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Adjoining one root need not split the polynomial: $Q (\sqrt[3]{2})$ does not split $x^{3} - 2$

Statement refuted

If a field is obtained by adjoining one root of a polynomial, then the polynomial splits over that field.

Facts & Assumptions

Given: The real root $α = \sqrt[3]{2}$ of $x^{3} - 2$ and the field $Q (α)$ .

[F1]

For $α = \sqrt[3]{2}$ and $ω = (- 1 + i \sqrt{3}) / 2$ , the roots of $x^{3} - 2$ are $α, α ω, α ω^{2}$ (The splitting field of $x^{3} - 2$ over $Q$ is $Q (\sqrt[3]{2}, ω)$ with $ω = (- 1 + i \sqrt{3}) / 2$ ).

[F2]

The real numbers form an ordered field, and the complex numbers have unique coordinates $a + b i$ over $R$ (The reals form a totally ordered field, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

Counterexample

technique · direct

1.1

The positive real number $α = \sqrt[3]{2}$ is a root of $x^{3} - 2$ , and every element of $Q (α)$ is real because it is obtained from rational numbers and $α$ by field operations inside $R$ .

1.2

In [F1], the imaginary part of $ω$ is $\sqrt{3} / 2 > 0$ , while direct multiplication gives the imaginary part of $ω^{2}$ as $- \sqrt{3} / 2 < 0$ . Since $α > 0$ , the roots $α ω$ and $α ω^{2}$ are both nonreal and hence neither lies in $Q (α)$ .

F1F2algebra

2.1

Therefore $Q (α)$ contains one root but not all roots of $x^{3} - 2$ , so the polynomial does not split there.

F1step 1.1step 1.2∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 73 results over 17 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

T. Judson, Abstract Algebra: Theory and Applications, Example 21.15 (standard reference, not scraped)

Adjoining one root need not split the polynomial: Q(23)Q(32​) does not split x3−2x3−2