Fields and Galois theory

Answer key — Fields and Galois theory

1. 1.

   $[\Q(\sqrt2) : \Q] = 2$. If $\sqrt3 = a + b\sqrt2$ with $a, b \in \Q$, squaring gives $3 = a^2 + 2b^2 + 2ab\sqrt2$, so $ab = 0$: either $\sqrt3 = b\sqrt2$ ($\sqrt{3/2}$ rational) or $\sqrt3 = a$ — both impossible. So $x^2 - 3$ stays irreducible over $\Q(\sqrt2)$ and $[\Q(\sqrt2, \sqrt3) : \Q(\sqrt2)] = 2$. By the tower law the degree is $2 \cdot 2 = 4$, with basis $1, \sqrt2, \sqrt3, \sqrt6$.

2. 2.

   $\alpha^2 = 5 + 2\sqrt6$, so $(\alpha^2 - 5)^2 = 24$, i.e. $\alpha^4 - 10\alpha^2 + 1 = 0$. To see this quartic is the minimal polynomial, note $\alpha^3 = 11\sqrt2 + 9\sqrt3$, so $\sqrt2 = \frac{\alpha^3 - 9\alpha}{2}$ and $\sqrt3 = \frac{11\alpha - \alpha^3}{2}$ lie in $\Q(\alpha)$. Hence $\Q(\alpha) = \Q(\sqrt2, \sqrt3)$ has degree $4$, and a monic rational polynomial of degree $4$ vanishing at $\alpha$ is its minimal polynomial. Its roots $\pm\sqrt2 \pm\sqrt3$ are permuted by the Galois group $(\Z/2)^2$.

3. 3.

   Constructible: $\sqrt2$, $\cos 72^\circ = \frac{\sqrt5 - 1}{4}$ and $\sqrt{1 + \sqrt2}$, all built from rationals by field operations and square roots. Not constructible: $\sqrt[3]{2}$, $\cos 20^\circ$ and $\cos(2\pi/7)$, each of degree $3$ over $\Q$; by the tower law a constructible number has degree a power of $2$. (Degree a power of $2$ is necessary but not sufficient in general; for the numbers here, the explicit square-root expressions settle the positive cases.)

4. 4.

   $\sigma\tau$ negates $\sqrt2$ and $\sqrt3$ and fixes $1$ and $\sqrt6$. So $a + b\sqrt2 + c\sqrt3 + d\sqrt6$ is fixed iff $b = c = 0$: the fixed field is $\Q(\sqrt6)$. The three subgroups of order $2$ correspond to the three quadratic subfields $\Q(\sqrt2), \Q(\sqrt3), \Q(\sqrt6)$, as the fundamental theorem predicts.

5. 5.

   Subfields of $\mathbb{F}_{p^n}$ correspond to divisors of $n$: $\mathbb{F}_{p^d}$ for each $d \mid n$, and no others. The divisors of $12$ are $1, 2, 3, 4, 6, 12$, so there are $6$ subfields. Galois theory explains this: $\operatorname{Gal}(\mathbb{F}_{2^{12}}/\mathbb{F}_2) \cong \Z/12\Z$, which has exactly one subgroup for each divisor of $12$.

6. 6.

   The roots are $\pm\sqrt[4]2, \pm i\sqrt[4]2$, so $L = \Q(\sqrt[4]2, i)$. By Eisenstein's criterion at $2$, $x^4 - 2$ is irreducible, so $[\Q(\sqrt[4]2) : \Q] = 4$. As $\Q(\sqrt[4]2) \subset \R$ and $i \notin \R$, $[L : \Q(\sqrt[4]2)] = 2$. By the tower law $[L : \Q] = 8$, and since $L/\Q$ is Galois, $|\operatorname{Gal}(L/\Q)| = 8$. The group is the dihedral group $D_4$: it acts on the four roots like the symmetries of the square they form in $\C$.

7. 7.

   Option (b). Abel–Ruffini (Ruffini 1799, Abel 1824) shows that no formula built from the coefficients by field operations and radicals solves every quintic. Galois's theory refines this: over a field of characteristic $0$, $f$ is solvable by radicals iff $\operatorname{Gal}(f)$ is a solvable group. $x^5 - 2$ (Galois group of order $20$, solvable) is solvable by radicals; $x^5 - 4x + 2$ (Galois group $S_5$, not solvable) is not.

8. 8.

   A finite field has prime characteristic $p$ and $p^n$ elements; with four elements, $p = 2$ and $1 + 1 = 0$. $\Z/4\Z$ has characteristic $4$ and the zero divisor $2$, and $\Z/2 \times \Z/2$ has zero divisors too. The field is $\mathbb{F}_2[x]/(x^2 + x + 1)$. In general $\Z/p^2\Z \ne \mathbb{F}_{p^2}$ — formalised in the lesson as zmod_sq_not_field.

9. 9.

   Option (a). finrank F K is $\dim_F K = [K : F]$, so the statement reads $[K : F][L : K] = [L : F]$. The proof takes a basis $(k_i)$ of $K$ over $F$ and a basis $(l_j)$ of $L$ over $K$ and shows the products $k_i l_j$ form a basis of $L$ over $F$.

10. 10.

    def mobius(d):
        result, q = 1, 2
        while q * q <= d:
            if d % q == 0:
                d //= q
                if d % q == 0:
                    return 0
                result = -result
            q += 1
        return -result if d > 1 else result
    
    def count_irreducible(p, n):
        total = sum(mobius(d) * p ** (n // d) for d in range(1, n + 1) if n % d == 0)
        return total // n
    

    The formula comes from $X^{p^n} - X = \prod_{d \mid n} \prod_{f \text{ irreducible, } \deg f = d} f$, whose degrees give $p^n = \sum_{d \mid n} d\,N(d)$; Möbius inversion turns this into Gauss's formula. Since $N(n) \ge \frac1n(p^n - \sum_{d \lt n} p^d) \gt 0$, irreducible polynomials of every degree exist — which gives another proof that $\mathbb{F}_{p^n}$ exists.