In this blog post, I'll discuss a proof of the Fundamental Theorem of Algebra that I came up with about a year ago.

The Fundamental Theorem of Algebra: $\bb{C}$ is an algebraically closed field.

Proof: Assume that $\bb{C}$ is not algebraically closed. Consider any non-trivial finite extension $\bb{K}/\bb{C}$. The field extension $\bb{K}/\bb{R}$ has degree $n > 2$. We may assume that $\bb{K}/\bb{R}$ is Galois after replacing $\bb{K}$ with its Galois closure over $\bb{R}$. Endow $\bb{K}$ with the structure of a smooth $n$-manifold using its $\bb{R}$-vector space structure.

Lemma 1: The following maps are smooth:

• The multiplication map $m: \bb{K} \times \bb{K} \to \bb{K}$.
• The Galois conjugation maps $\sigma: \bb{K} \to \bb{K}$ for $\sigma \in \mathrm{Gal}(\bb{K}/\bb{R})$.
• The norm map $N: \bb{K} \to \bb{R}$.
• The inverse map $i: \bb{K}^{\times} \to \bb{K}^{\times}$.

Thus, $\bb{K}^{\times}$ is a Lie group under the subspace topology.

Proof of Lemma 1: The multiplication map is $\R$-bilinear and the Galois conjugation maps are $\R$-linear, so they are both smooth. The norm map $N$ can be obtained from composing these functions appropriately and restricting the codomain, so it is smooth. Finally, the inverse map can be written as: $$i(\alpha) = \frac{1}{N(\alpha)}\prod_{\sigma \in \mathrm{Gal}(\bb{K}/\bb{R})} \sigma(\alpha)$$ making it smooth. Given that $m: \bb{K}^{\times} \times \bb{K}^{\times} \to \bb{K}^{\times}$ and $i: \bb{K}^{\times} \to \bb{K}^{\times}$ are smooth, $\bb{K}^{\times}$ must be a Lie group. $\square$

Lemma 2: The squaring map $F: \bb{K}^{\times} \to \bb{K}^{\times}$, given by $\alpha \mapsto \alpha^2$, is open.

Proof of Lemma 2: By the inverse function theorem, it suffices to show that for any $\alpha \in \bb{K}^{\times}$, the map $dF_{\alpha}: \bb{K} \to \bb{K}$ is an isomorphism of $\R$-vector spaces. Indeed, the derivative of the map $F: \alpha \mapsto \alpha^2$ is simply $dF_{\alpha}: x \mapsto 2\alpha x$, which is indeed an isomorphism of $\R$-vector spaces. $\square$

We now prove the Fundamental Theorem of Algebra. The homomorphism $F: \bb{K}^{\times} \to \bb{K}^{\times}$ induces a homomorphism $f: \bb{K}^{\times}/\bb{R}^+ \to \bb{K}^{\times}/\bb{R}^+$. We claim $f$ has the following properties:

• Open: This is equivalent to saying that the image of a saturated open set under $F$ is also saturated and open. Indeed, such an image must be saturated since $F: \bb{R}^+ \to \bb{R}^+$ is surjective and must be open by Lemma 2.
• Closed: This is because $\bb{K}^{\times}/\bb{R}^+ \cong S^{n-1}$ is compact Hausdorff.
• Surjective: The image of $f$ is non-empty, open, and closed. Since $S^{n-1}$ is connected, the image must be $S^{n-1}$.
• $2$-to-$1$: The kernel of $f$ corresponds to elements of $\bb{K}^{\times}$ whose squares are elements of $\bb{R}^+$. These are precisely the elements of $\bb{R}^{\times}$, so $\ker f = \bb{R}^{\times}/\bb{R}^+ \cong \{\pm 1\}$.

This means that the map $f: S^{n-1} \to S^{n-1}$ must be a double cover. However, for $n > 2$, the space $S^{n-1}$ is simply-connected, so we have the required contradiction. $\square$