Exercise 13.2.9

Proof.

(a)

Suppose that ${𝜃}_f(y)={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^6(y-{\beta }_i)$ is irreducible over $F$. Then ${𝜃}_f(y)$ is the minimal polynomial of ${\beta }_1=h({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)$ over $F$. Therefore $$[F({\beta }_1):F]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->{𝜃}_f(y)=6.$$

Since ${\beta }_1=h({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)\in F({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)=L$,

$$F\subset F({\beta }_1)\subset F({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)=L.$$

By the Tower Theorem,

$$[F({\beta }_1):F]\mid [L:F],$$

therefore

$$6\mid [L:F]=|\mathrm{Gal}(L∕F)|=|G|.$$

Since $6\nmid |\mathrm{AGL}(1,{𝔽}_5)|=20$, $G$ is not a subgroup of $\mathrm{AGL}(1,{𝔽}_5)$. By Theorem 13.2.2, since $G$ is a transitive subgroup of $S_5$, $G=A_5$ or $G=S_5$:

$$A_5\subset G.$$

(b)

Now suppose that $A_5\subset G$. Then $$G\supset \{{\tau }_1,\dots ,{\tau }_6\}=\{e,(123),(234),(345),(145),(125)\},$$

so ${\tau }_i\in G$ and the corresponding ${\sigma }_i$ are in $\mathrm{Gal}(L∕F)$. By Exercise 13.2.5, ${\sigma }_i({\beta }_1)={\beta }_i$, thus the orbit of ${\beta }_1$ under the action of $\mathrm{Gal}(L∕F)$ is ${\mathcal{}O}_{{\beta }_1}=\{{\beta }_1,\dots ,{\beta }_6\}$. This is sufficient to prove that $\mathrm{Gal}(L∕F)$ acts transitively on ${\beta }_1,\dots ,{\beta }_6$.

(c)

Let $p(y)$ be the minimal polynomial of ${\beta }_1$ over $F$. There exists ${\sigma }_i\in \mathrm{Gal}(L∕F)$ such that ${\sigma }_i({\beta }_1)={\beta }_i$, and since $p(y)\in F[y]$, $0={\sigma }_i(p({\beta }_1))=p({\sigma }_i({\beta }_1))=p({\beta }_i)$, so ${\beta }_i$ is a root of $p$, where $p$ is irreducible. Therefore $p$ is the minimal polynomial over $F$ of ${\beta }_1,\dots ,{\beta }_6$.

Under the hypothesis of Theorem 13.2.7 (and 13.2.6), $F\subset L$ is a separable extension, so ${\beta }_1$ is separable, therefore

$$p(x)=(x-{\gamma }_1)\cdots (x-{\gamma }_r),$$

where ${\gamma }_1,\dots ,{\gamma }_r$ are distinct. Since $p$ is the minimal polynomial over $F$ of ${\beta }_1,\dots ,{\beta }_6$, each ${\beta }_j$ is a ${\gamma }_i$ for some $i,1\le i\le r$, and since $p(y)$ divides ${𝜃}_f(y)$, each ${\gamma }_i$ is a ${\beta }_j$, so $\{{\gamma }_1,\dots ,{\gamma }_r\}=\{{\beta }_1,\dots ,{\beta }_6\}$, and ${\gamma }_1,\dots ,{\gamma }_r$ are the distinct roots of ${𝜃}_f(y)$.

Let $k_i$ the order of multiplicity of ${\beta }_i$ in ${𝜃}_f(y)$, so ${𝜃}_f(y)={(y-{\beta }_i)}^{k_i}{q}_i(y),q_i(y)\in L[y]$. Let $\sigma \in \mathrm{Gal}(L∕F)$ such that $\sigma ({\beta }_i)={\beta }_j$. Applying $\sigma$ to ${𝜃}_f(y)$, we obtain ${𝜃}_f(y)={(y-{\beta }_j)}^{k_i}(\sigma \cdot q_i)(y)$, so $k_j\ge k_i$, and similarly $k_i\ge k_j$, so the distinct ${\gamma }_i$ have the same order of multiplicity $m$ in ${𝜃}_f(y)$. Therefore

$${𝜃}_f(y)={(x-{\gamma }_1)}^m\cdots {(x-{\gamma }_r)}^m=p{(y)}^m.$$

Since $6=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({𝜃}_f(y))=m\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p(y))$, $m\mid 6$, so $m=1,2,3$ or $6$.

$m=6$ gives ${𝜃}_f(y)={(x-{\beta }_1)}^6$. Since the characteristic of $f$ is not 2, this is impossible by the proof of Theorem 13.2.6. It remains to prove the impossibility of $m=2$ or $m=3$.

(d)

If $m=2$, $${𝜃}_f(y)={(y^3+a{y}^2+\mathit{by}+c)}^2,a,b,c\in F.$$

By Proposition 13.2.5, this gives

$${(y^3+b_2{y}^2+b_{4}y+b_6)}^2-2^{10}\mathrm{\Delta }(f)y={(y^3+a{y}^2+\mathit{by}+c)}^2,$$

so

$$\begin{array}{llll}\hfill 2^{10}\mathrm{\Delta }(f)y& ={(y^3+b_2{y}^2+b_{4}y+b_6)}^2-{(y^3+a{y}^2+\mathit{by}+c)}^2& \hfill & \\ \hfill & =[(b_2-a)y^2+(b_4-b)y+(b_6-c)][2y^3+(b_2+a)y^2+(b_4+b)y+(b_6+c)]& \hfill & \end{array}$$

Therefore the coefficient in $y^5$ is $2(b_2-a)=0$. Since the characteristic is not 2,

$$b_2=a.$$

Using $b_2=a$, the coefficient in $y^4$ is $2(b_4-b)=0$, so

$$b_4=b,$$

and then the coefficient in $y^3$ is $2(b_6-c)$, so

$$b_6=c.$$

Therefore $2^{10}\mathrm{\Delta }(f)y=0$. Since the characteristic is not 2, $\mathrm{\Delta }(f)=0$, in contradiction with the assumed separability of $f$.

(e)

If $m=3$, there exist coefficients $a,b\in F$ such that $${𝜃}_f(y)={(y^2+\mathit{ay}+b)}^3={(y^3+b_2{y}^2+b_{4}y+b_6)}^2-2^{10}\mathrm{\Delta }(f)y.$$

$$\begin{array}{llll}\hfill 0={𝜃}_f(y)-{(y^2+\mathit{ax}+b)}^3& =-\left(3a-2b_2\right)y^5-\left(3a^2-b_2^2+3b-2b_4\right)y^4& \hfill & \\ \hfill & -\left(a^3+6\mathit{ab}-2b_2{b}_4-2b_6\right)y^3-\left(3a^{2}b+3b^2-b_4^2-2b_2{b}_6\right)y^2& \hfill & \\ \hfill & -\left(3a{b}^2-2b_4{b}_6+1024\mathrm{\Delta }(f)\right)y-b^3+b_6^2.& \hfill & \end{array}$$

We obtain $b_2,b_4,b_6$ with the equations corresponding to the coefficients of $y^5,y^4,y^3$:

$$\{\begin{array}{cc}0\hfill & =-3a+2b_2,\hfill \\ 0\hfill & =-3a^2+b_2^2-3b+2b_4,\hfill \\ 0\hfill & =-a^3-6\mathit{ab}+2b_2{b}_4+2b_6,\hfill \end{array}$$

which gives

$$b_2=\frac{3}{2}a,b_4=\frac{3}{8}{a}^2+\frac{3}{2}b,b_6=-\frac{1}{16}{a}^3+\frac{3}{4}\mathit{ab}.$$

If we substitute these values in the coefficient of $y^3$, we obtain

$$\begin{array}{llll}\hfill a^3+6\mathit{ab}-2b_2{b}_4-2b_6& =a^3+6\mathit{ab}-2\left(\frac{3}{2}a\right)\left(\frac{3}{8}{a}^2+\frac{3}{2}b\right)+2\left(\frac{1}{16}{a}^3+\frac{3}{4}\mathit{ab}\right)& \hfill & \\ \hfill & =0.& \hfill & \end{array}$$

The coefficient of $y^2$ gives $-\frac{3}{64}\left(a^4-8a^{2}b+16b^2\right)=-\frac{3}{64}{(a^2-4b)}^2=0$.

If we suppose that the characteristic is not $3$, then

$$a^2=4b.$$

The coefficient of $y$ gives

$$\begin{array}{llll}\hfill 0& =-\frac{3}{64}{a}^5+\frac{3}{8}{a}^{3}b-\frac{3}{4}a{b}^2-2^{10}\mathrm{\Delta }(f)& \hfill & \\ \hfill & =-\frac{3}{64}a{\left(a^2-4b\right)}^2-2^{10}\mathrm{\Delta }(f)& \hfill & \\ \hfill & =-2^{10}\mathrm{\Delta }(f)& \hfill & \end{array}$$

Therefore

$$\mathrm{\Delta }(f)=0.$$

Since $f$ is separable, this is a contradiction, so ${𝜃}_f(y)$ is irreducible.

It remains the case where the characteristic is 3. Then the equation

$${𝜃}_f(y)={(y^2+\mathit{ay}+b)}^3={(y^3+b_2{y}^2+b_{4}y+b_6)}^2-2^{10}\mathrm{\Delta }(f)y$$

gives the system

$$\{\begin{array}{cc}0\hfill & =2b_2,\hfill \\ 0\hfill & =b_2^2+2b_4,\hfill \\ 0\hfill & =-a^3+2b_2{b}_4+2b_6,\hfill \end{array}$$

Therefore $b_2=b_4=0$, so the initial equation gives

$$y^6+a^3{y}^3+b^3=y^6+2b_6{y}^3-2^{10}\mathrm{\Delta }(f)y+b_6^2,$$

and we have the same contradiction $\mathrm{\Delta }(f)=0$, and the same conclusion:

${𝜃}_f(y)$ is irreducible over $F$.

We give here the corresponding Sage instructions:

     y,b2,b4,b6,Delta,a,b,c = var(’y,b2,b4,b6,Delta,a,b,c’)
     u = (y^3+b2*y^2+b4*y+b6)^2 - 2^10*Delta*y - (y^2+a*y+b)^3
     u = u.expand().collect(y); u

$$-\left(3a-2b_2\right)y^5-\left(3a^2-b_2^2+3b-2b_4\right)y^4-\left(a^3+6\mathit{ab}-2b_2{b}_4-2b_6\right)y^3-b^3$$

$$-\left(3a^{2}b+3b^2-b_4^2-2b_2{b}_6\right)y^2+b_6^2-\left(3a{b}^2-2b_4{b}_6+1024\mathrm{\Delta }\right)y$$

     eq = [u.coefficient(y^i) for i in range(3,6)]
     solve(eq,b2,b4,b6)

$$\left[\left[b_2=\frac{3}{2}a,b_4=\frac{3}{8}{a}^2+\frac{3}{2}b,b_6=-\frac{1}{16}{a}^3+\frac{3}{4}\mathit{ab}\right]\right]$$

     v = u.coefficient(y^3)
     w = v.subs(b2 == 3/2*a, b4 == 3/8*a^2 + 3/2*b, b6 == -1/16*a^3 + 3/4*a*b)
     w.expand()

$$0$$

     s = u.coefficient(y^2)
     t = s.subs(b2 == 3/2*a, b4 == 3/8*a^2 + 3/2*b, b6 == -1/16*a^3 + 3/4*a*b)
     t.expand().factor()

$$-\frac{3}{64}{\left(a^2-4b\right)}^2$$

     p = u.coefficient(y)
     q = p.subs(b2 == 3/2*a, b4 == 3/8*a^2 + 3/2*b, b6 == -1/16*a^3 + 3/4*a*b)
     q.expand()

$$-\frac{3}{64}{a}^5+\frac{3}{8}{a}^{3}b-\frac{3}{4}a{b}^2-1024\mathrm{\Delta }$$

     q.expand().subs(b = a^2/4)

$$-1024\mathrm{\Delta }$$

We obtained $\mathrm{\Delta }(f)=0$.