Brian Bi
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Exercise 15.6.1 Since $$g \divides f$$, write $$f = gh$$ where $$h \in F[x]$$. Then $$f' = gh' + g'h$$. We are given that $$g \divides f'$$, so $$g \divides f' - gh' = g'h$$. By Theorem 12.3.10, $$F[x]$$ is a UFD, so $$g$$ is prime; therefore $$g \divides g'$$ or $$g \divides h$$. But $$g$$ can only divide $$g'$$ if $$g' = 0$$, which is not possible for $$\char F = 0$$ and $$g$$ irreducible. Therefore $$g \divides h$$. This implies $$g^2 \divides f$$.
1. Let $$a \in F$$. The extension $$F(\sqrt{a})$$ is $$F[x]/(x^2 - a)$$, which is only a field if $$x^2 - a$$ is irreducible, that is, $$a$$ is not already a square in $$F$$. The field $$F(\sqrt{a})$$ will have $$F$$-basis $$\{1, \sqrt{a}\}$$. Suppose $$(x + y\sqrt{a})^2 = z$$ where $$x, y, z \in F$$. Expanding, we obtain $$x^2 + ay^2 + 2xy\sqrt{a} = z$$, so that $$x^2 + ay^2 = z$$ and $$2xy = 0$$. Since $$\char F = 0$$, this implies $$x = 0$$ or $$y = 0$$. If $$y = 0$$, then $$x^2 = z$$, that is, $$z$$ is already a square in $$F$$. If $$x = 0$$, then $$y^2 = z/a$$. So the elements of $$F$$ that have square roots in $$F(\sqrt{a})$$ are precisely the elements that have square roots in $$F$$ and the elements $$z \in F$$ such that $$z/a$$ has a square root in $$F$$. The square root of $$z$$ in $$F(\sqrt{a})$$ will be $$\sqrt{z/a}\sqrt{a}$$ in the latter case.
2. Suppose $$a, b \in \Z$$ are two squarefree integers, not equal to 1. Then $$\Q(\sqrt{a})$$ and $$\Q(\sqrt{b})$$ are fields. Obviously $$a$$ has a square root in $$\Q(\sqrt{a})$$. Since $$b$$ doesn't have a square root in $$\Q$$, $$b$$ will have a square root in $$\Q(\sqrt{a})$$ iff $$b/a$$ has a square root in $$\Q$$. Suppose this is the case. Each prime $$p$$ occurs zero or one times in the prime factorization of $$a$$ and likewise for $$b$$, therefore in $$b/a$$ each prime occurs with exponent $$e \in \{-1, 0, 1\}$$. For $$b/a$$ to be a perfect square, each such exponent must be 0, so that $$b = \pm a$$. In the case that $$b = -a$$, $$b/a$$ is not a square in $$\Q$$. So $$b$$ has a square root in $$\Q(\sqrt{a})$$ iff $$b = a$$. This implies that $$\Q(\sqrt{a})$$ and $$\Q(\sqrt{b})$$ are nonisomorphic extensions when $$a, b$$ are distinct squarefree integers, not equal to 1.
Now in general a quadratic extension of $$\Q$$ can be formed by adjoining some square root $$\sqrt{a}$$ to $$\Q$$ (Proposition 15.3.3); and $$a$$ is not a square in $$\Q$$. There is some square $$q \in \Q$$ such that $$aq$$ is a squarefree integer, not equal to 1. Therefore $$\Q(\sqrt{a}) = \Q(\sqrt{aq})$$. So in fact the extensions generated by the square root of a squarefree integer other than 1 are all the quadratic extensions of $$\Q$$.
Exercise 15.6.3 The primitive root of unity $$\zeta_n = \exp(2\pi i/n)$$ generates the same extension field of $$\Q$$ as does $$\zeta_n^e$$ where $$e$$ is any integer relatively prime to $$n$$. So $$\Q[\sqrt{d}]$$ will contain a primitive $$n$$th root of unity iff it contains $$\zeta_n$$. If $$[\Q(\zeta_n) : \Q] > 2$$ then it is not possible to have $$\Q(\zeta_n) \subseteq \Q[\sqrt{d}]$$. According to Exercise 15.3.5, a primitive $$n$$th root of unity has degree 2 only when $$n \in \{3, 4, 6\}$$, so only these primitive roots can be contained in a quadratic number field. (We are disregarding $$1$$ and $$-1$$, which are primitive roots of unity of degree 1 and 2, respectively, and which are obviously contained in all quadratic number fields.) Now \begin{align*} \zeta_3 &= -\frac{1}{2} + i\frac{\sqrt{3}}{2} \in \Q[\sqrt{-3}] \\ \zeta_6 &= \frac{1}{2} + i\frac{\sqrt{3}}{2} \in \Q[\sqrt{-3}] \\ \zeta_4 &= i \in \Q[\sqrt{-1}] \end{align*} so the answer to the problem is $$d \in \{-1, -3\}$$.