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Morandi's Field and Galois Theorey, exercise 10.5c

Let $p$ be a prime, and suppose that either $F$ contains a primitive $p$th root of unity for $p$ odd, or that $F$ contains a primitive fourth root for $p=2$. If there is an $a\in F$ with $x^p-a$ irreducible over $F$, then $x^{p^2}-a$ is irreducible over $F$. (Hint: Use a norm argument)

My efforts: suppose $\alpha^{p^2}-a=0$, and let $K=F(\alpha),\beta=\alpha^p,L=F(\beta)$ Since $x^p-a$ is irreducible, we have $[L:F]=p$. If $[K:L]=p$, we have $[K:F]=p^2$ and therefore $x^{p^2}-a$ is irreducible, so we only need to show that $[K:L]=p$, which is equivalent to the irreducibility of $x^p-\beta$. Suppose not, since a $p$th primitive root is contained in $L$, we have $x^p-\beta$ splits in $L$, which means that $\alpha\in L$, therefore $\alpha$ is an $F$-polynomial of $\beta$.

I don't know how to proceed. Any help? Thanks!

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Yai0Phah
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  • To show that $x^p-\beta$ is irreducible, it suffices to show it has no root by a common theorem (see here). So, suppose that $x^p-\beta$ had a root $\gamma$ in $F(\alpha)$. Then, $\gamma^p=\beta$. Note then that $N_{L/F}(\beta)^p=N_{L/F}(\alpha^{p^2})=N_{L/F}(a)=a^p$. Thus, since $F$ contains the $p^{\text{th}}$ roots of unity, we see that $N_{L/F}(\beta)=\zeta a$ for some root of unity $\zeta$. Thus, $N_{L/F}(\gamma)^p=N_{L/F}(\beta)=\zeta a$. Thus, $(\zeta N_{L/F}(\gamma))^p=a$. Since $\zeta N_{L/F}(\gamma)\in F$, this is a contradiction. Does that make sense to you? – Alex Youcis Sep 30 '13 at 01:03
  • Forgot link: http://math.stackexchange.com/questions/266171/irreducibility-of-polynomial-if-no-root-capelli – Alex Youcis Sep 30 '13 at 01:05
  • @AlexYoucis There's a slight mistake. Judging from your argument, $(\zeta N_{L/F}(\gamma))^p$ should be $\zeta a$, not $a$. It could be corrected as follows: We could claim that $N_{L/F}(\beta)=(-1)^{p+1}a$, since the minimal polynomial of the linear map $x\mapsto\beta x$ over $F$ is $x^p-a$, so is the characteristic polynomial. – Yai0Phah Oct 02 '13 at 14:29
  • @AlexYoucis Could you please post your comment as an answer therefore I can accept it? – Yai0Phah Oct 02 '13 at 14:31
  • Frank, I will tonight. Have to proctor an exam now :) – Alex Youcis Oct 02 '13 at 14:32

1 Answers1

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Fix some algebraic closure $\overline{F}$ of $F$. Let $\alpha\in\overline{F}$ be a root of $T^{p^2}-a$. Then, $T^{p^2}-a$ is irreducible if and only if $[F(\alpha):F]=p^2$. But, let $\beta:=\alpha^p$. Note that since $\beta^p-a=0$, and by assumption $T^p-a$ is irreducible, we have that $[F(\beta):F]=p$. Thus, we see that $[F(\alpha):F]=p^2$ if and only if $[F(\alpha):F(\beta)]=p$.

But, to prove this it suffices to show that $T^p-\beta$ is irreducible in $F(\beta)[x]$. But, by Capelli's theorem, it suffices to show that $T^p-\beta$ has no root in $F(\beta)$. So, suppose that $\gamma\in F(\beta)$ is such that $\gamma^p=\beta$. Note then that

$$N_{F(\beta)/F}(\gamma)^p=N_{F(\beta)/F}(\gamma^p)=N_{F(\beta)/F}(\beta)$$

But, since the minimal polynomial of $\beta$ is $T^p-a$, we know that $N_{F(\beta)/F}$ is $(-1)^{p+1}a$.

Thus, if $p$ is odd, then $N_{F(\beta)/F}(\beta)=a$, and thus $N_{F(\beta)/F}(\gamma)$ is a root of $T^p-a$ in $F$, which is a contradiction.

If $p=2$, by assumption $\pm i\in F$, and so evidently $i\cdot N_{F(\beta)/F}(\gamma)$ and

$$(i N_{F(\beta)/F}(\gamma))^2=-1 (-1)^3 a=a$$

and so $i N_{F(\beta)/F}(\gamma)$ is a root of $T^p-a$ in $F$, which is also a contradiction.

I'm not sure why we needed $F$ to contain $\zeta_p$ for $p$ odd? Can anyone see an issue with the above?

Alex Youcis
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