LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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To prove the fundamental theorem of algebra, it suffices to split every real polynomial over $C$

Statement

Assume every nonconstant polynomial in $R [x]$ splits over $C$ . Then $C$ is algebraically closed.

Facts & Assumptions

Given: Every nonconstant polynomial in $R [x]$ splits over $C$ , and a nonconstant polynomial $g \in C [x]$ .

[L1]

A field is algebraically closed exactly when every nonconstant polynomial over it has a root in the field (An algebraically closed field: every nonconstant polynomial has a root in the field).

[L2]

A nonzero polynomial splits over a field extension when it is a nonzero scalar times a product of linear factors there (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

[F1]

The complex numbers form a field, and complex conjugation is a real-field automorphism of $C$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1givenF1algebra

Write $g = u + i v$ with $u, v \in R [x]$ , and put $\overline{g} : = u - i v$ . Then $g \overline{g} = u^{2} + v^{2} \in R [x] .$ Because $g$ is nonconstant, so is $g \overline{g}$ .

2.1givenL2step 1.1

By the standing hypothesis and [L2], the real polynomial $g \overline{g}$ splits over $C$ , so it has a complex root $z$ . In particular, $0 = (g \overline{g}) (z) = g (z) \overline{g} (z) .$

3.1F1step 2.1

Since $C$ is a field by [F1], step 2.1 implies either $g (z) = 0$ or $\overline{g} (z) = 0$ .

4.1F1step 3.1algebra

If $g (z) = 0$ then $g$ already has a root in $C$ . If $\overline{g} (z) = 0$ , then conjugating that equality and using the conjugation law from [F1] gives $g (\overline{z}) = 0$ . Thus $g$ has a root in $C$ in either case.

5.1L1step 4.1∎

The polynomial $g \in C [x]$ was arbitrary. Therefore every nonconstant polynomial over $C$ has a root in $C$ , and [L1] makes $C$ algebraically closed.

Depends on

Used by

The complex numbers are algebraically closed Theorem

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory, v5.10, Chapter 5 (standard reference, not scraped)
Keith Conrad, Applications of Galois Theory, Theorem 2.1 (standard reference, not scraped)

To prove the fundamental theorem of algebra, it suffices to split every real polynomial over CC