1
$\begingroup$

Let $ A $ be a ring with $ x^{2}+y^{2}+z^{2}=xy+yz+zx+xyz+1,\forall x,y,z\in A^{*} $.
Prove that $ A $ is a field with 2 elements.

If we put $ x=y=z=1 $ we obtain that $ 1+1=0 $.
If we put $y=z=1 $ we have that $ x^2=x $, which means $ A $ is a boolean ring.
That's all I did so far.

  • 2
    Does $A^*$ mean $A\setminus\{0\}$? (I would guess that it does from context, but this is rather nonstandard: normally $A^*$ would mean just the units of $A$.) – 2017-01-23
  • 0
    Yes, it's $A\setminus\{0\} $. – 2017-01-23

2 Answers 2

3

Hint: Try setting $x=y=z$.

(By the way, it is not correct to conclude that $|A|=2^n$ as you have done, since $A$ might be infinite. Also, the statement you are trying to prove is slightly incorrect, since the condition also vacuously holds if $A=\{0\}$. Your conclusion should be that $A$ is either $\{0\}$ or a field with two elements.)

  • 0
    If $x=y=z$ then $ x^3=1$. – 2017-01-23
  • 0
    @ztefelina And along with $x^2=x$, that gives you...? – 2017-01-23
  • 0
    $x=1$. Is this statement true? :) – 2017-01-23
  • 0
    I can't see how the statement holds if $\;A=\{0\}\;$ : I get that then $\;0=1\;$ , which, as far as I know, is widely agreed that does *not* hold in rings... – 2017-01-23
  • 0
    @DonAntonio: If $A=\{0\}$ then $A^*=\emptyset$ so the statement is vacuous (there are no $x,y,z$ for which it must hold). Also, $0=1$ is true in $A$ in that case (there is one element in $A$, which is both the additive and multiplicative identity). – 2017-01-24
  • 0
    @EricWofsey Thanks for the comment. I think that'd be a rather big stretch of the meaning of "vacuosly true" *in this case*, but being perfectly formal I must say I'd accept it. Perhaps the OP would want to add "let be **a non-zero** ring* to avoid that spooky object... – 2017-01-24
-1

$x^2 + x^2 + x^2 = x*x + x*x + x*x + x*x *x + 1 =x^2 + x^2 +x^2 + x^3 + 1$ so $x^3 + 1 = 0$ and $x^3 = -1$

$x^2 + x^2 + 1 = x*x + x*1 + 1*x + x*x *1 = x^2 + x + x + x^2 + 1$ so $x + x =0$ and$x = -x $ and therefore $x^3 = -1 = 1$

$x^2 = x^2 + ( 1-1) = x^2 + 1 + 1 = x^2 + 1^2 + 1^2 = x*1 + 1*1 + 1*x + x*1*1 + 1=x+x + x +1 + 1 = x$

So $1=x^3 = x(x^2) = xx = x^2 = x$

So $x =1$ for all $x \ne 0$.

Any ring with only two elements is a field.