Exercise 13.2.16

Proof.

(a)

If $a=0$, then $f=x^5+b$, so $f=x^5-{\alpha }^5$, where $\alpha$ is a root of $f$ in some extension of $F$. Since the characteristic of $F$ is 5, $f=x^5-{\alpha }^5={(x-\alpha )}^5$ is not separable, in contradiction with the hypothesis, so $a\ne 0$.

(b)

If $a=-1$, by Exercise 5.3.16 and 6.2.5, we know that $$x^5-x+b=(x-\alpha )(x-\alpha -1)(x-\alpha -2)(x-\alpha -3)(x-\alpha -4),$$

where $\alpha$ is a root of $f$ in some extension. Then $K=F(\alpha )$ is the splitting field of $f$ over $F$. By part (c) of exercise 6.2.5, we know also that

$$\phi \{\begin{array}{ccc}\hfill \mathrm{Gal}(L∕F)\hfill & \hfill \to \hfill & \hfill \mathbb{Z}∕5\mathbb{Z}\hfill \\ \hfill \sigma \hfill & \hfill \mapsto \hfill & \hfill \sigma (\alpha )-\alpha \hfill \end{array}$$

is a group isomorphism, so $\mathrm{Gal}(L∕F)$ is cyclic of order 5.

(c)

We search $\lambda$ such that $$x^5-x+b^{\prime }={\lambda }^{-5}({(\mathit{\lambda x})}^5+a(\mathit{\lambda x})+b)$$

for some $b^{\prime }$.

This is equivalent to

$${\lambda }^5{x}^5-{\lambda }^{5}x+{\lambda }^5{b}^{\prime }={\lambda }^5{x}^5+\mathit{a\lambda x}+b,$$

so $\mathit{a\lambda }=-{\lambda }^5,{\lambda }^4=-a$. Let $L$ a splitting field of $x^4+a$, and choose

$$\lambda =\sqrt[4]{-a}$$

a fixed root of $x^4+a$ in $L$.

The characteristic is 5, so $2^2=-1$, and

$$x^4+a=x^4-{\lambda }^4=(x^2+{\lambda }^2)(x^2-{\lambda }^2)=(x+2\lambda )(x-2\lambda )(x-\lambda )(x+\lambda )$$

splits completely over $F$. Since $\lambda \ne 0$, and since the characteristic is 5, the roots of $x^4+a$ are distinct. Therefore $L=F(\lambda )$ is the splitting field of the separable polynomial $x^4+a$ over $F$. Hence $F\subset L=F(\lambda )$ is a Galois extension.

So there exists $\lambda$ in some solvable Galois extension $L$ of $F$ such that $x^5-x+b^{\prime }={\lambda }^{-5}({(\mathit{\lambda x})}^5+a(\mathit{\lambda x})+b)$ with $b^{\prime }={(\sqrt[4]{-a})}^{-5}b$, where $\lambda ,b^{\prime }\in L$.

(d)

If $\beta$ is in some extension of $F$, $\beta$ is a root of $f$ if and only if ${\lambda }^{-1}\beta$ is a root of $x^5-x+b^{\prime }$. If $\alpha$ is a fixed root of $x^5-x+b^{\prime }$, then by part (b) the roots of $x^5-x+b^{\prime }$ are $\alpha ,\alpha +1,\alpha +2,\alpha +3,\alpha +4$, so the roots of $f$ are $${\beta }_0=\mathit{\lambda \alpha },{\beta }_1=\lambda (\alpha +1),{\beta }_2=\lambda (\alpha +2),{\beta }_3=\lambda (\alpha +3),{\beta }_4=\lambda (\alpha +4).$$

A splitting field of $f$ over $F$ is

$$K=F({\beta }_0,\dots ,{\beta }_4)=F(\mathit{\lambda \alpha },\lambda (\alpha +1),\dots ,\lambda (\alpha +4)).$$

Since $\lambda =\lambda (\alpha +1)-\mathit{\lambda \alpha }={\beta }_1-{\beta }_0\in K$, and $\alpha =(\mathit{\lambda \alpha })∕\lambda ={\beta }_0∕({\beta }_1-{\beta }_0)\in K$, $F(\lambda ,\alpha )\subset K$, and $\mathit{\lambda \alpha },\dots ,\lambda (\alpha +4)\in F(\lambda ,\alpha )$, so $K\subset F(\lambda ,\alpha )$:

$$K=F(\lambda ,\alpha )$$

is the splitting field of $f=x^5+\mathit{ax}+b$ over $F$.

Since $F(\lambda )=L\subset K$, $K=L(\alpha )$.

Since $f$ is irreducible over $F$, $5=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=[F({\beta }_0):F]\mid [K:F]$.

$\mathrm{Gal}(L∕F)$ is isomorphic to a subgroup of $S_4$, so $5\nmid [L:F]=|\mathrm{Gal}(L∕F)|$.

Since $[K:F]=[K:L][L:F]$, $5\mid [K:L]=[L[\alpha ]:L]$, where ${\alpha }^5-\alpha +b^{\prime }=0$. Therefore $x^5-x+b^{\prime }$ is irreducible over $L$ and by part (b),

$$\mathrm{Gal}(K∕L)\simeq \mathbb{Z}∕5\mathbb{Z}\text{ is cyclic}.$$

Since $F\subset L$ is a Galois extension,

$$\mathrm{Gal}(K∕F)∕\mathrm{Gal}(K∕L)\simeq \mathrm{Gal}(L∕F).$$

$\mathrm{Gal}(L∕F)$ is isomorphic to a subgroup of $S_4$, so is solvable, and $\mathrm{Gal}(K∕L)$ is cyclic, a fortiori solvable. Therefore $\mathrm{Gal}(K∕F)$ is solvable:

The Galois group of $f$ over $F$ is solvable.

(e)

Let $F={𝔽}_5({\sigma }_1,\dots ,{\sigma }_5)\subset {𝔽}_5(x_1,\dots ,x_5)$. The Galois group of $f=x^5-{\sigma }_1{x}^4+{\sigma }_2{x}^3-{\sigma }_3{x}^2+{\sigma }_{4}x-{\sigma }_5$ is $S_5$, and $S_5$ is not solvable. Therefore $f$ cannot be equivalent to a polynomial $x^5+\mathit{ax}+b$ whose Galois group over $F$ is solvable.