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I'm not sure how to go about finding the solution to this question.

Let R be a ring with identity such that $x^2 = 1_R$ for all $0_R \neq x\in R$. How many elements are in $R$?

I've just been playing around with squaring elements, like $$(x+1_R)^2 = x^2+2.x+1_R =2.x+2.1_R = 1_R.$$

But I'm not sure where to go with this. Any help?

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5  
Either $1_R+1_R=0$ or $(1_R+1_R)^2=1_R$ so $3\cdot 1_R=0$. –  Davide Giraudo Mar 5 '12 at 20:15

2 Answers 2

One possibility, of course, is $R=\{0\}$. Assume $1_R\neq 0$.

$R$ has no zero divisors: if $xy=0$ and $x\neq 0$, then $y = 1_Ry = xxy = x(0)=0$.

$R$ is commutative: if $x$ and $y$ are nonzero, then so is $xy$ by the above; hence $(xy)^2 = x^2y^2$; canceling from $xyxy=xxyy$ we get $yx=xy$.

(Of course, the ring satisfies $x^3=x$ for all $x$, so by a famous theorem of Jacobson, the ring is necessarily commutative; but we don't need to call in that heavy cannon to the fray).

Since every nonzero element is invertible, $R$ is a field. Since $x^2-1_R$ has $|R-\{0\}|$ solutions, we have $|R-\{0\}|\leq 2$, so $|R|\leq 3$.

If $1_R+1_R=0$, then $R$ is of characteristic $2$, so $R\cong \mathbb{F}_2$ and $|R|=2$. And, indeed, in this case the hypothesis holds.

If $1_R+1_R\neq 0$, then we get $4\cdot 1_R = 1_R$, hence $3\cdot 1_R=0$, so $R$ is of characteristic $3$, and therefore $R\cong\mathbb{F}_3$ and $|R|=3$. Again, the hypothesis holds for this ring.

In summary, $R$ has either $1$, $2$, or $3$ elements, and is either the trivial ring, $\mathbb{F}_2$, or $\mathbb{F}_3$.

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Thanks, this is very helpful, but why does $x^3=x$ imply that $R$ is commutative? –  098765 Mar 5 '12 at 20:28
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@098765: It is a common exercise in Ring Theory, and also a consequence of a famous Theorem of Jacobson, which states that a ring for which there exists $n\gt 0$ such that $x^n=x$ for all $x$ is always commutative. But I gave a simpler derivation that does not call upon such heavy artillery in an edit. –  Arturo Magidin Mar 5 '12 at 20:33
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@ArturoMagidin am pleased you revised your answer, because the proof that $x^3 = x$ implies R is commutative isn't trivial, and is probably beyond the stage at which the asker of this question is at. –  David Wheeler Mar 5 '12 at 20:53
    
Does your proof actually depend on commutativity? (mine does not). If not, then you can omit the proof of comutativity, since it's a trivial consequence of the final result. –  Bill Dubuque Mar 5 '12 at 22:42
    
@Bill: I use commutativity only to conclude that I have a field rather than a simply division ring; I believe it is equivalent to your "$x^{-1}=x\Rightarrow R$ field". –  Arturo Magidin Mar 6 '12 at 3:54

$\rm R=0\:$ works. Else $\rm\ x\ne 0$ $\Rightarrow$ $\rm x^2 =1$ $\Rightarrow$ $\rm x^{-1} = x $ $\Rightarrow$ $\rm R$ field, so $\rm\: x\ne 0\!\iff\!\! (x-1)(x+1) = 0$ $\iff$ $\rm x=\pm 1.\:$ But $\rm\: R\backslash0 = \{\pm1\}$ $\!\iff\!$ $\rm R\:\! \cong\:\! \mathbb Z/2\:$ or $\:\mathbb Z/3$.

More generally, the finite fields $\:\rm\mathbb F_p,\: \mathbb F_q,\ p,q\:$ prime, are axiomatized by the ring axioms plus $$\rm x^n =\: x,\quad n\: =\: 1 + lcm(p\!-\!1,q\!-\!1)$$ $$\rm q\:(x^p-x)\: =\: 0\: =\: p\:(x^q-x)$$ $$\rm pq\: =\: 0$$

Thus any identity true in both of these fields has a purely equational proof from the above axioms. This theorem extends similarly to any finite set of finite fields, for example see Stanley Burris and John Lawrence, Term rewrite rules for finite fields (1991). This result is very closely related to Jacobson's model-theoretic proof of commutativity of rings satisfying the identity $\rm\: x^{n_x} =\: x$.

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