A point $P\in\C$ is constructible if there is a sequence of finite points $P_0,\ldots, P_N = P$ such that $P_0 = 0, P_ 1 = 1$ and for all $n < N$ the point $P_{ n + 1}$ is in the intersection of two different circle-lines of type $\lr{P_α, P_β}$ with $0 ≤ α <β ≤ n$ or of type $ C (P_γ, | P_α - P_β |)$ with $0 ≤ α, β, γ ≤ n$.

Note that the two different circle-line whose intersection is considered can be of the same type (that is both circles, both lines or a line and a circle).
Given two points $P_1,P_2$ a ruler allows us to draw a line passing through $P_1$ and $P_2$. A compass allows us to draw a circle with center $P_1$ and radius $|P_1-P_2|$ or if three points are given $P_1,P_2,P_3$, then say a circle with center at $P_1$ and radius of length $|P_2-P_3|$ .

 Let $S = \{0,1\}\in\C$ or $\{(0, 0), (1, 0)\}\in\RR^2$. The figure below shows $n\in\C$ or $ (n, 0)\in\RR^2$ for all $n \in \Z$ are constructible from $S$. Similarly, $m+\ci n\in\C$ are constructible from $S$ for all $(m, n)\in \Z$

Since the set $S$ is a set of complex numbers, it is possible to construct a field $\Q(S)\subset\C$. For example if $S=\{\ci,\sqrt{2}\}$, then $\Q(S)=\Q(\ci,\sqrt{2})$

Proof
$r$ lies in the intersection of two circles or two lines or a line and circle, given by the general equations $$\begin{aligned} a_1z{\bar z}+b_1z+c_1{\bar z}+d_1=0\\ a_2z{\bar z}+b_2z+c_2{\bar z}+d_2=0 \end{aligned}$$ in the complex plane with $a_i,b_i\in\Q(S)$. For lines $a_i=0$, thus these become linear equations and degree of extension is $1$, else atleast one of the equations remains of degree $2$ giving a possible degree of extension as $2$ or $1$.

Consider a cube with side of unit length (volume $1$), then to double the cube we have to construct a side with length $\sqrt[3]{2}$ (volume $2$). But, $[\Q\sqrt[3]{2}:\Q]=3$ and not of the form $2^k$, thus it is impossible to construct $\sqrt[3]{2}$ with ruler and compass and hence impossible to double the cube.

Recall the trignometric identity $\cos 3\theta=4\cos^3\theta-3\cos\theta$ and we want to trisect $3\theta$, to get $\theta$ for example $3\theta=\pi/3$. Plug this in the above equation and set $2\cos \theta =X$ $$ \begin{align*} 4\cos^3\theta-3\cos\theta &=\frac{1}{2}\\ \frac{1}{2}(2 \cos\theta)^3-\frac{3}{2}(2\cos\theta)&=\frac{1}{2}\\ X^3-3X-1&=0 \end{align*} $$ By Rational Roots the only possible roots of $X^3-3X-1=0$ are $\pm 1$ and plugging either of them does not give $0$, thus it is an irreducible polynomial of degree $3$ which is not of the form $2^k$.

Recall that $2\cos(2\pi/p)=\exp(2\pi\ci/p)+\exp(-2\pi\ci/p)$ rearranging by setting $T=\exp(2\pi\ci/p)$ gives us a polynomial $$T^2-2\cos(2\pi/p)T+1=0\in \Q(\cos(2\pi/p))[X]\subset \RR[X]. $$ But, $\exp(2\pi\ci/p)\notin\RR$, hence the polynomial is irreducible and this gives the field extension of degree $2$, $\Q(\exp(2\pi\ci/p))\supset\Q(\cos(2\pi/p))$ . Since the Cyclotomic polynomial is irreducible of degree $p-1$ with root $\exp(2\pi\ci)$, this gives field extension $\Q(\exp(2\pi\ci/p))\supset\Q$ of degree $p-1$. Note that, $$\begin{align*} [\Q(\exp(2\pi\ci/p)):\Q(\cos 2\pi/p)][\Q(\cos 2\pi/p):\Q]& =[\Q(\exp(2\pi\ci/p)):\Q]\\ [\Q(\cos 2\pi/p):\Q]=\dfrac{p-1}{2} \end{align*} $$ By the theorem a $p$ gon with $p$ prime is constructible if $(p-1)/2=2^k$ or $p=2^r+1$ for some integer $r$.
 Recall the factorization $$ X^{2n+1}+1=(X+1)(X^{2n}-X^{2n-1}+\cdots+1). $$ Set $X=2^a$ and $b=2n+1$, that is $b$ odd, the following factorization is obtained $$ 2^{ab}+1=(2^a+1)((2^a)^{b-1}-(2^a)^{b-2}+\cdots+1). $$ Thus, $p=2^{ab}+1$ will factorize if $b$ is odd. In order to avoid the factorization of $p$ (for it to be prime) and noticing that the role of $a$ and $b$ can be interchanged, it has to be the case that $ab$ is a power of $2$, or $p=2^{2^r}+1$. Primes of this form are called Fermat Primes, for example $17=2^4+1$ is a Fermat prime and 17-gon is constructible. But, $2^{32}+1=(641)(6700417)$ is not prime.