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$x^{4} + 1$ factors over $R$ into two irreducible quadratics

Example

Over $R$ one has

x^{4} + 1 = (x^{2} + \sqrt{2} x + 1) (x^{2} - \sqrt{2} x + 1),

and each quadratic factor is irreducible.

Facts & Assumptions

Given: The polynomial $f (x) = x^{4} + 1$ .

[L1]

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

[L2]

The real numbers form an ordered field (The reals form a totally ordered field).

Verification

technique · direct

1.1

Using $(\sqrt{2})^{2} = 2$ from [L1], direct expansion gives $(x^{2} + \sqrt{2} x + 1) (x^{2} - \sqrt{2} x + 1) = x^{4} + 1.$

L1algebra

2.1

The quadratic $x^{2} + \sqrt{2} x + 1$ has discriminant $2 - 4 = - 2 < 0$ , so it has no real root; the same is true of $x^{2} - \sqrt{2} x + 1$ . Therefore both quadratic factors are irreducible over $R$ .

L2step 1.1algebra

2.2

Their roots in $C$ are $\frac{- \sqrt{2} \pm i \sqrt{2}}{2} and \frac{\sqrt{2} \pm i \sqrt{2}}{2},$ the four fourth roots of $- 1$ . Pairing each nonreal root with its conjugate recovers the two real quadratic factors from step 1.1.

L1step 1.1algebra

3.1

Hence $x^{4} + 1$ is a concrete polynomial that is irreducible over neither $R$ nor $C$ , but over $R$ it factors exactly into two irreducible quadratics.

step 2.1step 2.2∎

Depends on

Used by

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Sources

Thomas W. Judson, Abstract Algebra: Theory and Applications, polynomials and roots (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Chapter 5 (standard reference, not scraped)

x4+1x4+1 factors over RR into two irreducible quadratics