*This post assumes familiarity with some basic concepts in abstract algebra, specifically the terminology of field extensions, and the classical results in Galois theory and group theory.*

The fundamental theorem of algebra has quite a few number of proofs (enough to fill a book!). In fact, it seems a new tool in mathematics can prove its worth by being able to prove the fundamental theorem in a different way. This series of proofs of the fundamental theorem also highlights how in mathematics there are many many ways to prove a single theorem, and in re-proving an established theorem we introduce new concepts and strategies. And perhaps most of all, we accentuate the unifying beauty of mathematics.

**Problem:** Let $ p(x) \in \mathbb{C}[x]$ be a non-constant polynomial. Prove $ p(x)$ has a root in $ \mathbb{C}$.

**Solution:** Without loss of generality, we may assume $ p(x) \in \mathbb{R}[x]$ has *real* coefficients, since if $ p(x)$ has no roots, then neither does $ p(x)\overline{p(x)}$, where the bar represents complex conjugation. In particular, $ p(x)\overline{p(x)}$ is invariant under complex conjugation, and all such quantities are real.

Let $ F$ be the splitting field for $ p(x)$ over $ \mathbb{R}$, and embed $ F$ in some algebraic closure of $ \mathbb{R}$. Consider the extension $ F(i)$. We claim $ F(i)$ is Galois over $ \mathbb{R}$; indeed, it is the splitting field of the separable polynomial constructed as the square-free part of $ p(x)(x^2 + 1)$, i.e., the product of all linear factors of that polynomial.

Let $ G$ be the Galois group of this extension over $ \mathbb{R}$, and note that since $ i \in F(i)$, we have an intermediate field $ \mathbb{R} \subset F \subset F(i)$, so that the degree of the extension $ F \subset F(i)$ is 2; so 2 divides the degree of $ \mathbb{R} \subset F(i)$, and hence the order of the Galois group $ G$ (recall $ |\textup{Aut}_{k}(F)| = [F : k]$ for any Galois field extension).

Since $ 2 \big | |G|$, there is a Sylow 2-subgroup $ H \subset G$, which by Sylow’s theorem has odd index in $ G$. By the Galois correspondence, $ H$ corresponds to an intermediate field extension $ \mathbb{R} \subset E \subset F(i)$ of odd degree (the degree is equal to the index of the subgroup $ H$). Since every real polynomial of odd degree has a root in $ \mathbb{R}$ (recall the intermediate value theorem), we see that there are no irreducible polynomials of odd degree $ d > 1$, and hence there can be no separable extensions of degree $ d$ (in particular, every such extension is finite and separable, and all such extensions are simple: the minimal polynomial of the singly-adjoined element must be irreducible). Hence, the degree of $ E$ over $ \mathbb{R}$ must be 1, i.e. $ E = \mathbb{R}$ is a trivial extension.

Taking this back to the group, since the index $ [G:H] = [E:\mathbb{R}]$ by the Galois correspondence, we have $ [G:H] = 1$, and hence $ G=H$ has order $ 2^n$ for some $ n$. We note that since $ \mathbb{C} = \mathbb{R}(i) \subset F(i)$, $ F(i)$ is a Galois extension of $ \mathbb{C}$. In particular the automorphism group $ G’ = \textup{Aut}_{\mathbb{C}}(F(i))$ is a subgroup of $ G$, and hence has order $ 2^k$ for some $ k$ dividing $ n$. We will show that $ k=0$, which occurs precisely when the extension is trivial (again, by virtue of being a Galois extension).

If $ k > 0$, then we recall the corollary of Cauchy’s theorem for $ p$-groups, that $ G’$ has a subgroup of order $ 2^j$ for any $ j < k$, and in particular a subgroup of index 2. But such a subgroup corresponds to an intermediate field extension of degree 2. As with the case for odd extensions of $ \mathbb{R}$, we note that there are no irreducible polynomials of degree 2 over $ \mathbb{C}$: we have the sing-a-long quadratic formula which constructs the roots in $ \mathbb{C}$. So $ k=0$, and hence $ F(i) = \mathbb{C}$ is the splitting field of $ p(x)$, and contains all of its roots. $ \square$

Since 2 \big | |G|, there is a Sylow 2-subgroup H \subset G, which by Sylow’s theorem has odd index in G.

Why odd degree? Doesn’t Sylow say that the number of 2-Sylow-groups is odd, but nothing about the index?

The statement about index is part of the third Sylow theorem.

Wikipedia says:

Let np be the Number of p-Sylow-groups.

np divides m, which is the index of the Sylow p-subgroup in G.

np ≡ 1 mod p.

Since we don’t know how many of those groups exist, np could be 1 or could be 3. So m could be 1,2,3,6,9, or everything else?

All that matters for this proof is that the index is odd.

I think I got it. Since G is a group of order 2^n * m and H is a subgroup of order 2^n, its clear that [G/H]=m is not divided by 2, because all 2 factors were in the 2^n.

Thanks for your help.