Subgroups and corresponding subfields of Galois group of $x^5-5x^2-3$

I think I have found the correct Galois group of $$f(x) = x^5-5x^2-3$$ over $$\mathbb{Q}$$ to be $$D_5$$.

generators are $$σ=(1\ 2\ 4\ 5\ 3)$$ and $$τ=(2\ 3)(4\ 5)$$

I can write down the elements of the permutation group $$D_5=\{1,σ, σ^2,σ^3, σ^4,τ, σ τ,σ^2 τ, σ^3 τ , σ^4 τ\}$$

Let $$r_1,r_2,r_3,r_4,r_5$$ be roots of $$f$$ and $$r_1\in\Bbb R,r_2=\overline{r_3},r_4=\overline{r_5}$$.

Then $$D_5$$ has 6 proper subgroups: five of them are of order $$2$$, and one normal subgroup $$C_5$$.

So $$τ$$ acts by complex conjugation. So the fixed field of $$⟨τ⟩$$ has degree $$5$$ over $$\Bbb Q$$ and contains $$\Bbb Q[r_1]$$, but $$[\Bbb Q[r_1]:\Bbb Q]=5$$, so the fixed field of $$⟨τ⟩$$ is $$\Bbb Q[r_1]$$.

So the fixed field of $$⟨στ⟩,⟨σ^2τ⟩,⟨σ^3τ⟩,⟨σ^4τ⟩$$ are $$\Bbb Q[r_2],\Bbb Q[r_3],\Bbb Q[r_4],\Bbb Q[r_5]$$ in some order.

Since $$D_5\subseteq A_5$$, the discriminant is a square.

Since $$D_5$$ doesn't have a subgroup of index $$4$$, the splitting field $$K$$ of $$f$$ doesn't contain a 5th root of unity $$\zeta_5$$. Although $$\sum_j\zeta_5^jr_j$$ appears to be fixed by $$σ$$, it doesn't lie in $$K$$.

I'm not sure what is the fixed field of $$⟨σ⟩$$.

Related question: Prove that the Galois Group Is $D_{10}$

• @Justauser No, I think whole $D_5$ is contained in $A_5$ Commented May 15 at 8:49
• @Justauser $\sqrt{\Delta}\in\Bbb Q$ Commented May 15 at 8:50

The fixed field of $$⟨σ⟩$$ is $$\Bbb Q[\sum_{j=1}^5\sigma^j(r_1^2r_2)]$$.

f = QQbar['x'](x^5 - 5*x**2 - 3)
roots = [f.roots()[n-1][0] for n in [1,2,4,5,3]]
expr = sum([roots[i]**2*roots[(i+1)%5] for i in range(5)])
expr.minpoly()


We get the minimal polynomial of $$\sum_{j=1}^5\sigma^j(r_1^2r_2)$$ is$$x^2+5 x+100$$ Calculating the discriminant for this quadratic polynomial $$\sqrt{5^2-4\times100}=5\sqrt{-15}$$, we find $$\Bbb Q\left[\sum_{j=1}^5\sigma^j(r_1^2r_2)\right]=\Bbb Q[\sqrt{-15}]$$

To verify this:

Factor $$x^2+15$$ in $$K$$ by SageMath

# Define the polynomial ring and the polynomial
R.<x> = QQ[]
f = x^5 - 5*x^2 - 3

# Get the splitting field of the polynomial
K.<a> = f.splitting_field()

# Define the polynomial to be factored over the splitting field
g = K['x'](x^2 + 15)

# Factor the polynomial over the splitting field
g.factor()


Output: $$(x - \frac{1}{15} a^{5} - \frac{1}{3} a^{3}) \cdot (x + \frac{1}{15} a^{5} + \frac{1}{3} a^{3})$$

$$a$$ is a root of $$x^{10} + 10x^8 + 25x^6 + 3375$$.

To verify that it is splitting field (the Galois closure of root field):

f = QQ['x'](x^5 - 5*x**2 - 3)
K.<a>=f.root_field()
L.<b>=K.galois_closure()
L