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I'm a really confused about fields.

I know that it means $x$ is the reciprocal element of itself, and I can easily show that $1^2=1$ (not as trivial for $(-1)^2$ though), but I'm not sure how it helps me.

edit: oh... I can only approve one answer. Well Rankeya was first (by a very short time) so I guess I'll approve his though, I don't really have any idea what it means. Thanks to both Brian M. Scott and Rankeya for the help.

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    $\begingroup$ Dear @Nescio: You accept an answer that you feel is most helpful to you. It does not have to be the first answer that is posted. (But, make sure that you always accept answers if you feel you are satisfied with them. It encourages people to answer your questions, and also brings a sense of completeness/closure.) $\endgroup$
    – Rankeya
    Nov 3 '12 at 1:19
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A field is a domain, which in particular means that $ab = 0 \Rightarrow a = 0$ or $b = 0$. Write $x^2 = 1$ as $x^2 - 1 = 0$, and try to proceed from there.

Also, welcome to MSE!

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  • $\begingroup$ wow, that was so simple I feel stupid now... Thanks $\endgroup$
    – Nescio
    Nov 2 '12 at 20:26
  • $\begingroup$ It happens to the best of mathematicians. So, don't worry too much about it. $\endgroup$
    – Rankeya
    Nov 2 '12 at 20:27
  • $\begingroup$ Actually, I have been told by other mathematicians that it happens to the best of mathematicians. I often disbelieve them when they say this, and continue to feel stupid :) $\endgroup$
    – Rankeya
    Nov 2 '12 at 20:30
  • $\begingroup$ @Rankeya: The best way to believe it is to see enough people who are more experienced and smarter than you do the same. Then again, even if you do believe it, it doesn't mean it won't make you feel stupid when you it happens to you. $\endgroup$
    – tomasz
    Nov 2 '12 at 20:42
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HINT: $x^2=1$ if and only if $x^2-1=0$. In any field $x^2-1=(x-1)(x+1)$, so $x^2=1$ if and only if $(x-1)(x+1)=0$. Now prove that for any $a,b\in F$, $ab=0$ if and only if at least one of $a$ and $b$ is $0$.

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    $\begingroup$ +1 But you solved the problem for him. :) $\endgroup$
    – Mikasa
    Nov 2 '12 at 20:24
  • $\begingroup$ @Babak: He may not agree. :-) $\endgroup$
    – Brian M. Scott
    Nov 2 '12 at 20:25
  • $\begingroup$ still feeling stupid... Thanks for the quick response both of you. $\endgroup$
    – Nescio
    Nov 2 '12 at 20:27
  • $\begingroup$ @Andrey: You’re welcome. And don’t worry about it: it’s happened to all of us. $\endgroup$
    – Brian M. Scott
    Nov 2 '12 at 20:28

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