Prove that $R$ has unity

Question

Let $R$ be a finite ring such that $x^2=x$ for all $x$ in $R$. Prove that $R$ has unity.

I was able to show that it was commutative.

Proof:

$x^2=x$ $x^2-x = 0$ $x (x-1) =0 $

thus $x = 0$ or $x = 1$. Since $x\cdot 1 = x$. 1 is the unity of $R$. Thus $R$ has unity.

I feel like there is something wrong with my solution.

consider $3$ in $\mathbb{Z}/6\mathbb{Z}$. You have $3^2=3$ but $3$ is not $0$ or $1$. You can't assume that $R$ has no zero divisors. — Jonathan, Jul 03 '13 at 03:31

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

You're assuming that $R$ must be a domain. It is possible (or at least you haven't yet ruled it out) that two nonzero elements can multiply to zero. You also assume that $R$ has a unit element in order to prove it has one, which is of course illicit.

Here are some comments that might help. I will use 'ring' to mean "ring, possibly without identity." Rings with $x^2=x$ for all $x$ are called Boolean. You can show, by taking $(x+x)^2$, that all such rings must have characteristic two. This allows us to view such rings as vector spaces over $\mathbb Z_2$. In particular, a finite Boolean ring must have $2^n$ elements for some integer $n$.

For a complete solution, see here.

edited Apr 13 '17 at 12:21

Community

1

answered Jul 03 '13 at 03:31

Potato

41,411

hmm. Then how do I approach this problem? – mathpartyisfun Jul 03 '13 at 03:41
So x+x = (x+x)^2 then x+x = x + x + x + x x + x = 0 which shows that it has characteristic two. – mathpartyisfun Jul 03 '13 at 04:09
@mathpartyisfun Note $x+x=(x+x)^2=x+x+x+x$ implies $2x=0$ for all $x$ in $R$. – Potato Jul 03 '13 at 04:11
oh lol, I was editing my thing, then I saw your post. – mathpartyisfun Jul 03 '13 at 04:14
re: your last paragraph: not sure what "basis" you're talking about. the 4 element boolean ring I have in mind has $a=(1,0)$ and $b=(0,1)$ so that $ab=0$ and $a+b=(1,1)$ is the identity – Jonathan Jul 03 '13 at 04:19
@Potato.. yea what you wrote make sense. But this wasn't a True/False question. It specifically asks to prove that statement. – mathpartyisfun Jul 03 '13 at 04:21
why should there be a boolean ring ${0,a,b,a+b}$ with $ab=a+b$? – Jonathan Jul 03 '13 at 04:24
a (ab) = ab but a(a + b) = b (what you wrote)
So, ab cannot be equal to a+b ?
– mathpartyisfun Jul 03 '13 at 04:26
Ok, thanks, that's why. – Potato Jul 03 '13 at 04:27
it can be a ring, but why should it be boolean or have $ab=a+b$? – Jonathan Jul 03 '13 at 04:27
@Jonathan Because you set $-a=a$ and $-b=b$ (those are basis elements over $\mathbb Z_2$) and use $ab=a+b$ to define multiplication. But it turns out that's not well-defined. But my answer does show it suffices to give the multiplication just on basis elements. – Potato Jul 03 '13 at 04:28
i gave an example of an actual 4 element boolean ring, and it doesn't satisfy the arbitrarily imposed condition – Jonathan Jul 03 '13 at 04:30
@Jonathan You haven't ruled out the possibility of having two non-isomorphic 4-element Boolean rings. – Potato Jul 03 '13 at 04:32
I never said I had. hence my question "why should there be a boolean ring...". nevermind – Jonathan Jul 03 '13 at 04:33

Prove that $R$ has unity

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