# Questions in proof of Fundamental theorem of algebra using Galois Theory

I am self studying Galois theory from Hungerford. I have questions in the proof of Fundamental theorem of Algebra(Pg 266) and got struct on proof of Fundamental Theorem of Algebra.

i have question in 2 line of 2 nd paragraph of proof . Please note that ||.5.7 is Sylow 1 st Theorem.

Kindly note that Result 3.18 there is no extension fields of dimension 2 over the field of complex numbers. ( In case it is used to deduce these). and (B) is every polynomial in $$\mathbb{R}$$[x] of odd degree has a root in $$\mathbb{R}$$

Questions: 1.How does in 2nd line of 2nd paragraph $$Aut_{R} F$$ must have odd index whose fixed field has odd dimension?

and in last line of page 266 how does it implies degree must be 1?

In last paragraph how does Fundamental theorem of Galois Theory implies $$E_0$$ is an extension of $$\mathbb{C}$$ with dimension $$[Aut_{C} F : J]=2$$ ?

I know i have asked 3 questions but I am really struck on this and would be thankful for your help.

• The index of a $p$-Sylow subgroup cannot be divisible by $p$, this follows from the definition.
– Mark
Sep 25, 2020 at 20:15

A Sylow $$p$$-subgroup of a group $$G$$ has order $$p^{n_p}$$, where $$p^{n_p}$$ is the largest power of $$p$$ dividing $$|G|$$. In your setting, $$|\mathrm{Aut}_{\Bbb{R}}\, F| = 2^n \cdot \text{[product of powers of odd primes]} \text{,}$$ so the index is a (possibly empty) product of powers of odd primes, so is odd.
As noted, the index of $$E$$ over $$\Bbb{R}$$ is the same as this index, so is odd.
This makes the minimal polynomial have odd degree. Hypothesis (B) says this polynomial has a root in $$\Bbb{R}$$, so no extension (by $$u$$) is needed and the root is real, so its minimal polynomial (over $$\Bbb{R}$$) is linear. (Every element of the base field is a root of a linear polynomial. Let $$u$$ be such an element; $$x - u$$ is the polynomial.)
The previous paragraph ends with the assertion that $$\mathrm{Aut}_{\Bbb{C}}\, F$$ has order $$2^m$$, $$0 \leq m$$. This paragraph starts by assuming $$m > 0$$. There is a $$2$$-subgroup of order $$2^{m-1}$$ properly contained in a Sylow $$2$$-subgroup (of order $$2^m$$). The Fundamental theorem show a correspondence between the degrees of the fields (also called the dimensions of the extensions) and the indices of the Galois groups. So if the index of $$J$$ is $$2$$, the degree over the fixed field is also $$2$$.