Simple question about Galois theory

Question

Let $f (x) $ be a polynomial over $\mathbb R $, and call $E $ the splitting field of $f $. Take a $2-$subgroup of $\mathrm {Gal}(E/F) $, call it $P $, and consider the field $F=\mathrm {Inv}P$. Since $[F:\mathbb R]=|G:P|$, necessarily $[F:\mathbb R]$ is an odd number.

Given these informations, I must show that $F=\mathbb R$. In order to help us, the professor wrote to notice that $F/\mathbb R$ is a separable extension, and then apply "some properties of real numbers". This hint makes me think that this exercise is quite elementary, however I have no idea of how should I proceed. The fact that $P $ is the subgroup of transformation of order $2^n$ suggest me that the involved property of $\mathbb R$ is that the squares are exactly the positive numbers, but can't think of a way to link this to the exercise. Can you give me a bigger hint? Thank you in advance

If $|F:\Bbb R|$ is odd, each element of $F$ is a solution of an equation of odd degree over the reals. — Angina Seng, Dec 26 '19 at 16:36
I think the goal is to prove that for any non-real $a$ algebraic $\Bbb{R}(a)$ is algebraically closed just from the axiom that odd degree real polynomials have a real root — reuns, Dec 26 '19 at 16:38
@LordSharktheUnknown Actually I thought of that but I didn't come to a conclusion — Dr. Scotti, Dec 26 '19 at 16:44
I share the hunch reuns described. The story continues here, but is really standard textbook stuff. — Jyrki Lahtonen, Dec 26 '19 at 18:04

WhatsUp · Accepted Answer · 2019-12-26T17:58:07.567

5

The property of real numbers you want to use is the following:

Lemma: Every polynomial of odd degree over $\Bbb R$ has a root in $\Bbb R$, hence is reducible if the degree is strictly larger than $1$.

and as corollary:

Corollary: A finite extension of $\Bbb R$ of odd degree is necessarily $\Bbb R$ itself.

Proof of Lemma: simply consider the limits $\lim\limits_{x \rightarrow +\infty} f(x)$ and $\lim\limits_{x\rightarrow -\infty}f(x)$, and use the continuity of $f$.

Proof of Corollary: simply consider the minimal polynomial of any element in the extension.

edited Dec 26 '19 at 17:58

answered Dec 26 '19 at 16:55

WhatsUp

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Sorry I don't understand the corollary; for example the splitting field of $x^3-2$ over $\mathbb R$ has degree $3$ and it is not $\mathbb R$, isn't it? – Dr. Scotti Dec 26 '19 at 17:54
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No, it does not have degree $3$, as it has a root $\sqrt[3]2$ in $\Bbb R$. For the corollary, if $F/\Bbb R$ is any extension of odd degree, then you may look at any element $x\in F$ and the subextension $\Bbb R[x]/\Bbb R$ should still have odd degree. But the degree of $\Bbb R[x]/\Bbb R$ is simply the degree of the minimal polynomial of $x$, which is an irreducible polynomial. By the Lemma, the degree must be $1$. This means that any element $x\in F$ is indeed in $\Bbb R$. – WhatsUp Dec 26 '19 at 17:57
Of course you are right; thank you now it is all clear – Dr. Scotti Dec 26 '19 at 18:55

Severin Schraven · Answer 2 · 2019-12-26T16:41:58.113

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We can pick $E\subseteq \mathbb{C}$. Note that $[\mathbb{C}:\mathbb{R}]=2$. Now use the tower law $$2=[\mathbb{C}:\mathbb{R}]=[\mathbb{C}: F]\cdot [F:\mathbb{R}]$$

edited Dec 26 '19 at 16:41

answered Dec 26 '19 at 16:39

Severin Schraven

19,179

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No, this is part of an exercise whose aim is to demonstrate that every splitting field of a polynomial is contained in $\mathbb C $, so I can't proceed like that. – Dr. Scotti Dec 26 '19 at 16:41
Thank you anyway – Dr. Scotti Dec 26 '19 at 16:42
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Sorry, I didn't know that you cannot use this fact :( – Severin Schraven Dec 26 '19 at 16:48
Maybe, Dorian, you should state the question so we know what you can and what you cannot use. – Gerry Myerson Dec 26 '19 at 17:04

Simple question about Galois theory

2 Answers2