Worked solution: Wantzel's algebraic impossibility proof for doubling the cube (1837)

Frequently asked questions

Could the rational root test (used for the angle-trisection cubic) work here instead of Eisenstein's criterion?

Yes — since $x^3-2$ has degree $3$ , checking that none of the rational candidates $\pm1,\pm2$ is a root also proves irreducibility, exactly as in the trisection proof. Eisenstein's criterion is used here simply because it settles the question in one line and generalizes to $\sqrt[n]{p}$ for any prime $p$ and any $n$ .

Source

Pierre Laurent Wantzel, 1837

Recherches sur les moyens de reconnaître si un problème de géométrie peut se résoudre avec la règle et le compas

Step 3 of 5: Eisenstein's criterion shows

x^3-2

is irreducible

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In plain words

Eisenstein's criterion is a quick divisibility test: pick a prime number and check that it divides every coefficient of the polynomial except the leading one, and that its square does not divide the constant term. If such a prime exists, the polynomial cannot be split into smaller integer-coefficient pieces — a bit like how a padlock with one very specific, non-repeating key cannot be picked apart into two simpler locks.

For $x^3-2$ the prime $2$ does exactly this job almost by accident: the polynomial's only 'hidden' coefficients (besides the leading $1$ ) are $0, 0, -2$ , and $2$ divides all of them but $4$ does not divide $-2$ .

P(x) = x^3 - 2,\quad 2\mid 0,\ 2\mid(-2),\ 2\nmid 1,\ 4\nmid(-2) \implies P \text{ irreducible}

Detailed analysis

Eisenstein's criterion states: for an integer-coefficient polynomial $a_nx^n+\cdots+a_1x+a_0$ , if some prime $p$ divides every coefficient $a_0,\ldots,a_{n-1}$ but not the leading coefficient $a_n$ , and $p^2$ does not divide $a_0$ , then the polynomial is irreducible over $\mathbb{Q}$ .

Apply this to $P(x)=x^3-2$ , with coefficients $a_3=1,\ a_2=0,\ a_1=0,\ a_0=-2$ . Take $p=2$ : it divides $a_2=0$ , $a_1=0$ , and $a_0=-2$ , but does not divide the leading coefficient $a_3=1$ ; and $p^2=4$ does not divide $a_0=-2$ . All conditions hold, so $P(x)=x^3-2$ is irreducible over $\mathbb{Q}$ (this reasoning appears already in Wantzel 1837, §III, in an equivalent form predating Eisenstein's later general statement of the criterion in 1850).

Since $P$ is the minimal polynomial of $\sqrt[3]{2}$ and has degree $3$ , $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$ . This is exactly the number needed for the final step: it must now be compared against the list of allowed degrees $1,2,4,8,\ldots$ from Step 2.

Terms in this step

Eisenstein's criterion: A sufficient test for irreducibility over $\mathbb{Q}$ : if a prime $p$ divides every coefficient of a polynomial except the leading one, and $p^2$ does not divide the constant term, the polynomial cannot factor into lower-degree integer-coefficient polynomials.

Knowledge used in this step

Rings and fields