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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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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