## Return to Answer

2 edited body

As in one of the previous posts, consider the projective space $CP^n$ of nonzero all polynomials

$$c_nT^n + c_{n-1} T^{n-1} + \cdots + c_1 T + c_0$$ considered up to nonzero scalar multiple. We'll show directly that any such polynomial admits a factorization into linear factors.

Consider the map $\phi: CP^1 \times \cdots \times CP^1 \to CP^n$ given by

$$([\alpha_1:\beta_1],\dots,[\alpha_n:\beta_n]) \mapsto \prod_{i=1}^n (\alpha_i T - \beta_i)$$

In other words, this map sends a set of roots to the polynomial which has precisely those roots. It suffices for us to show that $\phi$ is surjective.

If the points $[\alpha_i:\beta:i]$ [\alpha_i:\beta_i]$are distinct, it is easy to check that the differential$d\phi$is nonzero. Hence the polynomial$T^n - 1$(for example) is a regular value of$\phi$with exactly$n!$preimages (here we've used the fact that polynomials factor uniquely into irreducibles). Thus the map$\phi$has positive degree. It is a fact that any map of positive degree between compact connected complex manifolds of the same dimension is surjective. (Proof: Since any such map is orientation preserving, the number of preimages of any regular value must be exactly equal to the degree - not just up to multiplicity. Hence the image contains the set of regular values, which is dense by Sard's theorem. But the image is also closed since it is the image of a compact set, hence the map is surjective.) We conclude that$\phi$is surjective. In other words, every polynomial of degree$n$has a factorization into linear factors. 1 [made Community Wiki] As in one of the previous posts, consider the projective space$CP^n$of nonzero all polynomials $$c_nT^n + c_{n-1} T^{n-1} + \cdots + c_1 T + c_0$$ considered up to nonzero scalar multiple. We'll show directly that any such polynomial admits a factorization into linear factors. Consider the map$\phi: CP^1 \times \cdots \times CP^1 \to CP^n$given by $$([\alpha_1:\beta_1],\dots,[\alpha_n:\beta_n]) \mapsto \prod_{i=1}^n (\alpha_i T - \beta_i)$$ In other words, this map sends a set of roots to the polynomial which has precisely those roots. It suffices for us to show that$\phi$is surjective. If the points$[\alpha_i:\beta:i]$are distinct, it is easy to check that the differential$d\phi$is nonzero. Hence the polynomial$T^n - 1$(for example) is a regular value of$\phi$with exactly$n!$preimages (here we've used the fact that polynomials factor uniquely into irreducibles). Thus the map$\phi$has positive degree. It is a fact that any map of positive degree between compact connected complex manifolds of the same dimension is surjective. (Proof: Since any such map is orientation preserving, the number of preimages of any regular value must be exactly equal to the degree - not just up to multiplicity. Hence the image contains the set of regular values, which is dense by Sard's theorem. But the image is also closed since it is the image of a compact set, hence the map is surjective.) We conclude that$\phi$is surjective. In other words, every polynomial of degree$n\$ has a factorization into linear factors.