# Multiplicative Selfinverse in Fields

I assume there are only two multiplicative self inverse in each field with characteristice bigger than $2$ (the field is finite but I think it holds in general). In a field $F$ with $\operatorname{char}(F)>2$ a multiplicative self inverse $a \in F$ is an element such that

$$a \cdot a = 1.$$

I think in each field it is $1$ and $-1$. Any ideas how to proof that?

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This doesn't hold in characteristic 2, only 1 will be a self inverse in that case (note 1=-1). – Nate Iverson Jun 13 '12 at 15:43
Good point! Thanks! – joachim Jun 13 '12 at 15:48
Is 1=-1 for every field with characteristic 2? – joachim Jun 13 '12 at 16:16
@joachim: Yes, because a field $F$ has characteristic 2 precisely when $1_F+1_F=0_F$, where $0_F$ and $1_F$ are the additive and multiplicative identities of $F$, respectively. Thus $1_F=-1_F$. – Zev Chonoles Jun 13 '12 at 17:32

The equation $x^2-1$ is degree $2$ and thus can have at most two solutions in any field. So checking that $1$ and $-1$ satisfy this is enough to know that they are the only self-inverse elements. (As Nate points out, in the field of characteristic $2$ they are also equal to each other, so there is only one self-inverse element in this case).

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Since $x^2-1=(x-1)(x+1)$, you don't even need to worry about polynomials: $(x-1)(x+1)=0$ in a field implies that one factor is 0.... This proof actually works in an integral domain.... – N. S. Jun 13 '12 at 15:47
Also, might be worth pointing, that outside integral domains this is not true: in M_2 the diagonal matrix $A$ with 1 and -1 on the diagonal satisfies this relation, and so does any matrix of the form $PAP^{-1}$. It is easy to show that there are infinitely many such matrices.... – N. S. Jun 13 '12 at 15:53

Hint $\rm\ x^2\! =\! 1\!\iff\! (x\!-\!1)(x\!+\!1) = 0\! \iff\! x = \pm1,\:$ by $\rm\:ab=0\:\Rightarrow\: a=0\:\ or\:\ b=0\:$ in a field.

This may fail if the latter property fails, i.e. if nontrivial zero-divisors exist. Consider, for example, $\rm\ x^2 = 1\:$ has $4$ roots $\rm\:x = \pm1, \pm 3\:$ in $\rm\:\mathbb Z/8 =$ integers mod $8,\:$ i.e. $\rm\:odd^2 \equiv 1\pmod 8$.

Rings satsifying the latter property (no zero-divisors) are called (integral) domains. They are characterized by a generalization of the above, viz. a ring $\rm\: D\:$ is a domain $\iff$ every nonzero polynomial $\rm\ f(x)\in D[x]\$ has at most $\rm\ deg\ f\$ roots in $\rm\:D.\:$ For the simple proof see my post here, where I illustrate it constructively in $\rm\: \mathbb Z/m\:$ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree,$\:$ we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd$. The quadratic case of this result is at the heart of many integer factorization algorithms, which try to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m,\:$ e.g. a square root of $1$ that is not $\:\pm 1$.

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