9
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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, 2012 at 1:19

2 Answers 2

Reset to default
16
$\begingroup$

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!

$\endgroup$
4
  • $\begingroup$ wow, that was so simple I feel stupid now... Thanks $\endgroup$
    – Nescio
    Nov 2, 2012 at 20:26
  • $\begingroup$ It happens to the best of mathematicians. So, don't worry too much about it. $\endgroup$
    – Rankeya
    Nov 2, 2012 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, 2012 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, 2012 at 20:42
12
$\begingroup$

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 $0$.

$\endgroup$
4
  • 2
    $\begingroup$ +1 But you solved the problem for him. :) $\endgroup$
    – Mikasa
    Nov 2, 2012 at 20:24
  • $\begingroup$ @Babak: He may not agree. :-) $\endgroup$
    – Brian M. Scott
    Nov 2, 2012 at 20:25
  • $\begingroup$ still feeling stupid... Thanks for the quick response both of you. $\endgroup$
    – Nescio
    Nov 2, 2012 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, 2012 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.