Start with a finite field F_q (meaning there are a finite number of elements, actually exactly q elements). Meaning, any element K of F_q, when raised to the q power, becomes itself again. K^q = K. Make sure q is equal to a prime raised to the some n power, so q = p^n.
To repeat, K = K^q when you do arithmetic in this field.
The Frobenius map FM takes in an element and raises it to the q power. Now, no matter what element R of F_q we input to FM, we get output R^q.
That’s useless.
We already know that R = R^q. So the input equals the output!
BUT! If you take the algebraic closure of F_q, say F_qc, then in general a random element Y of F_qc, when plugged in to FM, will produced Y^q such that Y DOES NOT EQUAL Y^q!
In fact, FM is an element of the Galois group G(F_qc/F_q) and repeated applications of FM generates the entire Galois group.
