Boolean rings with identity can only take 2 elements?

[pivoxa15]

Using the theorem that in any boolean ring a+a=0 for all a in boolean ring R.

Then 0 is in R. Make the multiplicative identity 1 is also in it. Therefore R can only take 0 and 1 and no more because 1+1=0. 0+0=0. 1+0=1 always. So 2 or other elements can never occur.

[matt grime]

What? That makes no sense. Mainly because you do not ask a question at all.

[Dick]

Ok, so '2' isn't in the ring. Why does that mean there is nothing else in the ring? There's no rule that says you can generate everything in a ring by adding 1's.

[pivoxa15]

I was looking for a confirmation that my claim is correct.

So far we know 0 and 1 is in R. Take all 4 combinations of these two elements with respect to the 2 operations. So 8 additions and multiplications all together, all of which gives 0 or 1 so R contains only two elements 0 and 1. The only non trivial one is 1+1 which has to be 0 as by the theorem in the OP.

[Dick]

Your 'claim' is not only incorrect. It's ridiculous. So all rings are generated by 0 and 1? You had better add that to the axiom list, because its not there yet, last I heard.

[matt grime]

Here is a meta-answer that shows the claim to be silly. If there is a unique boolean ring, why did you ask about boolean ringS?

[pivoxa15]

Remember I am talking about a boolean wring with multiplicative identity.

I am claiming there is only one unique boolean ring with multiplicative identity.

Or with even this consideration taken into account, it is not correct? If so why?

If the boolean ring dosen't have multiplicative identity than other combinations may be possible.

[HallsofIvy]

Using the theorem that in any boolean ring a+a=0 for all a in boolean ring R.

Then 0 is in R. Make the multiplicative identity 1 is also in it. Therefore R can only take 0 and 1 and no more because 1+1=0. 0+0=0. 1+0=1 always. So 2 or other elements can never occur.
I don't see why those prevent other elements. What about a set {0, 1, a, b} with operation tables:
addition:
0 1 a b
0 0 1 a b
1 1 0 b a
a a b 0 1
b b a 1 0

multiplication
0 1 a b
0 0 0 0 0
1 0 1 a b
a 0 a b 1
b 0 b 1 a

[matt grime]

Remember I am talking about a boolean wring with multiplicative identity.

and?

I am claiming there is only one unique boolean ring with multiplicative identity.

but that is patently silly.

Just write down something to produce a counter example. It is trivial to produce such example. Hell, diagonal matrices with 0s and 1s on the diagonals gives infinitely many counter examples without having to think at all.