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The minimal polynomial of $\sqrt{2 + \sqrt{3}}$ over $Q$ is $x^{4} - 4 x^{2} + 1$

Example

Let $a = \sqrt{2 + \sqrt{3}}$ , with the nonnegative real square roots supplied by the completeness of $R$ (The Cauchy-sequence reals have the least-upper-bound property) and Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ . Its minimal polynomial over $Q$ is $p (x) = x^{4} - 4 x^{2} + 1,$ so $[Q (a) : Q] = 4$ .

Facts & Assumptions

Given: The real number $a = \sqrt{2 + \sqrt{3}}$ .

[F1]

Eisenstein's criterion proves irreducibility over $Q$ under its prime divisibility hypotheses (Eisenstein criterion over the integers).

[F2]

The polynomial-ring universal property gives substitution homomorphisms such as $f (x) \mapsto f (x + 1)$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[F3]

A root of a nonzero polynomial is algebraic (Algebraic and transcendental elements and algebraic extensions), and the minimal polynomial of an algebraic element is the monic irreducible polynomial generating its evaluation kernel (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F4]

Its degree equals the degree of the associated simple extension (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[F5]

The real numbers are a complete ordered field, so the displayed nonnegative square roots exist and square to their radicands (The Cauchy-sequence reals have the least-upper-bound property, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Verification

technique · direct

1.1

From $a^{2} = 2 + \sqrt{3}$ one obtains $(a^{2} - 2)^{2} = 3$ , hence $p (a) = a^{4} - 4 a^{2} + 1 = 0$ .

F5algebra

1.2

Substitution from [F2] gives $p (x + 1) = x^{4} + 4 x^{3} + 2 x^{2} - 4 x - 2$ , which satisfies [F1] at $2$ and is therefore irreducible.

F1F2algebra

2.1

Substitution by $x - 1$ is inverse to substitution by $x + 1$ , so a factorization of $p$ would produce one of $p (x + 1)$ . Thus $p$ is irreducible.

F2step 1.2algebra

3.1

Since $p$ is monic, irreducible, and annihilates $a$ , [F3] identifies it as the minimal polynomial; [F4] gives degree $4$ .

F3F4step 1.1step 2.1∎

Depends on

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Sources

J. S. Milne, Fields and Galois Theory (standard reference, not scraped)

The minimal polynomial of 2+32+3​​ over QQ is x4−4x2+1x4−4x2+1