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$Q (\sqrt{2}) / Q$ has degree two

Example

The extension $Q (\sqrt{2}) / Q$ has degree $2$ and basis $(1, \sqrt{2})$ . Its coordinate arithmetic includes

$(a + b \sqrt{2}) (c + d \sqrt{2}) = (a c + 2 b d) + (a d + b c) \sqrt{2} .$

Facts & Assumptions

Given: The positive real number $s = \sqrt{2}$ .

[L1]

If an algebraic element has irreducible minimal polynomial of degree $n$ , its simple extension has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[L2]

Eisenstein's criterion proves suitable primitive integer polynomials irreducible over $Q$ (Eisenstein criterion over the integers).

[L3]

Every nonnegative real has a unique nonnegative square root (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

1.1givenL2L3

By [L3], $s^{2} = 2$ . The polynomial $t^{2} - 2$ is Eisenstein at $2$ , so [L2] makes it irreducible over $Q$ .

2.1step 1.1L1

Apply [L1] to obtain the basis $(1, s)$ and degree $2$ .

3.1step 1.1algebra∎

Multiplying $(a + b s) (c + d s)$ and reducing $s^{2}$ to $2$ gives the displayed coordinate formula.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

A. W. Knapp, Basic Algebra, 2nd ed., Chapter IX, Section 1 (standard reference, not scraped)

Q(2)/QQ(2​)/Q has degree two