Worked solution: Wantzel's algebraic impossibility proof via field extensions (1837)

Step 5 of 7: No rational root

\implies

irreducible over

\mathbb{Q}

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In plain words

If a fraction is going to be a root of a polynomial with whole-number coefficients, that fraction cannot be just anything — its top and bottom are forced to divide the polynomial's first and last coefficients. For our cubic those coefficients are $1$ and $-1$ , so there are only two candidates to check by hand: $x=1$ and $x=-1$ .

Neither works, and for a cubic that is already the whole story: a degree- $3$ polynomial that factors at all must split off a linear piece, which always hands you a rational root. No rational root means no factorization is possible.

P(x)=x^3-3x-1,\quad P(1)=-3,\ P(-1)=1 \implies P \text{ irreducible over } \mathbb{Q}

Detailed analysis

By the rational root theorem, any rational root $p/q$ (in lowest terms) of an integer-coefficient polynomial must have $p$ dividing the constant term and $q$ dividing the leading coefficient. For $P(x)=x^3-3x-1$ the leading coefficient is $1$ and the constant term is $-1$ , so $q=\pm1$ and $p\in\{1,-1\}$ : the only candidates are $x=1$ and $x=-1$ .

Checking both: $P(1)=1-3-1=-3\neq0$ and $P(-1)=-1+3-1=1\neq0$ . So $P$ has no rational root. Because $P$ has degree $3$ , if it factored over $\mathbb{Q}$ into nontrivial pieces, one factor would have to be linear, and a linear factor $qx-p$ forces $p/q$ to be a root — which has just been excluded. Hence $P$ is irreducible over $\mathbb{Q}$ .

An irreducible polynomial of degree $3$ satisfied by a number $\alpha$ is (up to a constant factor) the minimal polynomial of $\alpha$ , so $[\mathbb{Q}(\alpha):\mathbb{Q}]$ equals its degree. Here $\alpha = 2\cos 20^\circ$ is a root of $P$ , so $[\mathbb{Q}(2\cos 20^\circ):\mathbb{Q}] = 3$ — the last ingredient the final step needs.

Terms in this step

Rational root theorem: For an integer-coefficient polynomial, every rational root $p/q$ in lowest terms has numerator $p$ dividing the constant term and denominator $q$ dividing the leading coefficient — a short list of candidates to check.
Minimal polynomial: The lowest-degree monic polynomial with coefficients in $\mathbb{Q}$ that a given number $\alpha$ satisfies; its degree equals $[\mathbb{Q}(\alpha):\mathbb{Q}]$ .

Knowledge used in this step

Prime numbers