## 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 $(a+b)^p = \sum_{k=0}^p \binom{p}{k} a^kb^{p-k} = a^p + b^p$.