Is "$x^2=1\implies x=\pm 1$" true on any field? why?

Question

The only elements $x\in\Bbb F_p$ with $x^2=1$ are $\pm1$, right? Is that true for any field? How do I see that it's true?

Answer 1

First, in any ring $1^2=1$ and $(-1)^2=1$.

Next, in any integral domain a polynomial of degree $N$ can have at most $N$ roots (a consequence of Bézout's theorem and nonexistence of zero divisors). So, a polynomial $x^2-1$ can have at most two roots.

If the characteristic is not $2$, we have $-1 \neq 1$, so these are the two roots. If the characteristic is $2$, we note that now $x^2 - 1 = x^2+1 = x^2 - 2x + 1 = (x-1)^2$, so $1$ is a root of multiplicity $2$.

Answer 2

Assume that $G=\langle g \rangle$ is a finite cyclic group with even order, $|G|=2n$.
Then we may list the elements of the group in the following way: $$ G=\left\{1,g,g^2,\ldots, g^{2n-1}\right\} $$ and see how the map $\varphi:x\mapsto x^2$ acts on $G$: it is a $2$-to-$1$ map, mapping $G$ into the subgroup of squares, also known as the set of quadratic residues: $$ \varphi(G) = \left\{1,g^2,g^4,\ldots,g^{2n-2}\right\} $$ If now we take some $h\in G$, only two things may happen: $h=g^{2m+1}$, so $h$ is not a quadratic residue, or $h=g^{2m}$, and in such a case $h=(g^m)^2 = (g^{m+n})^2$, i.e. $h$ has two distinct square roots.

If $\mathbb{F}_{p^k}$ is a finite field, $\mathbb{F}_{p^k}^*$ is a cyclic group with $p^k-1$ elements. Assuming $p\neq 2$, the above considerations apply: $1$ is for sure a quadratic residue, since $1=1^2$, with two distinct square roots, that cannot be anything else than $\pm 1$.

Answer 3

Hint $\ (x\!-\!1)(x\!+\!1) = 0\,\Rightarrow\, x\!-\!1 = 0\,$ or $\,x\!+\!1 = 0\,$ by the field (or domain) axiom that $\, ab = 0\,\Rightarrow\, a=0\,$ or $\,b = 0,\,$ i.e. no zero-divisors exist, i.e. nonzero elts are cancellable (regular).

More generally, by iterating the Factor Theorem we can show that a nonzero polynomial $\,f\,$ over a field (or domain) $D$ has no more roots than its degree $\,n.\,$ Indeed if $\,f\,$ has $\,\ge n\,$ distinct roots $\,r_i$ then inductively applying the Factor Theorem (as in the Theorem below) shows $\,f = c(x\!-\!r_1)\cdots (x\!-\!r_n),\,$ so $\, r\ne r_i\,\Rightarrow\, f(r)= c(r\!-\!r_1)\cdots (r\!-\!r_n) \ne 0\,$ by all factors $\ne 0,\,$ and $D$ a domain. Thus $\,f\,$ has at most $\,n\,$ roots. Below we show how the inductive step works.

Bifactor Theorem $\ $ Let $\rm\,a,b\in R,\,$ a commutative ring, and $\rm\:f\in R[x]\:$ a polynomial over $\,\rm R.\,$
If $\rm\ \color{#0a0}{a\!-\!b}\ $ is $\,\color{#c00}{\rm cancelable}\,$ in $\rm\,R\,$ (i.e. not a zero-divsor) $ $ then

$$\rm f(a) = 0 = f(b)\ \iff\ f\, =\, (x\!-\!a)(x\!-\!b)\ h\ \ for\ \ some\ \ h\in R[x]$$

Proof $\,\ (\Leftarrow)\,$ clear. $\ (\Rightarrow)\ $ Applying Factor Theorem twice, while canceling $\rm\: \color{#0a0}{a\!-\!b},$

$$\begin{eqnarray}\rm\:f(b)= 0 &\ \Rightarrow\ &\rm f(x)\, =\, (x\!-\!b)\,g(x)\ \ for\ \ some\ \ g\in R[x]\\ \rm f(a) = (\color{#0a0}{a\!-\!b})\,g(a) = 0 &\color{#c00}\Rightarrow&\rm g(a)\, =\, 0\,\ \Rightarrow\,\ g(x) \,=\, (x\!-\!a)\,h(x)\ \ for\ \ some\ \ h\in R[x]\\ &\Rightarrow&\rm f(x)\, =\, (x\!-\!b)\,g(x) \,=\, (x\!-\!b)(x\!-\!a)\,h(x)\end{eqnarray}$$

Remark $\ $ The theorem may fail when $\rm\ a\!-\!b\ $ is not cancelable (i.e. is a zero-divisor), e.g.

$$\rm mod\ 8\!:\,\ f(x)=x^2\!-1\,\Rightarrow\,f(3)\equiv 0\equiv f(1)\ \ but\ \ x^2\!-1\not\equiv (x\!-\!3)(x\!-\!1)\equiv x^2\!-4x+3\quad$$

It will prove instructive to examine the above proof in this special case to see how it fails.

As interesting application of this is that we can quickly find a factor of factor $\rm m>1\,$ by a gcd calculation if we are given a polynomial with more roots mod $\rm\, m\,$ than its degree, e.g. see this answer where I show how it works for low degree polynomials. This idea underlies some factorization algorithms.