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The splitting field of $x^{4} + 2 x^{2} - 8$ over $Q$ is $Q (\sqrt{2}, i)$

Example

The polynomial $x^{4} + 2 x^{2} - 8$ has roots $\pm \sqrt{2}$ and $\pm 2 i$ , so its splitting field over $Q$ is $Q (\sqrt{2}, i)$ .

Facts & Assumptions

Given: The polynomial $x^{4} + 2 x^{2} - 8 \in Q [x]$ .

[F1]

The real number $2$ has a square root $\sqrt{2}$ with square $2$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[F2]

The complex numbers form a field containing $i$ with $i^{2} = - 1$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[F3]

A splitting field is the field generated over the base by all roots of the polynomial (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

Verification

technique · direct factorisation

1.1

Over $Q$ , direct multiplication gives $x^{4} + 2 x^{2} - 8 = (x^{2} - 2) (x^{2} + 4)$ . By [F1] and [F2], the first factor has roots $\pm \sqrt{2}$ and the second has roots $\pm 2 i$ .

F1F2algebra

2.1

All four roots lie in $Q (\sqrt{2}, i)$ . Conversely, the field generated by them contains $\sqrt{2}$ and $i = (2 i) / 2$ , so it is exactly $Q (\sqrt{2}, i)$ . The claim follows from [F3].

F3step 1.1∎

Depends on

Used by

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Sources

T. Judson, Abstract Algebra: Theory and Applications, Section 21.2 (standard reference, not scraped)

The splitting field of x4+2x2−8x4+2x2−8 over QQ is Q(2,i)Q(2​,i)