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I'm studying the Bhattacharya's algebra book, and I have the following doubts:

1-What this $g(x)$ has to do with $f(x)$? I mean why proving the splitting field of $g(x)$ over $\mathbb R$ is $\mathbb C$ we prove the theorem?

2- Why $|G|=2^mq$?

3- Why $[E:L]=2^m$

Please I need help, any one is welcome.

Thanks a lot

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up vote 4 down vote accepted

(1) Note that $g(x)$ contains $f(x)$ as one of its factors. So if $g$ splits into linear factors, so does $f$.

(2) $|G|$ is a non-negative integer. The expression $2^m q$ is just its prime factorization, except you're only caring about the specific exponent of the prime 2, and you don't care about what primes divide $q$. This uses nothing about Galois theory.

(3) This is because $L$ is the subfield corresponding to the Sylow 2-subgroup $H \subset G$, which by definition has order $2^m$. One consequence of the Galois correspondence is that the order of subgroups correspond to indices of field extensions.

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I think it's the contrary, if $f$ splits into linear factors, so does $g$. Thank you very much for your answer – user42912 Jan 4 '13 at 1:37
Rafael, while what you say is true, the whole point of the proof is to show that $g$ splits into linear factors. You do this because that's how you know $f$ splits into linear factors. – user54535 Jan 4 '13 at 2:02

(1) By introducing the factor $ x^{2} + 1 $, we can ensure that the field extension $ E $ contains a root of $ -1 $, which we denote by $ \sqrt{-1} $. Hence, given now that $ \sqrt{-1} \in E $, the next step naturally is to show that $ E = \mathbb{R}[\sqrt{-1}] $. The reason for introducing the conjugate of $ f(x) $ is that it allows us to assume, without loss of generality, that $ f $ is a polynomial with real coefficients.

(2) As mentioned by User 24601, every positive integer can be written uniquely as $ 2^{m} q $, where $ m \in \mathbb{N}_{0} $ and $ q $ is a positive odd integer.

(3) Once again, as mentioned by User 24601, this has to do with the Galois correspondence between subgroups of $ \text{Gal}(E/\mathbb{R}) $ and subfields of $ E $ that contain $ \mathbb{R} $.

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