Tag Archives: Frobenius homomorphism

The order of the Frobenius map on a finite field

Let \varphi denote the Frobenius map x \mapsto x^p on \mathbb{F}_{p^n}. Prove that \varphi is an automorphism and compute its order in \mathsf{Aut}(\mathbb{F}_{p^n}).


Recall that \varphi is a homomorphism. Moreover, if \alpha \in \mathsf{ker}\ \varphi, then \alpha^p = 0. Since fields contain no nontrivial zero divisors, we have \alpha = 0 (using induction if you want). So the kernel of \varphi is trivial, and thus \varphi is injective. Since \mathbb{F}_{p^n} is finite, \varphi is surjective, and so is a field isomorphism.

Next, we claim that \varphi^t(\alpha) = \alpha^{p^t} for all \alpha and all t \geq 1, and will show this by induction. The base case certainly holds, and if \varphi^t(\alpha) = \alpha^{p^t}, then \varphi^{t+1}(\alpha) = \varphi(\varphi^t(\alpha)) = \varphi(\alpha^{p^t}) = (\alpha^{p^t})^p = \alpha^{p^{t+1}} as desired.

Now \varphi^n(\alpha) = \alpha^{p^n} = \alpha, since the elements of \mathbb{F}_{p^n} are precisely the roots of x^{p^n}-x. So we have \varphi^n = 1.

If \varphi^t = 1, then we have \alpha^{p^t} - \alpha = 0 for all \alpha, so that each \alpha is a root of x^{p^t}-x. So x^{p^n-1}-1 divides x^{p^t-1}-1, and so p^n-1 divides p^t-1 (by this previous exercise) and then n divides t (by this previous exercise). In particular, n \leq t.

So n is the order of \varphi in \mathsf{Aut}(\mathbb{F}_{p^n}).

f(x)ᵖ = f(xᵖ) over ZZ/(p)

Prove that f(x)^p = f(x^p) for all f(x) \in \mathbb{F}_p[x].


Let f(x) = \sum c_ix^i. Remember that the elements of \mathbb{F}_p are precisely the roots of x^p-x; in particular, \alpha^p = \alpha for all \alpha \in \mathbb{F}_p.

Then f(x)^p = (\sum c_ix^i)^p = \sum (c_ix^i)^p = \sum c_i^p(x^i)^p = \sum c_i (x^p)^i = f(x^p) as desired.

The Frobenius endomorphism in ZZ/(p)

Prove that (a+b)^p \equiv a^p + b^p \mod p for all integers a and b.


We proved this in greater generality in this previous exercise. (See part 4.)

Basic properties of the characteristic of a ring

The characteristic \mathsf{char}(R) of a ring R is the smallest positive integer n such that \sum_{i=1}^n 1 = 0; if no such n exists, then the characteristic is zero. For instance, \mathbb{Z}/(n\mathbb{Z}) is a ring of characteristic n for each positive integer n and \mathbb{Z} is a ring of characteristic 0.

  1. Prove that the map \varphi : \mathbb{Z} \rightarrow R defined by \varphi(0) = 0, \varphi(k+1) = \varphi(k)+1 for k \geq 0, and \varphi(k) = -\varphi(-k) for k < 0 is a ring homomorphism whose kernel is n\mathbb{Z}, where n is the characteristic of R. (This explains the use of the terminology “characteristic zero” instead of the archaic phrase “characteristic infinity” if no finite sum of 1 is zero.)
  2. Determine the characteristics of \mathbb{Q}, \mathbb{Z}[x], and (\mathbb{Z}/(n))[x].
  3. Prove that if p is a prime and R is a commutative ring of characteristic p then (a+b)^p = a^p + b^p for all a,b \in R.

  1. We begin by showing that \varphi(a+b) = \varphi(a) + \varphi(b) for nonnegative b by induction. For the base case b = 0, we have \varphi(a+0) = \varphi(a) = \varphi(a) + 0 = \varphi(a) + \varphi(0). For the inductive step, suppose that for some b \geq 0, for all a, \varphi(a+b) = \varphi(a)+\varphi(b). If a+b > 0, then \varphi(a+(b+1)) = \varphi((a+b)+1) = \varphi(a+b)+1 = \varphi(a) + \varphi(b) + 1 = \varphi(a) + \varphi(b+1). Thus the conclusion holds for all positive b. Suppose now that b < 0. If a \geq 0, then \varphi(a+b) = \varphi(b+a) = \varphi(b)+\varphi(a) = \varphi(a) + \varphi(b). If a < 0, then a+b < 0, and we have \varphi(a+b) = -\varphi(-a-b) = -(\varphi(-a) + \varphi(-b)) = -\varphi(-a) - \varphi(-b) = \varphi(a)+\varphi(b). Thus for all integers a and b, \varphi(a+b) = \varphi(a) + \varphi(b).

    To show that \varphi(ab) = \varphi(a)\varphi(b), we again proceed for nonnegative b by induction. For the base case, note that \varphi(a0) = \varphi(0) = 0 = \varphi(a) \cdot 0 = \varphi(a)\varphi(0). For the inductive step, suppose the result holds for all a for some b \geq 0. Then \varphi(a(b+1)) = \varphi(ab+a) = \varphi(ab)+\varphi(a) = \varphi(a)\varphi(b) + \varphi(a) = \varphi(a)(\varphi(b)+1) = \varphi(a)\varphi(b+1). Thus by induction the result holds for all nonnegative b. Now suppose b < 0. Then \varphi(ab) = \varphi((-a)(-b)) = \varphi(-a)\varphi(-b) = (-\varphi(a))(-\varphi(b)) = \varphi(a)\varphi(b). Thus \varphi is a ring homomorphism.

    Now we show that \mathsf{ker}\ \varphi = n\mathbb{Z}, where n = \mathsf{char}\ R. (\subseteq) Suppose a \in \mathsf{ker}\ \varphi; then \varphi(a) = 0. Suppose now that n does not divide a. Then we have a = nq+r for some r with 0 < |r| < n by the division algorithm. Note moreover that \varphi(n) = 0 by definition, and that n is minimal with this property. Now 0 = \varphi(a) = \varphi(qn+r) = \varphi(r), a contradiction. Thus n divides a, and a \in n\mathbb{Z}. (\supseteq) Suppose a \in n\mathbb{Z}. Then a = nb for some b. Note that \varphi(n) = 0 by definition, so that \varphi(a) = \varphi(nb) = 0. Thus a \in \mathsf{ker}\ \varphi.

    1. Consider \mathbb{Q}. We claim that \varphi = \iota, the inclusion map. To see this, note that \varphi(0) = 0 = \iota(0), that for a \geq 0, \varphi(a+1) = \varphi(a)+1 = \iota(a)+1 = a+1 = \iota(a+1), and that for a < 0, \varphi(a) = -\varphi(-a) -\iota(-a) = -(-a) = a = \iota(a).

      Since \iota = \varphi is injective, \mathsf{ker}\ \varphi = 0. Thus \mathsf{char}\ \mathbb{Q} = 0.

    2. Consider \mathbb{Z}[x]. We claim that \varphi = \iota, the inclusion map. The proof of this proceeds exactly as in the previous example.
    3. Consider (\mathbb{Z}/(n))[x]. Note that \varphi(n) = \overline{n} = 0. Thus n\mathbb{Z} \subseteq \mathsf{ker}\ \varphi. Now let a \in \mathsf{ker}\ \varphi and suppose n does not divide a. By the division algorithm, we have a = qn+r where 0 < |r| < n. Then 0 = \varphi(a) = \varphi(qn+r) = \varphi(r), a contradiction since n is minimal. Thus n divides a, and \mathsf{ker}\ \varphi \subseteq n\mathbb{Z}. Thus \mathsf{char}\ \mathbb{Z}/n\mathbb{Z} = n.
  2. Let R be a commutative ring with characteristic p, and let a,b \in R. Note that p divides p! but not k! or (p-k)! when 0 < k < p. Thus p divides \binom{p}{k} for 0 < k < p. Using the Binomial Theorem, we have (a+b)^p = \sum_{k=0}^p \binom{p}{k} a^kb^{p-k} = a^p + b^p.