Алгебра | Константин Чепуркин. Лекция 3
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1:40:00 Пусть 0=1 и a', a" два элемента из кольца, тогда из дистрибутивности следует 0(a'−a")=0 и a'−a"=1(a'−a")=0(a'−a")=0 ⇒ a'=a"=0=1. Докажем 0a=a0=0. a+0a=1a+0a=(1+0)a=1a=a ⇒ 0a=0, аналогично a0=0. Из последнего следует (−1)a=a(−1)=−a. a+(−1)a=1a+(−1)a=(1+(−1))a=0a=0.
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