Raising element of field to characteristic power

Question

I came across this in a set of notes.

Let $K$ be a field of characteristic $p$ and let $\lambda\in K$. Then $$\lambda^{p-1}=1.$$

I've never seen this before. Is it correct?

Answer 1

The statement you gave is "known" (i.e. when stated correctly, see the comments or the link) and is called Fermat's little theorem.

Hope that helps,

Answer 2

The relevant fact here is that, for any field $F$ of characteristic $p$, there is a unique field homomorphism $\mathbf{F}_p \to F$, and that the nonzero elements of (the image of) $\mathbf{F}_p$ are precisely the $(p-1)$-th roots of unity in $F$.

$\mathbf{F}_p$ means the field of $p$ elements, which is isomorphic to the integers modulo $p$.

Answer 3

As stated, it isn't true-it is known (but I shall not prove it here) that in a field, $F$, that a polynomial of degree $n$ in $F[x]$ has at most $n$ roots.

If $\text{char}(F) = p$, we certainly have: $x^p - x$ is a polynomial of degree $p$ in $F[x]$, and this factors in the prime field of $F$ (which is isomorphic to $\Bbb Z_p$) as:

$x^p - x = x(x - 1)\cdots(x - (p - 1))$ (which should be clear since $\Bbb Z_p - \{0\}$ is a group of order $p - 1$, and thus every non-zero element has (multiplicative) order dividing $p-1$, from basic group theory).

So if $F \neq \Bbb Z_p$ (or more properly, is (strictly) larger than its prime subfield), we cannot have any other roots of $x^p - x = x(x^{p-1} - 1)$, so no roots of $x^{p-1} - 1$ other than the ones in the prime subfield.

Concretely, if $F$ is a field of $4$ elements, say $\{0,1,a,1+a\}$, of characteristic $2$, then $a \in F$, but $a = a^1 = a^{2-1} \neq 1$.