^3+a^3)=1 \) sau \( ab-b^2a^2=1 \) c.c.t.d.

Reciproc analog.