In field ($F, +, \cdot$) , how can I prove $x^2 =1\implies x=1,-1$

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.

• 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.) Nov 3, 2012 at 1:19

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!

• wow, that was so simple I feel stupid now... Thanks Nov 2, 2012 at 20:26
• It happens to the best of mathematicians. So, don't worry too much about it. Nov 2, 2012 at 20:27
• 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 :) Nov 2, 2012 at 20:30
• @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. Nov 2, 2012 at 20:42

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 $0$.

• +1 But you solved the problem for him. :) Nov 2, 2012 at 20:24
• @Babak: He may not agree. :-) Nov 2, 2012 at 20:25
• still feeling stupid... Thanks for the quick response both of you. Nov 2, 2012 at 20:27
• @Andrey: You’re welcome. And don’t worry about it: it’s happened to all of us. Nov 2, 2012 at 20:28