Exercise 5.3.7

Proof. Let $F$ be a field of characteristic $p$, and $f\in F[x]$ irreducible over $F$, with degree $n\ge 1$.

(a)

We suppose first that $f^{\prime }\ne 0$.

Let $d=f\wedge f^{\prime }$. Then $d$ divides $f$ (and $d$ is monic), and $f$ is irreducible over $F$, thus $d=1$ or $d=\mathit{\lambda f},\lambda \in F^{\text{∗}}$. If $d=\mathit{\lambda f}$, then $f={\lambda }^{-1}d$ divides $f^{\prime }$. As $f^{\prime }\ne 0$, $f^{\prime }=\mathit{qf}$ implies that $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=n\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f^{\prime })\le n-1$ : this is a contradiction.

Thus $d=f\wedge f^{\prime }=1$, and then Proposition 5.3.2 shows that $f$ is separable.

(b)

Suppose now that $f^{\prime }=0$, where $f=\underset{i=0}{\overset{n}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_i{x}^i$. Then $0=f^{\prime }=\underset{i=1}{\overset{n}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}i{a}_i{x}^{i-1}$, and consequently $i{a}_i=0,i=1,\cdots ,n$. If $p\nmid i,a_i=0$, thus $$f=\underset{0\le i\le n,p\mid i}{\sum }{a}_i{x}^i=\underset{k=0}{\overset{\lfloor n∕p\rfloor }{\sum }}{a}_{\mathit{kp}}{x}^{\mathit{kp}}.$$

If we write $g_1=\underset{k=0}{\overset{\lfloor n∕p\rfloor }{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{a}_{\mathit{kp}}{x}^k$, then $f=g_1(x^p)$.

(c)

If $g_1$ was reducible, $$g_1=\mathit{uv},g_1,g_2\in F[x],1\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u),1\le \mathit{deg}(v).$$

But then $f=g_1(x^p)=u(x^p)v(x^p)$, where $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u(x^p))=p\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u)\ge 1,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(v(x^p))\ge 1$, and so $f$ would be reducible, which contradicts the hypothesis on $f$.

(d)

If $g_1^{\prime }\ne 0$, part (a) shows that $g_1$ is separable, and then the wanted conclusion is obtained with $e=1$. Otherwise the arguments of parts (b) and (c) shows that there exits an irreducible polynomial $g_2\in F[x]$ such that $g_1=g_2(x^p)$, so $f=g_2(x^{p^2})$, and so on. While $g_i^{\prime }\ne 0$, we can build a sequence $g_1,\cdots ,g_k$ such that $g_i=g_{i+1}(x^p)$, where $g_i$ irreducible over $F$.

This sequence is necessarily finite, since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g_{i+1})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g_i)∕p<\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g_i)$.

Thus there exists an integer $e\ge 1$ such that $f=g_1(x^p),g_1=g_2(x^p),\cdots ,g_{e-1}=g_e(x^p)$, and $g_e^{\prime }\ne 0$, and so $g=g_e$ is separable.

If we take the induction hypothesis $f=g_k(x^{p^k})$, for $k<e$, (verified for $k=1$) then $f=g_{k+1}({(x^{p^k})}^p)=g_{k+1}(x^{p\cdot p^k})=g_{k+1}(x^{p^{k+1}})$.

Hence $f=g_e(x^{p^e})=g(x^{p^e})$.

Conclusion: if $F$ is a field of characteristic $p$, and if $f\in F[x]$ is irreducible over $F$, then there exists an integer $e\ge 1$ and an irreducible separable polynomial $g\in k[x]$ such that $f=g(x^{p^e})$.