Abstract Algebra: Theory and Applications

Let $\varphi :\mathbb{Z}\to F$ be the ring homomorphism defined by $\varphi (n)=n\cdot 1\text{.}$ Since the characteristic of $F$ is $p\text{,}$ the kernel of $\varphi$ must be $p\mathbb{Z}$ and the image of $\varphi$ must be a subfield of $F$ isomorphic to ${\mathbb{Z}}_p\text{.}$ We will denote this subfield by $K\text{.}$ Since $F$ is a finite field, it must be a finite extension of $K$ and, therefore, an algebraic extension of $K\text{.}$ Suppose that $[F:K]=n$ is the dimension of $F\text{,}$ where $F$ is a $K$ vector space. There must exist elements ${\alpha }_1,\dots ,{\alpha }_n\in F$ such that any element $\alpha$ in $F$ can be written uniquely in the form

where the $a_i$’s are in $K\text{.}$ Since there are $p$ elements in $K\text{,}$ there are $p^n$ possible linear combinations of the ${\alpha }_i$’s. Therefore, the order of $F$ must be $p^n\text{.}$

We will prove this lemma using mathematical induction on $n\text{.}$ We can use the binomial formula (see Chapter 2, Example 2.4) to verify the case for $n=1\text{;}$ that is,

If $0<k<p\text{,}$ then

must be divisible by $p\text{,}$ since $p$ cannot divide $k!(p-k)!\text{.}$ Note that $D$ is an integral domain of characteristic $p\text{,}$ so all but the first and last terms in the sum must be zero. Therefore, $(a+b{)}^p=a^p+b^p\text{.}$

Now suppose that the result holds for all $k\text{,}$ where $1\le k\le n\text{.}$ By the induction hypothesis,

Therefore, the lemma is true for $n+1$ and the proof is complete.

Let $f(x)$ be separable. Then $f(x)$ factors over some extension field of $F$ as $f(x)=(x-{\alpha }_1)(x-{\alpha }_2)\cdots (x-{\alpha }_n)\text{,}$ where ${\alpha }_i\ne {\alpha }_j$ for $i\ne j\text{.}$ Taking the derivative of $f(x)\text{,}$ we see that

Hence, $f(x)$ and $f^{\prime }(x)$ can have no common factors.

To prove the converse, we will show that the contrapositive of the statement is true. Suppose that $f(x)=(x-\alpha {)}^{k}g(x)\text{,}$ where $k>1\text{.}$ Differentiating, we have

Therefore, $f(x)$ and $f^{\prime }(x)$ have a common factor.

Let $f(x)=x^{p^n}-x$ and let $F$ be the splitting field of $f(x)\text{.}$ Then by Lemma 22.5, $f(x)$ has $p^n$ distinct zeros in $F\text{,}$ since $f^{\prime }(x)=p^n{x}^{p^n-1}-1=-1$ is relatively prime to $f(x)\text{.}$ We claim that the roots of $f(x)$ form a subfield of $F\text{.}$ Certainly 0 and 1 are zeros of $f(x)\text{.}$ If $\alpha$ and $\beta$ are zeros of $f(x)\text{,}$ then $\alpha +\beta$ and $\alpha \beta$ are also zeros of $f(x)\text{,}$ since ${\alpha }^{p^n}+{\beta }^{p^n}=(\alpha +\beta {)}^{p^n}$ and ${\alpha }^{p^n}{\beta }^{p^n}=(\alpha \beta {)}^{p^n}\text{.}$ We also need to show that the additive inverse and the multiplicative inverse of each root of $f(x)$ are roots of $f(x)\text{.}$ For any zero $\alpha$ of $f(x)\text{,}$ we know that $-\alpha$ is also a zero of $f(x)\text{,}$ since

provided $p$ is odd. If $p=2\text{,}$ then

If $\alpha \ne 0\text{,}$ then $({\alpha }^{-1}{)}^{p^n}=({\alpha }^{p^n}{)}^{-1}={\alpha }^{-1}\text{.}$ Since the zeros of $f(x)$ form a subfield of $F$ and $f(x)$ splits in this subfield, the subfield must be all of $F\text{.}$

Let $G$ be a finite subgroup of $F^{\ast }$ of order $n\text{.}$ By the Fundamental Theorem of Finite Abelian Groups (Theorem 13.4),

where $n=p_1^{e_1}\cdots p_k^{e_k}$ and the $p_1,\dots ,p_k$ are (not necessarily distinct) primes. Let $m$ be the least common multiple of $p_1^{e_1},\dots ,p_k^{e_k}\text{.}$ Then $G$ contains an element of order $m\text{.}$ Since every $\alpha$ in $G$ satisfies $x^r-1$ for some $r$ dividing $m\text{,}$ $\alpha$ must also be a root of $x^m-1\text{.}$ Since $x^m-1$ has at most $m$ roots in $F\text{,}$ $n\le m\text{.}$ On the other hand, we know that $m\le |G|\text{;}$ therefore, $m=n\text{.}$ Thus, $G$ contains an element of order $n$ and must be cyclic.