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

[mad mathematician]

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?

Thanks!

 

[martinbn]

In a Boolean ring $a^2=a$, so $2x=(2x)^2=4x^2=4x$, therefore $2x=0$.

 

[mad mathematician]

Yes, I forgot about it.
Got confused between rings,algebras and lattices...