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[parent] proof of fundamental theorem of algebra (Proof)

If $ f(x) \in \mathbb{C}[x]$ let $ a$ be a root of $ f(x)$ in some extension of $ \mathbb{C}$. Let $ K$ be a Galois closure of $ \mathbb{C}(a)$ over $ \mathbb{R}$ and set $ G = \operatorname{Gal}(K/\mathbb{R})$. Let $ H$ be a Sylow 2-subgroup of $ G$ and let $ L = K^H$ (the fixed field of $ H$ in $ K$). By the Fundamental Theorem of Galois Theory we have $ [L:\mathbb{R}] = [G:H]$, an odd number. We may write $ L = \mathbb{R}(b)$ for some $ b \in L$, so the minimal polynomial $ m_{b,\mathbb{R}}(x)$ is irreducible over $ \mathbb{R}$ and of odd degree. That degree must be 1, and hence $ L = \mathbb{R}$, which means that $ G = H$, a 2-group. Thus $ G_1 = \operatorname{Gal}(K/\mathbb{C})$ is also a 2-group. If $ G_1 \ne 1$ choose $ G_2 \le G_1$ such that $ [G_1:G_2] = 2$, and set $ M = K^{G_2}$, so that $ [M:\mathbb{C}] = [G_1:G_2] = 2$. But any polynomial of degree 2 over $ \mathbb{C}$ has roots in $ \mathbb{C}$ by the quadratic formula, so such a field $ M$ cannot exist. This contradiction shows that $ G_1 = 1$. Hence $ K = \mathbb{C}$ and $ a \in \mathbb{C}$, completing the proof.



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Cross-references: proof, contradiction, field, quadratic formula, polynomial, degree, odd, irreducible, minimal polynomial, odd number, fundamental theorem of Galois theory, fixed field, Galois closure, extension, root
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This is version 2 of proof of fundamental theorem of algebra, born on 2002-11-16, modified 2006-10-15.
Object id is 3603, canonical name is ProofOfFundamentalTheoremOfAlgebra.
Accessed 6596 times total.

Classification:
AMS MSC: 12D99 (Field theory and polynomials :: Real and complex fields :: Miscellaneous)
 30A99 (Functions of a complex variable :: General properties :: Miscellaneous)
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