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There is a field $\Bbb F$ such that the equation $x^2=1$ have more than two solutions (for some $x\in\Bbb F$)?

This question comes suddenly to my mind.

I know that if $\Bbb F=\Bbb C$ then the fundamental theorem of algebra states that the equation $x^2=1$ have two solutions, but this would imply that for an arbitrary field $x^2=1$ have, at most, two solutions?

Im sorry if this question is trivial but at this moment Im unable to achieve a conclusion from the axioms that define a field.

Masacroso
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    In any field, a polynomial of degree $n$ can have at most $n$ roots. – lulu Jul 04 '17 at 21:46
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    Well, in a division ring (skew field) it can occur, e.g. among the quaternions, $x^2=1$ has infinitely many solutions – Berci Jul 04 '17 at 21:46
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    If $x_1^2=1$ and $x_2^2=1$ then $0=x_1^2-x_2^2=(x_1-x_2)(x_1+x_2)$. From there you get that either $x_1=x_2$ or $x_1=-x_2$. – Bettybel Jul 04 '17 at 21:47
  • @Berci: are you sure about that? – Robert Lewis Jul 04 '17 at 22:18
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    @RobertLewis: No.. but $x^2=-1$ has infinitely many solutions for sure, all purely imaginary unit quaternions. At first I thought we just multiply all solutions by $i$ to get solutions for $x^2=1$, but that was wrong because of noncommutativity.. – Berci Jul 04 '17 at 22:32
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    Extensive discussion of this at : https://math.stackexchange.com/questions/122898/why-are-the-solutions-of-polynomial-equations-so-unconstrained-over-the-quaterni , which also discusses why $x^2=k$ can have uncountably many solutions in a skew field. – MJD Jul 05 '17 at 01:34

6 Answers6

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No: over an integral domain, a polynomial of degree $n$ has at most $n$ roots.

This is a consequence of the fact that a polynomial $p$ is divisible by $X-\alpha$ if and only if $p(\alpha)=0$.

Bernard
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If $x^2=1$ then $(x-1)(x+1)=0$, and since $F$ is a field either $x-1=0$ or $x+1=0$.

In general, if $f(x)$ is a polynomial over a field $F$ of degree $n$, then $f$ has at most $n$ roots in $F$.

carmichael561
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No, in any field $x^2 = 1$ iff $x^2-1 = 0$ iff $(x-1)(x+1) = 0$ and in a field (no zero-divisors) this means that either $x-1 = 0$ (so $x=1$) or $x+1 = 0$ (so $x=-1$).

Henno Brandsma
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Hint : $x^2 = 1 \iff (x-1)(x+1) = 0$ and a field is an integral domain

krirkrirk
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If $\alpha$ solves

$x^2 = 1, \tag{1}$

so that

$\alpha^2 = 1, \tag{2}$

or

$\alpha^2 -1 = 0, \tag{3}$

we have

$(\alpha - 1)(\alpha + 1) = 0; \tag{4}$

if now

$\alpha \ne 1, \tag{5}$

so that

$\alpha -1 \ne 0, \tag{6}$

then since we are operating in a field,

$\alpha + 1 = 0, \tag{7}$

or

$\alpha = -1. \tag{8}$

At most two possibilities. And that's all.

Robert Lewis
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In any non zero ring $R$ with $1$, which has no nonzero zero divisors, $x^2=1$ has exactly $1$ root if $char(R)=2$ and exactly 2 roots if $char(R)\neq2$.

algen
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