It is Albert "Modern Higher Algebra", chapter 7, section 9, exercises 5 and 6.
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Let $K$ be a field of degree $n$ over $F$ of characteristic $p$ such that every quantity $\alpha$ of $K$ is a root of a corresponding equation $x^{p^t}=a$ in $F$. Prove that $n$ is a power of $p$ and that there exists an integer $e$, called the exponent of $K$ over $F$, such that $k^{p^e}$ is in $F$ for every $k\in K$, and $e$ is the least integer with this property.
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Let $Z>K>F$ where $K$ is as in Ex. 5 and $Z=K(\xi)$ is a simple separable extension of degree $n$ over $F$. Prove that $Z=K\times F(\xi_0)$ where $\xi_0=\xi^{p^e}$ and $e$ is the exponent of $K$ over $F$. Prove in fact that $Z=K(\xi)$ and $Z_0=F_0(\xi_0)$ are equivalent (but not necessarily equivalent over $F$) under the correspondence $a\longleftrightarrow a^{p^e}$ for every $a$ of $Z$, where $F_0$ is a certain subfield of $F$.
Question
I do not know how to prove that $Z$ is generated by $K$ and $F(\xi_0)$, or to prove that $\xi\in K(\xi_0)$, and I do not know how to use the isomorphism between $Z$ and $Z_0$ to conclude that $[Z:F]=[K:F][F(\xi_0):F]$.
Notation
If $K$ is finite field extension of $F$ and $A$, $B$ are subfields of $K$ containing $F$, we say $K=A\times B$ if $K$ is the subfield of $K$ generated by $A\cup B$ and $[K:F]=[A:F][B:F]$.
My attempt
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Induction on $[K:F]=n$. If $k\in K$, $k\notin F$, then let $u$ be the least such that $k^{p^u}\in F$, so let $a=k^{p^u}$, then the polynomial $x^{p^u}-a$ is irreducible in $F[x]$. Indeed, if $x^{p^u}-a=g(x)h(x)$ where $g(x),h(x)\in F[x]$, then, in $K[x]$: $$g(x)h(x)=x^{p^u}-a=x^{p^u}-k^{p^u}=(x-k)^{p^u},$$ so $g(x)=(x-k)^s$ where $s\leq p^u$, but $g(x)\in F[x]$, so $k^s\in F$, so, if $d=\mathrm{gcd}(s,p^u)$, then $d=p^v$ for some $v\leq u$, and $k^d\in F$, so $k^{p^v}\in F$, so $u\leq v$, so $p^u=d\mid s$, so $p^u\leq s$, so $s=p^u$, and therefore: $$g(x)=(x-k)^{p^u}=x^{p^u}-a.$$ Therefore $[F(k):F]=p^u$. The hypothesis is certainly valid for $K$ over $F(k)$ and $[K:F(k)]=n/p^u$, so, by the induction hypothesis, $n/p^u$ is a power of $p$, so that $n$ is also a power of $p$, and there is an $f$ such that $l^{p^f}\in F(k)$ for every $l\in K$. Also, for every $m\in K$, then there are $a_i\in F$, $i=0,\dots,p^u-1$, such that $m=\sum a_ik^i$, so: $$m^{p^u}=\left(\sum a_ik^i\right)^{p^u}=\sum a_i^{p^u}(k^{p^u})^i\in F.$$ Therefore for every $l\in K$ we have $l^{p^f}\in F(k)$, so $l^{p^{f+u}}=(l^{p^f})^{p^u}\in F$.
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Because the characteristic of $F$ is $p$, it is easy to see that: $$F_0=\{k^{p^e}:k\in K\}$$ is a subfield of $F$ and $a\longrightarrow a^{p^e}$ is an injective homomorphism from $Z$ to $Z_0$. Also, for every $b\in Z_0$, there are $b_i\in F_0$ such that $b=\sum b_i\xi_0^i$, so there are $a_i\in K$ such that $b_i=a_i^{p^e}$, so: $$b=\sum b_i\xi_0^i=\sum a_i\xi^{p^e\cdot i}=\left(\sum a_i \xi^i\right)^{p^e}.$$ For $Z=K\times F(\xi_0)$, I have to show that: (a) $Z$ is generated by $K$ and $F(\xi_0)$; (b) $[Z:F]=[K:F][F(\xi_0):F]$. The item (a) is equivalent to prove that $\xi\in K(\xi_0)$, and I do not know how to do this. The item (b) is equivalent to $[Z:K]=[F(\xi_0):F]$, and, by the isomorphism between $Z$ and $Z_0$, this is equivalent to $[Z_0,F_0]=[F(\xi_0):F]$, and this is equivalent to prove that the minimal polynomial $P_0(x)$ of $\xi_0$ in $F_0$ has the same degree as the minimal polynomial $P(x)$ of $\xi_0$ in $F$. Obviously $\mathrm{deg}P(x)\leq\mathrm{deg}P_0(x)$. But I do not know how to prove the other inequality.
Best Answer
$K/F$ is a finite extension. For $\alpha \in K$ we are told that $\alpha^{p^t}\in F$ for some $t$, thus $\alpha$ is a root of $x^{p^t}-\alpha^{p^t}=(x-\alpha)^{p^t}$, its $F$-minimal polynomial must be of the form $h(x)=(x-\alpha)^m$, $(-1)^m h(0) = \alpha^m$ is in $F$ so $\alpha^{\gcd(m,p^t)}=\alpha^{p^r}$ is in $F$, repeating we get that the minimal polynomial is of the form $x^{p^t}-\alpha^{p^E}$ where $E$ is the least integer such that $\alpha^{p^E} \in F$.
Since $K/F$ is finitely generated there is a least $e$ such that the $p^e$-th powers of the generators are in $F$ and since $\beta\to \beta^{p^e}$ is a ring homomorphism then that $e$ is the least $e$ such that $K^{p^e}\subset F$.
Given $Z/K$ finite separable, thus $Z=K(\xi)$. With $g(x) = \sum_{n=0}^N a_n x^n$ the $K$-minimal polynomial of $\xi$ which is separable (coprime with $g'$) we get that $g^{p^e}(x)=\sum_{n=0}^N a_n^{p^e} x^n$ is separable, in $F[x]$, and $g^{p^e}(\xi^{p^e}) = g(\xi)^{p^e}=0$. Thus $F(\xi^{p^e})/F$ is separable.
For each $K$-conjugate $\sigma(\xi)$ distinct from $\xi$ we have that $\sigma(\xi^{p^e})$ is distinct from $\xi^{p^e}$. Thus $\xi$ and $\xi^{p^e}$ have the same number of $K$-conjugates, their $K$-minimal polynomials have the same degree, ie. $[K(\xi):K]=[K(\xi^{p^e}):K]$ and $K(\xi)=K(\xi^{p^e})$.
We also obtained that $[K(\xi^{p^e}):F]=[F(\xi^{p^e}):F][K:F]$ thus $K(\xi^{p^e}) = F(\xi^{p^e})\otimes_F K$.
The extensions $K(\xi)/K$ and $K^{p^e}(\xi^{p^e})/K^{p^e}$ are isomorphic.