Show that a Boolean ring is a commutative ring.

Question

I have a question about (https://math.stackexchange.com/q/10279)'s proof to this. I also asked as a comment but I am unsure whether it will be replied to since the post was made 8 years ago?

My question to their proof is: How does $x+xy+yx+y=x+y$ imply $xy=yx$? Doesn't $xy=yx$ imply $xy-yx=0$, and so how is $xy+yx=0$?

Timothy Wagner's answer to the question you cite shows that every Boolean ring is of characteristic $2$, so $xy+xy=0$ — saulspatz, Apr 03 '18 at 22:57
Possible duplicate of How to show that every Boolean ring is commutative? — Robert Frost, Apr 04 '18 at 07:05

cansomeonehelpmeout · Accepted Answer · 2018-04-03T23:04:21.693

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In a Boolean ring $(-1)^2=-1$, so $1=-1$. In particular $$xy-yx=xy+yx$$

edited Apr 03 '18 at 23:04

answered Apr 03 '18 at 22:57

cansomeonehelpmeout

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I see this now! Thanks so much for making me understand in a clear easy manner. – Apr 03 '18 at 23:02
@numericalorange My pleasure! – cansomeonehelpmeout Apr 03 '18 at 23:04

Show that a Boolean ring is a commutative ring.

1 Answers1