Prove $x^{p^k}-a$ is irreducible in a field $K$ of characteristic $p>0$ for all $k \geq 0$, where $a$ is not a $p^{th}$ power in $K$.

Question

I've been trying to prove the above by induction on $k$. I think I've proved it for $k=1$ (I put the proof below for completeness), but I don't see to go from assuming that $f(x)=x^{p^n}-a$ is irreducible to showing that $f(x)=x^{p^{n+1}}-a$ is also irreducible. I'd appreciate any hint towards this. Thanks in advance.

Let $\alpha \in \bar{K}$(the algebraic closure of $K$) be a root of $f(x)=x^p-a$. Then $\alpha$ is the unique $p^{th}$ root of $a$ in $\bar{K}$. Assuming $f$ reducible, we can express $f$ as a product of irreducible polynomials. The element $\alpha$ is a root of each factor. As these irreducible factors are also monic, they must be the minimum polynomial of $\alpha$ over $K$, namely, $f(x)=(f^{\alpha}_K)^n$ in $K[x]$ for some $n \leq p$.

Let $f^{\alpha}_K(x)=x^d+k_{d-1}x^{d-1}+...+k_0$ where $d \leq p$ and $k_{d-1},...,k_0 \in K$. Then $x^p-\alpha = (f^{\alpha}_K)^n = (x^d+k_{d-1}x^{d-1}+...+k_0)^n = x^{dn} +$ other terms in lower powers of $x$. Hence $nd=p$, so either $n=1$ and $d=p$ or $n=p$ and $d=1$.

If $n=p$, by the corresponding condition on $d$, $f^{\alpha}_K(x)=x+k$ for some $k \in K$. Thus \begin{align*} (f^{\alpha}_K(x))^n &= (f^{\alpha}_K(x))^p \\ &=(x+k_0)^p \\ &=x^p+(k_0)^p \end{align*} where we used that the Frobenius map is a field homomorphism in fields of characteristic $p$. The equality $f^{\alpha}_K(\alpha)=0$ implies $\alpha^p=-(k_0)^p$, but this contradicts that $a$ isn't a $p^{th}$ power in $K$. If $n=1$, $f$ is equal to an irreducible polynomial and so must be irreducible itself.

Answer 1

Over some larger field, $X^{p^k}-a=(X-b)^{p^k}$. So if $X^{p^k}-a$ factors over $K$ it has a factor of the form $(X-b)^r$ in $K$ for some $r$ with $1\le r\le p^k-1$. That entails $b^r\in K$. But as $b^{p^k}\in K$ we get $b^g\in K$ for $g=\gcd(r,p^k)$. Then $g=p^l$ where $0\le l\le k-1$, and from $b^{p^l}\in K$ we deduce $b^{p^{k-1}}\in K$. That's a $p$-th root of $a$.

Answer 2

By hypothesis, $a \notin K^p$ and I am going to prove by induction on $k$. The problem is trivial in case $k=0$ so let assume that it holds true for $k-1$ and we want to prove $f_k(x)=x^{p^k}-a$ is irreducible. Let $\phi(x)$ be a monic irreducible factor of $f_n(x)=x^{p^k}-a$ in $K[x]$ (recall that polynomial ring over a field is UFD) and $\phi^h$ be the highest power of $\phi$ in the factorization of $f_k$. Therefore, we can write $$f_k = \phi^h \varphi \ \text{in which} \ \mathrm{gcd}(\phi,\varphi)=1$$ taking the derivatives of both sides we obtain $$h\phi'\varphi + \phi \varphi'=0$$ after dividing by $\phi^{h-1}$. Since $\mathrm{gcd}(\phi,\varphi)=1$ the above implies that $\varphi'=0$ and hence it forces $h\phi'=0$ so we can write $\varphi(x) = \varphi_1(x^p),(\phi^h(x))=\phi_1(x^p)$ for some $\varphi_1,\phi_1 \in K[x]$. Rewrite the equality $f_k = \phi^h \varphi$ $$x^{p^k}-a = \phi_1(x^p)\varphi_1(x^p)$$ and because $K$ has characteristic $p>0$, the morphism $x \mapsto x^p$ is injective so by replacing $x^p$ with $x$ we deduce that $f_{k-1}(x)=x^{p^{k-1}}-a = \phi_1(x)\varphi_1(x)$. By our induction hypothesis, $f_{k-1}$ is irreducible so $\varphi_1(x) \equiv 1$ because both of $f_{k-1},\phi_1$ are monic. Thus, $$f_k(x) = x^{p^k}-a = (\phi(x))^h$$ Now if $h$ is a multiple of $p$ then $(\phi(x))^h \in K^p[x]$ and therefore $f_k(x) \in K^p[x]$ but this is a contradiction since $a \notin K^p$. Hence $h$ is not a multiple of $p$ and this implies $\phi'(x)=0 \Rightarrow \phi(x) \in K^p[x]$. By setting $\phi(x)=\phi_2(x^p)$ we find out that $$f_{k-1} = \phi_2^h \Rightarrow h=1$$ since $f_{k-1}$ is irreducible. Finally, we conclude that $f_k = \phi$ which is of course irreducible by our assumption.