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Looking for some pointers on how to approach this problem:

Let $F$ be a field consisting of exactly three elements $0$, $1$, $x$. Prove that $x + x = 1$ and that $x x = 1$.

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3 Answers 3

up vote 4 down vote accepted

Hint 1: We know that $1x=x$ and $0x=0$. But $x$ must have a multiplicative inverse, since $x\neq 0$. So the multiplicative inverse has to be <fill in the blank>

Hint 2. Note that $1+x$ must be either $0$, $1$, or $x$. Can it be $1$? Then we would have $1+x=1$, which implies <fill in the blank>. Can it be $x$? What does that leave for $x+x$?

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Did you mean $\,x+x\,$ or $\,1+x\,$ in your final sentence? Either way, it is not clear how you intend to finish the proof. Perhaps a further hint would help to clarify that for the OP. –  Bill Dubuque Jul 8 '12 at 16:40
    
@Bill: I meant $x+x$. The values of $x+0$, $x+1$, and $x+x$ must be, in some order, $0$, $1$, and $x$. We know $x+0$ is $x$. If we know that $x+1$ is, then that tells you what $x+x$ must be. –  Arturo Magidin Jul 8 '12 at 17:40
    
The multiplication part is straightforward, however I'm scratching my head with the addition part. If 1 + x = 1, that implies x = 0, which is a contradiction. If 1 + x = x, then that implies 1 = 0 which is also a contradiction. The only remaining choice is then x + 1 = 0, which implies x is equivalent to -1? If that's the case, then I don't see how x + x could be 1. Is there a property of fields that I'm missing or don't understand? –  BrandonK. Jul 9 '12 at 1:27
    
@BrandonK. "$-1$" just means "the element which added to $1$ yields $0$". What you need to note that is $x+0$, $x+1$, and $x+x$ must give you the three results $0$, $1$, and $x$, in some order (think about the addition table: every row and every column has to contain each and every element exactly once). Since you know that $0+x$ is $x$; and you know that $1+x$ is $0$, that means that $x+x$ has to be whatever is left. P.S. Don't accept an answer at least until you are sure it has actually answered your question! –  Arturo Magidin Jul 9 '12 at 1:36
    
Thanks for the help. I proved x+x=1 by contradicting both x+x=x (implying x=0) and x+x=0 (since x+1=0, x+x=0=x+1 which implies that x=1). –  BrandonK. Jul 9 '12 at 5:33

Write down the addition and multiplication tables. Much of them are known immediately from the properties of 0 and 1, and there's only one way to fill in the rest so that addition and multiplication by nonzero are both invertible.

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Suppose that $xx=x$. Then because $x\neq 0$, we can multiply both sides by $x^{-1}$ (whatever it may be) to get $x=1$. But this contradicts $x\neq 1$, so we cannot have $xx=x$.

Suppose that $xx=0$. Then because $F$ is a field, it is in particular an integral domain, so for any $a,b\in F$ we have $ab=0\implies a=0$ or $b=0$. Thus, $xx=0$ implies $x=0$, which contradicts $x\neq 0$.

Thus, the only remaining possibility is that $xx=1$.

Similar reasoning can be used for $x+x$.

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How along these lines does similar reasoning rule out $x+x=0$? There isn't a counterpart to the integral domain argument for $xx=0$. –  alex.jordan Jul 8 '12 at 5:59
    
I starting thinking about that right after I wrote it :) I think the easiest approach is Arturo's second hint, but one could also go with $x=-x$ implies $1=-1$ which implies the field should have a cardinality that is a power of 2. I guess I was just being vague for the purpose of leaving it to the OP. –  Zev Chonoles Jul 8 '12 at 6:03
    
I've been trying to reason this out without using Arturo's or Hurkyl's answers (which I believe both make use of one-to-one-ness of addition by a fixed value, and are great). If $x+x=0$, then $x(1+1)=0$, so because of the integral domain property, $1+1=0$. Now what is $x+1$? (Here we are using experience to tell us it must be a fourth element.) Not $x$ (or $1$), or else $1$ (resp $x$) is $0$. So $x+1=0$, which in turn is $1+1$, which implies $x=1$. So $x+1$ is none of the three, a contradiction. (In hindsight, maybe this is what Arturo is getting at.) –  alex.jordan Jul 8 '12 at 6:14
    
The reasoning for $\,x + x\,$ is not so trivial, and not what I would call "similar" - try it. –  Bill Dubuque Jul 8 '12 at 16:57

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