CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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Every nonconstant polynomial in $C [x]$ splits into linear factors

Statement

Every nonconstant polynomial in $C [x]$ splits over $C$ .

Facts & Assumptions

Given: A nonconstant polynomial $f \in C [x]$ .

[L1]

The field $C$ is algebraically closed (The complex numbers are algebraically closed).

[L2]

A field is algebraically closed exactly when every nonconstant polynomial over it splits (A field is algebraically closed exactly when every nonconstant polynomial splits, equivalently when it has no nontrivial finite extension).

Proof

technique · direct

1.1

By [L1], the field $C$ is algebraically closed.

2.1

Applying [L2] to the field $C$ and the given polynomial $f$ shows that $f$ splits over $C$ .

step 1.1L2

3.1

Since $f$ was arbitrary, every nonconstant polynomial in $C [x]$ splits over $C$ .

step 2.1∎

Depends on

Used by

Dependency tree · two levels

23 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, Corollary 5.7 (standard reference, not scraped)
Keith Conrad, Applications of Galois Theory, Theorem 2.1 (standard reference, not scraped)

Every nonconstant polynomial in C[x]C[x] splits into linear factors