Brian Bi

## Section 15.5. Constructions with Ruler and Compass

Exercise 15.5.1 \begin{align*} \cos 15^\circ &= \cos(45^\circ - 30^\circ) \\ &= \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \\ &= \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2} \\ &= \frac{\sqrt{6} + \sqrt{2}}{4} \end{align*}

Exercise 15.5.2

1. Observe that $$\cos 72^\circ = \cos 288^\circ = 2\cos^2 144^\circ - 1 = 2(2 \cos^2 72^\circ - 1)^2 - 1$$. So, if we put $$x = \cos 72^\circ$$ then we have $$x = 2(2x^2 - 1)^2 - 1$$ which can be expanded to yield a quartic equation for $$x$$. However, two of the roots $$x_1, x_2$$ of this quartic are obviously just the roots of $$x = 2x^2 - 1$$, which lie in a quadratic extension $$\Q(\sqrt{d})$$ of the rationals. Dividing out $$(x - x_1)(x - x_2)$$ from the quartic yields a quadratic polynomial with coefficients in $$\Q(\sqrt{d})$$, so the remaining two roots lie in some quadratic extension of this field. One of these four roots is $$\cos 72^\circ$$, so by Theorem 15.5.6, the latter is constructible. This implies that the regular pentagon is constructible.
2. It is possible to devise the construction by writing out the explicit form of $$\cos 72^\circ$$, which is $$\frac{\sqrt{5} - 1}{4}$$, and applying Lemma 15.5.11, constructing first 5, then $$\sqrt{5}$$, then $$\sqrt{5} - 1$$, and finally $$(\sqrt{5} - 1)/4$$. More elegant constructions are given here.

Exercise 15.5.3 The regular nonagon is not constructible. To see this, use the same argument used to show that the regular heptagon is not constructibe. Namely, let $$\zeta = e^{2\pi i/9}$$ and note that the minimal polynomial of $$\zeta$$ is $$x^6 + x^3 + 1$$ (Exercise 12.4.18). If the regular nonagon were constructible, then $$\cos 2\pi/9$$ and $$\sin 2\pi/9$$ would be constructible and would lie in a field extension $$K$$ of the rationals with degree $$2^k$$, and $$\zeta = \cos 2\pi/9 + i\sin 2\pi/9$$ would lie in $$K(i)$$, which has degree $$2^{k+1}$$, but this contradicts the fact that the degree of $$\zeta$$ is 6.

Exercise 15.5.4 Yes. If the side length of the triangle is $$s$$, then its area is $$\frac{\sqrt{3}}{4}s^2$$ so the square with the same area would have side length $$\frac{\sqrt[4]{3}}{2}s$$. The number $$\frac{\sqrt[4]{3}}{2}$$ is constructible, and we are given $$s$$, so $$\frac{\sqrt[4]{3}}{2}s$$ is constructible.

Exercise 15.5.5 I skipped this problem because I have no idea what it's asking.

Exercise 15.5.6 Say that a complex number $$z$$ is $$\C$$-constructible if $$z \in K$$ where $$K$$ is a subfield of $$\C$$ and there is a sequence of quadratic field extensions $$\Q = F_0 \subseteq F_1 \subseteq \ldots \subseteq F_n = K$$. (We use this term to avoid confusion with the concept of real constructibility as defined on page 453.) We claim that a point $$(x, y)$$ is constructible iff the complex number $$z = x + yi$$ is $$\C$$-constructible. We will make use of a simple lemma:

Suppose \begin{gather*} \Q = F_0 \subseteq F_1 \subseteq \ldots \subseteq F_m \subseteq \\ \Q = K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n \subseteq \end{gather*} where each field extension $$F_j/F_{j-1}$$ and $$K_j/K_{j-1}$$ has degree 2. Then there is a tower of fields \begin{gather*} \Q = L_0 \subseteq L_1 \subseteq \ldots \subseteq L_p \end{gather*} where each extension $$L_j/L_{j-1}$$ is quadratic and $$L_p$$ contains both $$F_m$$ and $$K_n$$ as subfields.

Let $$L_j = F_j$$ for $$j = 0, 1, \ldots, m$$. Since each extension $$K_j/K_{j-1}$$ is quadratic, it is generated by a single element $$\alpha_i$$, and $$K_j = \Q(\alpha_1, \ldots, \alpha_j)$$. Let $$L_{m+j} = L_m(\alpha_1, \ldots, \alpha_j)$$ for $$j = 1, \ldots, n$$. Then $$F_{j-1} = \Q(\alpha_1, \ldots, \alpha_{j-1}) \subseteq F_m(\alpha_1, \ldots, \alpha_{j-1}) = L_{m+j-1}$$. The degree of $$\alpha_j$$ over $$L_{m+j-1}$$ is then at most its degree over $$F_{j-1}$$, so it is at most 2. Also, $$L_{m+n} = F_m(\alpha_1, \ldots, \alpha_n)$$, so it contains both $$F_m$$ as a subfield and $$\Q(\alpha_1, \ldots, \alpha_n) = K_n$$. The tower of fields \begin{equation*} \Q = L_0 \subseteq L_1 \subseteq \ldots \subseteq L_{m+n} \subseteq \C \end{equation*} then satisfies the desired criteria, after links with degree 1 are removed from the chain.

We now show that if the point $$(x, y)$$ is constructible, then $$x + yi$$ is $$\C$$-constructible. Since $$(x, y)$$ is constructible, both of its coordinates are constructible. We can write $$x \in F_m, y \in K_n$$ with \begin{gather*} \Q = F_0 \subseteq F_1 \subseteq \ldots \subseteq F_m \subseteq \R \\ \Q = K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n \subseteq \R \end{gather*} where $$[F_j : F_{j-1}] = 2$$ for each $$j = 1, \ldots, m$$ and $$[K_j : K_{j-1}] = 2$$ for each $$j = 1, \ldots, n$$. Applying the Lemma, we construct a field $$L_p$$ generated by successive quadratic extensions of $$\Q$$ that contains both $$x$$ and $$y$$. The field $$L_p(i)$$ is then a field generated by successive quadratic extensions of $$\Q$$ that contains $$x + yi$$. Therefore $$x + yi$$ is $$C$$-constructible.

Now, suppose $$z = x + yi \in K$$ where $$\Q = F_0 \subseteq F_1 \subseteq \ldots \subseteq F_n = K \subseteq \C$$ and $$[F_i : F_{i-1}] = 2$$ for each $$i = 1, \ldots, n$$. We will prove by induction that $$(x, y)$$ is constructible. The base case is $$n = 0$$; here $$z$$ is rational and the point $$(x, y)$$ is just the point $$(z, 0)$$, so it is constructible. Now, for the inductive case, assume $$n$$ is the smallest possible for the given $$z$$, so that $$z \notin F_{n-1}$$. Since $$[F_n : F_{n-1}] = 2$$, there exist $$a, b \in F_{n-1}$$ with $$z^2 + az + b = 0$$. By the induction hypothesis, the points $$(\Re(a), \Im(a))$$ and $$(\Re(b), \Im(b))$$ are constructible, so the real numbers $$\Re(a), \Re(b), \Im(a), \Im(b)$$ are constructible. Successive applications of the Lemma imply that these four real numbers belong to a common field $$L$$ which is generated by successive quadratic extensions from $$\Q$$ and $$L$$ can be chosen so that $$L \subseteq \R$$. Using the explicit formula for the square roots of a complex number in terms of its real and imaginary parts, we can solve the quadratic $$z^2 + az + b = 0$$ for $$x$$ and $$y$$ in terms of the real and imaginary parts of $$a$$ and $$b$$ using the four basic operations and square roots. This implies that $$x$$ and $$y$$ are constructible real numbers, so $$(x, y)$$ is constructible.

This result shows that $$\C$$-constructibility of complex numbers is the same as constructibility of the real and imaginary parts, and that the criterion for $$\C$$-constructibility is analogous to the criterion for constructibility of real numbers. We can therefore simply speak of complex numbers as being constructible or not, without any ambiguity.