Exercise 13.2.10

Proof. If ${𝜃}_f(y)$ has a root $\beta \in F$, then by Theorem 13.2.6(b), $G$ is conjugate to a subgroup of $\mathrm{AGL}(1,{𝔽}_5)$.

Conversely, assume that $G$ is conjugate to a subgroup of $\mathrm{AGL}(1,{𝔽}_5)$. Relabeling the roots, we may assume that $⟨(12345)⟩\subset G\subset \mathrm{AGL}(1,{𝔽}_5)$, and by Theorem 13.2.6(b), ${𝜃}_f(y)$ has a root ${\beta }_1\in F$, so ${𝜃}_f(y)=(y-{\beta }_1)g(y)$.

(a)

Write $\rho =(12345)$ and $\tilde{\rho }\in \mathrm{Gal}(L∕F)$ the corresponding automorphism. Then $$\tilde{\rho }({\alpha }_1)={\alpha }_2,\dots ,\tilde{\rho }({\alpha }_4)={\alpha }_5,\tilde{\rho }({\alpha }_5)={\alpha }_1,$$

and ${\sigma }_i\in \mathrm{Gal}(L∕F)$ corresponds to ${\tau }_i$.

We name the left coset representatives of $\mathrm{AGL}(1,{𝔽}_5)$ given in $S_5$:

$${\tau }_1=e,{\tau }_2=(123),{\tau }_3=(234),{\tau }_4=(345),{\tau }_5=(145),{\tau }_6=(125).$$

Note that these cosets representatives verify $\rho {\tau }_1{\rho }^{-1}={\tau }_1=e$, and

$$\rho {\tau }_2{\rho }^{-1}={\tau }_3,\cdots ,\rho {\tau }_5{\rho }^{-1}={\tau }_6,\rho {\tau }_6{\rho }^{-1}={\tau }_2.$$

By definition, $h_i={\tau }_i\cdot h,i=1,\dots ,6$, where $h=u^2$ and $u$ is given in (13.17), and

$${\sigma }_i({\beta }_1)=({\tau }_i\cdot h)({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)={\beta }_i$$

(see Exercise 13.2.5). Since $\rho \in \mathrm{AGL}(1,5)$, $\rho \cdot h=h$, therefore, for $2\le i\le 5$

$$(\rho {\tau }_i)\cdot h=(\rho {\tau }_i{\rho }^{-1})\cdot (\rho \cdot h)={\tau }_{i+1}\cdot h,$$

and $(\rho {\tau }_6)\cdot h={\tau }_2\cdot h$.

If $\tilde{\phi }\in \mathrm{Gal}(L∕F)$ corresponds to some $\phi \in S_5$, then $\tilde{\phi }({\alpha }_i)={\alpha }_{\phi (i)},i=1,\dots ,5$, so

$$\tilde{\phi }(h({\alpha }_1,\dots ,{\alpha }_5))=h({\alpha }_{\phi (1)},\dots ,{\alpha }_{\phi (5)})=(\phi \cdot h)({\alpha }_1,\dots ,{\alpha }_5).$$

Since $\tilde{\rho }\circ {\sigma }_i\in \mathrm{Gal}(L∕F)$ corresponds to $\rho {\tau }_i$,

$$\begin{array}{llll}\hfill \tilde{\rho }({\beta }_i)& =\tilde{\rho }({\sigma }_i({\beta }_1))& \hfill & \\ \hfill & =(\tilde{\rho }\circ {\sigma }_i)(h({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5))& \hfill & \\ \hfill & =[(\rho {\tau }_i)\cdot h]({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)& \hfill & \\ \hfill & =[(\rho {\tau }_i{\rho }^{-1})\cdot (\rho \cdot h)]({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)& \hfill & \\ \hfill & =({\tau }_{i+1}h)({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)& \hfill & \\ \hfill & ={\beta }_{i+1},i=2,3,4,5,& \hfill & \end{array}$$

and similarly $\tilde{\rho }({\beta }_6)={\beta }_2$. So the images of ${\beta }_i$ by the automorphism $\tilde{\rho }$ corresponding to $\rho =(12345)$ are given by

$${\beta }_2\mapsto {\beta }_3\mapsto {\beta }_4\mapsto {\beta }_5\mapsto {\beta }_6\mapsto {\beta }_2,{\beta }_1\mapsto {\beta }_1,$$

therefore $\mathrm{Gal}(L∕F)$ acts transitively on ${\beta }_2,\dots ,{\beta }_6$.

(b)

Let $p(y)\in F[y]$ be the minimal polynomial of ${\beta }_2$ over $F$. Since $\tilde{\rho }\in \mathrm{Gal}(L,F)$, $p({\beta }_3)=p(\tilde{\rho }({\beta }_2))=\tilde{\rho }(p({\beta }_2))=0$, so ${\beta }_3$, and similarly ${\beta }_4,{\beta }_5,{\beta }_6$ are roots of $p$, so $p$ is the minimal polynomial of ${\beta }_2,\dots ,{\beta }_6$.

${𝜃}_f(y)=(y-{\beta }_1)g(y)$, therefore $g(y)\in F[y]$.

$F\subset L$ is a separable extension, so ${\beta }_2$ is separable, therefore

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

where ${\gamma }_1,\dots ,{\gamma }_r$ are distinct. As $p$ is the minimal polynomial of ${\beta }_2$ over $F$, and $g({\beta }_2)=0$, with $g\in F[y]$, $p(y)$ divides ${𝜃}_f(y)$, so each ${\gamma }_i$ is a ${\beta }_j$, and each ${\beta }_j,2\le j\le j$, is a root of $p$ so is a ${\gamma }_i$. Therefore $\{{\gamma }_1,\dots ,{\gamma }_r\}=\{{\beta }_2,\dots ,{\beta }_6\}$, and ${\gamma }_1,\dots ,{\gamma }_r$ are the distinct roots of $g$.

Since $\mathrm{Gal}(L∕F)$ acts transitively on ${\beta }_2,\dots ,{\beta }_6$, then the distinct roots ${\gamma }_1,\cdots ,{\gamma }_r$ have the same order of multiplicity $m$ (as in Exercise 9). Therefore $g(y)=p{(y)}^m$, so

$${𝜃}_f(y)=(y-{\beta }_1)p{(y)}^m.$$

Since $5=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=m\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(p)$, $5\mid m$, so $m=1$ or $m=5$. We need to rule out $m=5$.

(c)

If $m=5$, $${𝜃}_f(y)=(y-{\beta }_1){(y-{\beta }_2)}^5.$$

Then, with some formal computations,

$$\begin{array}{llll}\hfill 0& ={(y^3+b_2{y}^2+b_{4}y+b_6)}^2-2^{10}\mathrm{\Delta }(f)y-(y-{\beta }_1){(y-{\beta }_2)}^5& \hfill & \\ \hfill & =\left(2b_2+{\beta }_1+5{\beta }_2\right)y^5+\left(b_2^2-5{\beta }_1{\beta }_2-10{\beta }_2^2+2b_4\right)y^4+2\left(5{\beta }_1{\beta }_2^2+5{\beta }_2^3+b_2{b}_4+b_6\right)y^3& \hfill & \\ \hfill & -\left(10{\beta }_1{\beta }_2^3+5{\beta }_2^4-b_4^2-2b_2{b}_6\right)y^2+\left(5{\beta }_1{\beta }_2^4+{\beta }_2^5+2b_4{b}_6-1024\mathrm{\Delta }(f)\right)y-{\beta }_1{\beta }_2^5+b_6^2.& \hfill & \end{array}$$

The coefficients of $y^5,y^4,y^3$ give

$$\begin{array}{llll}\hfill 0& =2b_2+{\beta }_1+5{\beta }_2,& \hfill & \\ \hfill 0& =b_2^2-5{\beta }_1{\beta }_2-10{\beta }_2^2+2b_4,& \hfill & \\ \hfill 0& =10{\beta }_1{\beta }_2^2+10{\beta }_2^3+2b_2{b}_4+2b_6,& \hfill & \end{array}$$

so

$$\begin{array}{llll}\hfill b_2& =-\frac{1}{2}{\beta }_1-\frac{5}{2}{\beta }_2,& \hfill & \\ \hfill b_4& =-\frac{1}{8}{\beta }_1^2+\frac{5}{4}{\beta }_1{\beta }_2+\frac{15}{8}{\beta }_2^2,& \hfill & \\ \hfill b_6& =-\frac{1}{16}{\beta }_1^3+\frac{5}{16}{\beta }_1^2{\beta }_2-\frac{15}{16}{\beta }_1{\beta }_2^2-\frac{5}{16}{\beta }_2^3.& \hfill & \end{array}$$

Substituting these values in the equation, we obtain

$$\begin{array}{llll}\hfill 0& =a{y}^2+\mathit{by}+c,& \hfill & \end{array}$$

where

$$\begin{array}{llll}\hfill a& =\frac{5}{64}({\beta }_1^4-4{\beta }_1^3{\beta }_2+6{\beta }_1^2{\beta }_2^2-4{\beta }_1{\beta }_2^3+{\beta }_2^4)=\frac{5}{64}{({\beta }_1-{\beta }_2)}^4,& \hfill & \\ \hfill b& =\frac{1}{64}{\beta }_1^5-\frac{15}{64}{\beta }_1^4{\beta }_2+\frac{25}{32}{\beta }_1^3{\beta }_2^2-\frac{35}{32}{\beta }_1^2{\beta }_2^3+\frac{45}{64}{\beta }_1{\beta }_2^4-\frac{11}{64}{\beta }_2^5-1024\mathrm{\Delta }(f),& \hfill & \\ \hfill c& =\frac{1}{256}{\beta }_1^6-\frac{5}{128}{\beta }_1^5{\beta }_2+\frac{55}{256}{\beta }_1^4{\beta }_2^2-\frac{35}{64}{\beta }_1^3{\beta }_2^3+\frac{175}{256}{\beta }_1^2{\beta }_2^4-\frac{53}{128}{\beta }_1{\beta }_2^5+\frac{25}{256}{\beta }_2^6.& \hfill & \\ \hfill & & \hfill \end{array}$$

Since $a=0$, if the characteristic is not 5, then ${\beta }_1={\beta }_2$, so ${𝜃}_f(y)={(y-{\beta }_1)}^6$. But the proof of Theorem 13.2.6 shows that this implies that $\mathrm{\Delta }(f)=0$, and this is a contradiction. Thus $m=1$ and $g$ is irreducible over $F$. This proves Proposition 13.2.7. in characteristic $\ne 5$.

Sage instructions for part (e):

     x,y,beta1,beta2,Delta,b2,b4,b6 = var(’x,y,beta1,beta2,Delta,b2,b4,b6’)
     p = (y^3 + b2*y^2 + b4*y + b6)^2 - 2^10*Delta*y - (y-beta1)*(y-beta2)^5
     p = p.expand().collect(y)
     R.<y>=QQ[]
     l = [p.coefficient(y,i) for i in range(5,-1,-1)]
     eq = l[:3]
     solve(eq,b2,b4,b6)
     q=p.subs(b2 == -1/2*beta1 - 5/2*beta2,
           b4 == -1/8*beta1^2 + 5/4*beta1*beta2 + 15/8*beta2^2,
           b6 == -1/16*beta1^3 + 5/16*beta1^2*beta2 -15/16*beta1*beta2^2 - 5/16*beta2^3)
     q=q.expand().collect(y)
     q.coefficient(y,2).factor()
      

$$\frac{5}{64}{\left({\beta }_1-{\beta }_2\right)}^4$$

It remains the case where the characteristic is 5. Then the equations $b=0,c=0$ become,

$$\begin{array}{llll}\hfill 0& =\frac{1}{64}({\beta }_1^5-{\beta }_2^5)-2^{10}\mathrm{\Delta }(f),& \hfill & \\ \hfill 0& =\frac{1}{256}{\beta }_1({\beta }_1^5-{\beta }_2^5).& \hfill & \end{array}$$

If ${\beta }_1\ne 0$, then ${\beta }_1^5-{\beta }_2^5=0$, hence $2^{10}\mathrm{\Delta }(f)=0$. Since the characteristic is not 2, we obtain $\mathrm{\Delta }(f)=0$, which is impossible since $f$ is separable by hypothesis.

If ${\beta }_1=0$, it remains a unique equation

$$0={\beta }_2^5+2^{16}\mathrm{\Delta }(f),$$

that is

$$0={\beta }_2^5+\mathrm{\Delta }(f).$$

In this case

$${𝜃}_f(y)=y{(y-{\beta }_2)}^5=y(y^5-{\beta }_2^5)=y^6+\mathrm{\Delta }(f)y.$$

Since

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

this condition is equivalent to $b_2=b_4=b_6=0$.

I don’t see an immediate contradiction ...

In fact, with this analysis, we can show a counterexample of Proposition 13.2.7:

Let $f=x^5-x+1\in {𝔽}_5[x]$, where $c_1=c_2=c_3=0,c_4=-1,c_5=-1$. Then $f$ is irreducible, separable, and the Sage instructions given in Exercise 6, where we replace the field $\mathbb{Q}$ by ${𝔽}_5$ show that $b_2=b_4=b_6=0$, and

$$\begin{array}{llll}\hfill \mathrm{\Delta }(f)& =-1,& \hfill & \\ \hfill {𝜃}_f(y)& =y^6-y.& \hfill & \end{array}$$

(We verify this directly on the formulas (13.24) : almost all the terms of $B_2,B_4,B_6$ contain ${\sigma }_1,{\sigma }_2$ or ${\sigma }_3$, if not, the coefficient is a multiple of 5, so $b_2=b_4=b_6=0$.)

Therefore $\beta =0$ is a root of ${𝜃}_f$, and ${𝜃}_f(y)=(y-\beta )g(y)$, where $g(y)=y^5-1={(y-1)}^5$ is not irreducible over ${𝔽}_5$.

The second equivalence of Proposition 13.2.7 is false if the characteristic of $F$ is 5.

Note : Here we know explicitly the roots of $f$ in ${𝔽}_{5^5}[x]$. If $\alpha$ is a root of $f$ in this field, the roots of $f$ are $\alpha ,\alpha +1,\alpha +2,\alpha +3,\alpha +4$ (see Ex. 6.2.5). If we substitute (with the help of Sage) these roots in $h_1,\dots ,h_6$, we obtain $0,1,1,1,1,1$, therefore

$${𝜃}_f(y)=\underset{i=1}{\overset{6}{\prod }}(y-h_i({\alpha }_1,\dots ,{\alpha }_5))=y{(y-1)}^5=y^6-y.$$

This is a confirmation of the previous results.