Brian Bi

## Section 16.6. Galois Theory

Note from Brian: My solutions to these problems don't seem very Galois-theoretic. If you know of solutions that actually use the material covered in this section, feel free to email me.

Exercise 16.6.1 The polynomial $$x^3 + x + 1$$ is irreducible over $$\Q$$, so $$[\Q(\alpha) : \Q] = 3$$. Since $$\sqrt{-31}$$ has degree 2 over $$\Q$$, it can't be in this field. On the other hand, according to (16.2.8), the discriminant of $$x^3 + x + 1$$ is $$-31$$, so $$\sqrt{-31} = (z_1 - z_2)(z_1 - z_3)(z_2 - z_3)$$ where $$z_1, z_2, z_3$$ are the roots of $$x^3 + x + 1$$, where $$z_1$$ is real, $$\Im(z_2) > 0$$, and $$\Im(z_3) < 0$$. (We label the roots in this way only to get the correct sign.) Therefore $$\sqrt{-31}$$ is in the splitting field $$K$$.

Exercise 16.6.2 It can be shown fairly easily that $$\sqrt{3} \notin \Q(\sqrt{2})$$ and with a bit more algebra that $$\sqrt{5} \notin \Q(\sqrt{2}, \sqrt{3})$$, so the adjoinment of each square root induces a quadratic extension, and $$[K : \Q] = 8$$. There are eight automorphisms of $$K$$ over $$\Q$$ of the form $$A^a B^b C^c$$ where $$A(\sqrt{2}) = -\sqrt{2}, A(\sqrt{3}) = \sqrt{3}, A(\sqrt{5}) = \sqrt{5}$$, and similarly $$B$$ and $$C$$ each change the sign of one of the square roots, and each exponent $$a, b, c$$ is 0 or 1; this may be easily verified. So $$\Gal(K/\Q)$$ has order at least 8, and since $$[K : \Q] = 8$$, we conclude that $$\Gal(K/\Q)$$ has order exactly 8. The three automorphisms $$A, B, C$$ each have order 2, so $$\Gal(K/\Q) \cong C_2^3$$.

Exercise 16.6.3 We first observe that the claim is not generally true in characteristic 2 (finding a counterexample is left as an exercise for the reader). However, as of section 16.4 all fields are assumed to have characteristic zero, so we proceed under that assumption.

According to Proposition 15.3.3, a quadratic extension is generated by the square root of some element. There are two cases:

• Case 1: $$K$$ is generated over $$L$$ by the square root of an element of $$F$$. Write $$L = F(\alpha), K = L(\beta)$$, where $$\alpha^2, \beta^2 \in F$$. Let $$\zeta = \alpha + \beta$$. Then $$\zeta$$ is a root of the polynomial $$(x - \alpha - \beta)(x - \alpha + \beta)(x + \alpha - \beta) (x + \alpha + \beta) = x^4 - 2(\alpha^2 + \beta^2)x^2 + (\alpha^2 - \beta^2)^2$$, which is of the desired form.

Furthermore, we have $$\zeta^3 = (\alpha^2 + 3\beta^2)\alpha + (\beta^2 + 3\alpha^2)\beta$$, so that \begin{align*} \alpha &= \frac{(\beta^2 + 3\alpha^2)\zeta - \zeta^3}{2(\alpha^2 - \beta^2)} \\ \beta &= \frac{(\alpha^2 + 3\beta^2)\zeta - \zeta^3}{2(\beta^2 - \alpha^2)} \end{align*} This establishes that $$\alpha \in F(\zeta)$$, so in fact $$L \subseteq F(\zeta)$$; also, $$\beta \in F(\zeta)$$, so $$K = L(\beta) \subseteq F(\zeta)$$, therefore $$K = F(\zeta)$$.

• Case 2: Writing $$L = F(\alpha)$$ as before, with $$\alpha^2 \in F$$, $$K$$ is generated over $$L$$ by $$\beta$$ where $$\beta^2 \in L$$ but, unlike in case 1, $$\beta^2 \notin F$$. Write $$\beta^2 = p\alpha + q$$, where $$p, q \in F$$, and $$p \ne 0$$, so that $$\beta^2 \notin F$$ as assumed. Then $$\beta$$ is a root of the polynomial $$(x^2 - q)^2 = p^2 \alpha^2$$, which can also be written $$x^4 - 2qx^2 + (q^2 - p^2 \alpha^2) = 0$$, which is of the desired form. We also have that $$\alpha = p^{-1}(\beta^2 - q)$$, therefore $$F(\beta)$$ contains $$\alpha$$, therefore $$F(\beta)$$ is all of $$K$$.

In both cases, since the root of the quartic described generates the extension $$K/F$$ of degree 4, the quartic is irreducible.