Graduate A part in a solution for exercise 1.24 from Atiyah-MacDonald

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The discussion revolves around a specific equality in a solution to exercise 1.24 from Atiyah-MacDonald, specifically the expression ##a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c)##. A participant seeks clarification on the addition of ##2ac + 2ab + 2abc = 2a \wedge (b \vee c)## in the proof. It is noted that in a Boolean ring, the property ##a^2 = a## leads to the conclusion that ##2x = 0##, which resolves the confusion. The conversation highlights the distinction between rings, algebras, and lattices, indicating a common point of confusion among participants. Understanding these concepts is crucial for grasping the solution's validity.
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In the part below where the author of this solution, wants to show that: ##a\vee (b\wedge c)=(a\vee b)\wedge (a\vee c)##, in the third equality he adds ##2ac+2ab+2abc=2a\wedge(b\vee c)##; can someone explain to me why can we do this here?
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In a Boolean ring ##a^2=a##, so ##2x=(2x)^2=4x^2=4x##, therefore ##2x=0##.
 
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Yes, I forgot about it.
Got confused between rings,algebras and lattices... :oldbiggrin:
 

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