# Show that $a^{p^n}=a\mod p$

My book says that for elements $$\alpha$$ in $$\mathbb F_p$$, where $$p$$ is prime, it holds that $$\alpha^{p^n}=\alpha,$$ because of Fermat's little theorem, which says that $$a^p=a\mod p.$$ Of course it's clear that $$\alpha^p=\alpha$$, but I don't see why this would hold for arbitrary powers of $$p$$. We have $$\alpha^{p^n}=\alpha^n,$$ but this isn't necessarily equal to $$\alpha$$, is it?

For context: my book uses this argument, to show that the roots of $$X^{p^n}-X$$ in $$\overline{F_p}$$ form a field of $$p^n$$ elements, and they argue that $$F_p$$ is a subset of the set of roots of $$X^{p^n}-X$$.

• $\alpha^{p^2} \equiv \alpha^{pp} \equiv (\alpha^p)^p \equiv \alpha^p \equiv \alpha \pmod{p}$. Mar 29, 2019 at 2:52

Fixed points stay fixed under iteration, by an obvious induction:

if $$\,f(x) = x\,$$ then $$\, \color{#c00}{f^{\large n}(x) = x}\,\Rightarrow\, f^{\large n+1}(x) = f(\color{#c00}{f^{\large n}(x)}) = f(\color{#c00}x) = x$$

OP is the special case $$\, f(x) := x^{\large p}\,$$ so $$\,f^{\large n}(x) = x^{\large p^{\Large n}}\pmod {p}$$

Observe that, in any ring $$R$$, for $$\alpha \in R$$ and $$m \in \Bbb N$$ we have

$$\alpha^{m^2} = (\alpha^m)^m; \tag 1$$

$$\alpha^{m^3} = (\alpha^{m^2})^m; \tag 2$$

and in general, for $$n \in \Bbb N$$,

$$\alpha^{m^n} = \alpha^{m^{n - 1}m} = (\alpha^{m^{n - 1}})^m. \tag 3$$

Now suppose that $$R$$ is such that

$$\alpha^m = \alpha; \tag 4$$

then from (1) and (2),

$$\alpha^{m^2} = (\alpha^m)^m = \alpha^m = \alpha, \tag 5$$

$$\alpha^{m^3} = (\alpha^{m^2})^m = \alpha^m = \alpha; \tag 6$$

thus if

$$\alpha^{m^j} = \alpha, \tag 7$$

then

$$\alpha^{m^{j + 1}} = \alpha^{m^jm} = (\alpha^{m^j})^m = \alpha^m = \alpha, \tag 7$$

and by induction it may be concluded that

$$\alpha^{m^n} = \alpha \tag 8$$

holds for all $$n \in \Bbb N$$.

Now taking

$$m = p \in \Bbb P, \tag 9$$

and

$$R = \Bbb F_p \tag{10}$$

we have

$$\alpha^p = \alpha; \tag{11}$$

thus it follows that

$$\alpha^{p^n} = \alpha, \; \forall n \in \Bbb N. \tag{12}$$

• Simpler: fixed points stay fixed on iteration - see my answer. Mar 30, 2019 at 2:11