I need help with the last step of the proof of why a $20$ degree angle is not constructible. I understand we try and trisect a $60$ degree angle.
We use the triple angle formula and get
$$
\cos(\theta) = 4\cos^3\left(\frac\theta3\right) – 3\cos\left(\frac\theta3\right)
$$
Let $\theta = 60, \beta = 20$ and we get
$$
8\alpha^3 – 6\alpha – 1 = 0
$$
We use the rational root test and find that the only roots can be $\left\{\pm1,\pm1/2,\pm1/4,\pm1/8\right\}$, none of which statisfy the equation. Therefore it has no rational roots and is irreducible in $\mathbb{Q}$.
Now the part I don't understand or rather I want to make sure I understand correctly is,
Since get $8\alpha^3 – 6\alpha – 1$ is irreducible,
$$
[F(\alpha):F] = \deg(m_a) = 3 = \deg(\alpha)
$$
Why does the degree of the minimal polynomial equal the degree of the field extension?
Then from this since $3 \neq 2^k$ for some $k$ its not constructible.
Best Answer
Take some field $K$ and $\alpha\in L$ where $L$ is some field containing $K$. Suppose that $\alpha$ is algebraic over $K$, i.e. there is some polynomial $f(X)\in K[X]$ such that $f(\alpha)=0$. We now consider the minimal polynomial $p:=\min_K(\alpha)$ which is a monic irreducible polynomial of smallest degree possible such that $p(\alpha)=0$.
We now consider the simple extension $K(\alpha)$. Recall that $[K(\alpha):K]=\dim_K K(\alpha)$. All elements of the subfield $K\subset K(\alpha)$ can be realised as scalar multiplies of $1$ and hence the only elements of interest for giving a $K$-basis are the powers of $\alpha$. Consider the family $1,\alpha,\dots,\alpha^i,\dots$; we know that there has to be some positive integer $n$ such that $\alpha^n$ can be expressed as $K$-linear combination of lower powers (say for $n\ge\deg p$ using that $p(\alpha)=0$).
Hence the dimension of $K(\alpha)$ as $K$-vector space is at most $\deg p$. Now, suppose that there is some $m<\deg p$ such that $\{1,\alpha,\dots,\alpha^{m-1}\}$ is already linearily dependent over $K$. In this case there has to be some non-trivial relation $a_m\alpha^{m-1}+\cdots+a_0=0$ which (after rescaling) gives a polynomial expression in $\alpha$ of strictly smaller degree than $p$ equal to $0$. But this cannot be by the minimality of $p$. Hence $K(\alpha)$ is precisly of degree $\deg p$ over $K$.