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$Q (\sqrt{2}, \sqrt{3})$ has degree four and equals $Q (\sqrt{2} + \sqrt{3})$

Example

The biquadratic field satisfies

$[Q (\sqrt{2}, \sqrt{3}) : Q] = 4,$

has basis $(1, \sqrt{2}, \sqrt{3}, \sqrt{6})$ , and is simple:

$Q (\sqrt{2}, \sqrt{3}) = Q (\sqrt{2} + \sqrt{3}) .$

Facts & Assumptions

Given: The positive roots $u = \sqrt{2}$ , $v = \sqrt{3}$ , and $a = u + v$ .

[L1]

Degrees multiply in finite towers (Tower law for finite extensions: $[L : F] = [L : K] [K : F]$ ).

[L2]

Products of bases form a basis in a tower (Products of bases form a basis in a tower of finite extensions).

[L3]

The field $F (a)$ is the smallest subfield containing $F$ and $a$ (Field extensions, generated subrings $F [S]$ , generated subfields $F (S)$ , and simple extensions).

[L4]

Positive square roots exist uniquely in $R$ (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

The polynomial $t^{2} - 2$ is irreducible over $Q$ . Also $v \notin Q (u)$ : if $v = r + s u$ with rationals $r, s$ , squaring and using uniqueness of the coordinates $(1, u)$ gives $2 r s = 0$ and $r^{2} + 2 s^{2} = 3$ ; either case would make $3$ or $3 / 2$ a rational square, contradicted by comparing the parity of prime exponents in numerator and denominator.

givenL4algebra

1.2

Since $(u + v) (v - u) = v^{2} - u^{2} = 1$ , one has $a^{- 1} = v - u$ . Therefore $v = (a + a^{- 1}) / 2$ and $u = (a - a^{- 1}) / 2$ both lie in $Q (a)$ .

givenalgebra

2.1

Hence both steps in $Q \subset Q (u) \subset Q (u, v)$ have degree $2$ . By [L1] the total degree is $4$ , and [L2] gives the product basis $(1, u, v, u v)$ .

step 1.1L1L2

3.1

Thus $Q (u, v) \subseteq Q (a)$ , while the reverse inclusion follows from $a = u + v$ and [L3]. The fields are equal.

step 1.2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 44 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

Q(2,3)Q(2​,3​) has degree four and equals Q(2+3)Q(2​+3​)