I'm reading a proof of the necessary condition for a real number to be constructible, and it seems to leave out a few details that I can't really fill in. This is what I understand so far.
We have to prove that:
If the point $(p, q)$ is constructible with straightedge and compass starting from the points $(0, 0)$ and $(0, 1)$, then both $p$ and $q$ belong to field extensions whose degrees over $\Bbb Q$ are powers of $2$.
We prove this by induction on the number of steps required to construct the point. For $0$, just set $F=\Bbb Q$.
If we assume that every point constructible in $n$ steps or less satisfies the property, then let $(x, y)$ be a point constructible in $n+1$ steps. It'll take $4$ (not necessarily distinct) points to construct a new point (two for each circle/line), and all $8$ of the coordinates $x_1,...x_8$must belong to fields whose degrees over $\Bbb Q$ are all powers of $2$. We can express $x$ and $y$ as roots of either linear or quadratic polynomials with coefficients in $\Bbb Q(x_1,...x_8)$. Thus $x$ and $y$ both belong to fields of degree at most $2$ over $\Bbb Q(x_1,...x_8)$.
Now the theorem seems almost in reach, through multiplicativity of degrees, but then I realized... how do I know that $(\Bbb Q(x_1,...x_8)/\Bbb Q)$ is a power of $2$? All I know is that $(\Bbb Q(x_1)/\Bbb Q),...(\Bbb Q(x_8)/\Bbb Q)$ are each powers of $2$, but I don't see how that tells me anything about $\Bbb Q(x_1,...x_8)$.
What am I missing?
You prove by induction that the degrees of the fields $K_n = \Bbb Q(x_1,y_1,\ldots,x_n,y_n)$ is a power of $2$. Then you deduce that $(\Bbb Q(x_n)/\Bbb Q)$ and $(\Bbb Q(y_n)/\Bbb Q)$ are also a power of $2$ because it has to divide $(K_n/\Bbb Q)$
If $(x_n,y_n)$ is obtained at the intersection of $2$ lines then $x_n,y_n$ are solution of a system of linear equations with coefficients in $K_{n-1}$, and so $K_n = K_{n-1}$. If $(x_n,y_n)$ is obtained at the intersection of a circle with a line, or of two circles, then $x_n,y_n$ are roots of degree $2$ polynomials with coefficient in $K_{n-1}$, and there is also a linear relation $f(x_n,y_n,1) = 0$ where the coefficients are in $K_{n-1}$ so either $K_n = K_{n-1}$, either $K_n$ is an extension of degree $2$ of $K_{n-1}$