Exercise 14.2.13

Proof. (a) Since $F$ has characteristic different from 2 or 3, $f^{\prime }=6x^5+3b{x}^2\ne 0$. By hypothesis, $f$ is irreducible, thus any factor of $f$ is associate to $f$ or $1$. Since $f^{\prime }\ne 0$, $\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,f^{\prime })$ divides $f^{\prime }$, thus $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,f^{\prime }))\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f^{\prime })=5$, and $\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,f^{\prime })$ is a factor of $f$, which cannot be associate to $f$, therefore $\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,f^{\prime })=1$. This proves that $f$ is separable. (b) If $\alpha$ is a root of $f$ in $L$, then $\lambda ={\alpha }^3\in L$ is a root of $g=x^2+\mathit{bx}+c$. Let $\mu =-b-\lambda \in L$. Then $\mu$ is a root of $g$:

Moreover $\lambda \ne \mu$, otherwise $b=-2\lambda ,c=-{\lambda }^2-\mathit{b\lambda }={\lambda }^2$, and $g={(x-\lambda )}^2$, where $\lambda =-b∕2\in F$, so that $f=g(x^3)={(x^3-\lambda )}^2$ would not be irreducible over $f$. Therefore

Write $K=F(\lambda ,\mu )$. Then $F\subset K\subset L$ is an intermediate field which is a splitting field of $g$ over $F$. If $\lambda \in F$, then $\mu \in F$ and $f=g(x^3)=(x^3-\lambda )(x^3-\mu )$ would not be irreducible over $F$. Therefore $\lambda \notin F,\mu \notin F$, $g$ is irreducible over $F$. $K$ is the splitting field of the separable polynomial $g$, thus $F\subset K$ is a Galois extension, where $F⊊K\subset L$.

Moreover,

where $f_1=x^3-\lambda ,f_2=x^3-\mu \in K[x]$. Therefore three roots $\alpha ,{\alpha }^{\prime },{\alpha }^{″}$ of $f$ are the roots of $x^3-\lambda$, and three other roots $\beta ,{\beta }^{\prime },{\beta }^{″}$ of $f$ are the roots of $x^3-\mu$ :

This gives the blocks

If $\sigma \in \mathrm{Gal}(L∕F)$, then $g=\sigma \cdot g=(x-\sigma (\lambda ))(x-\sigma (\mu ))$, thus $\{\lambda ,\mu \}=\{\sigma (\lambda ),\sigma (\mu )\}$: $\sigma$ fixes $\lambda ,\mu$, or exchanges $\lambda ,\mu$. Since $\alpha ,\beta ,\gamma$ are the roots of $x^3-\lambda$, $\sigma (\alpha ),\sigma (\beta ),\sigma (\gamma )$ are the roots of $x^3-\sigma (\lambda )$, thus $\sigma (R_1)=R_1$ or $\sigma (R_1)=R_2$, and similarly $\sigma (R_2)=R_1$ or $R_2$. This proves that $f$ is imprimitive, with blocks $R_1,R_2$, and $\mathrm{Gal}(L∕F)$ is isomorphic to a subgroup of $S_2\wr S_3$(Corollary 14.2.10), whose order is $2{(3!)}^2=72$ (see Ex. 6 (c)). (c) We don’t know if $F$ or $K$ contains the cubic roots of unity, but $L$ does: since $\alpha ,{\alpha }^{\prime }$ are two distinct roots of $x^3-\lambda$ (where $\alpha \ne 0$, otherwise $\lambda =0\in F$), then ${(\frac{{\alpha }^{\prime }}{\alpha })}^3=1$. If we write $\omega =\frac{{\alpha }^{\prime }}{\alpha }$, then $\omega \in L$ and ${\omega }^3=1,\omega \ne 1$, thus $1,\omega ,{\omega }^2$ are three distinct roots of $x^3-1$, and $1+\omega +{\omega }^2=0$, so that $[K(\omega ):K]=1$ or $2$.

Then the six roots of $f$ are

Therefore

Consider the chain of fields

$\text{∙}$ Since $\lambda ,\mu$ are the roots of the irreducible polynomial $f$, $K=F(\lambda ,\mu )$ is a quadratic extension of $F$, so $[K:F]=2$.

$\text{∙}$ Since ${\omega }^2+\omega +1=0$, $[K(\omega ):K]\le 2$.

$\text{∙}$ Since ${\alpha }^3-\lambda =0$, where $\lambda \in K\subset K(\omega )$, $[K(\omega ,\alpha ):K(\omega )]\le 3$.

$\text{∙}$ Since ${\beta }^3-\mu =0$, where $\mu \in K\subset K(\omega ,\alpha )$, $[K(\omega ,\alpha ,\beta ):K(\omega ,\alpha )]\le 3$.

By the Tower Theorem,

Therefore,

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