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I need help for the following problem(what is the key-idea that problem):

Problem: Let $f(x)$ be a monic polynomial with rational coefficients. Assume $f(x)$ is irreducible in $\mathbb{Q}[x]$ and the Galois-group of $f(x)$ over $\mathbb{Q}$ is a group of order 99. What is the degree of $f(x)$?

Solution(my attempt): Let $\alpha$ be a root of $f \Longrightarrow \big[ \mathbb{Q}(\alpha) : \mathbb{Q} \big] = deg(f)$.

Let $K$ be the splitting field of $f$ over $\mathbb{Q} \Longrightarrow \Big|Aut_{\mathbb{Q}}(K) \Big| = 99$.

How can we compute the degree of the polynomial $f$?

Arturo Magidin
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    The Galois group of an irreducible polynomial of degree $n$ is a transitive subgroup of $S_n$. Because you were asked this question, I assume you have been given enough tools (of group theory) to narrow it down. Hard to give a more precise hint unless you actually tell us something about your group theory background. Sylow theory? Group actions? – Jyrki Lahtonen Dec 23 '23 at 19:13

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Note that a group of order $99=3^2\times 11$ is abelian: by Sylow's Theorems, there is a unique Sylow $11$-subgroup and a unique Sylow $3$-subgroup, and both are necessarily abelian by order considerations.

The degree of $f$ must divide the order of the Galois group. In addition, because the Galois group permutes the roots of $f$ and is completely determined by the action on those roots, it must be a transitive abelian subgroup of $S_n$, where $\deg(f)=n$.

So we know that $n$ divides $99 =3^2\times 11$. And that $99$ divides $n!$.

However, transitive abelian subgroups of $S_n$ have order $n$, so that would seem to require that $n=99$.

Now, does this make sense? Yes!

The degree can definitely be $99$, since by the Primitive Element Theorem the extension is simple, and if the extension is $\mathbb{Q}(a)$, then the minimal polynomial of $a$ is monic, irreducible, and of degree $99$. So from a purely "meta" perspective, if this answer is going to have a definite answer, that answer must be $99$... and the fact about abelian transitive subgroups seals the deal.

Arturo Magidin
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    In the interest of sealing the deal we can construct an explicit example as follows. Let $\zeta$ be a complex root of unity of order $19\cdot23$. A basic fact is that the extension $\Bbb{Q}(\zeta)/\Bbb{Q}$ is Galois with an abelian Galois group $G\simeq \Bbb{Z}_{23\cdot19}^*$ of order $\phi(19\cdot23)=2^2\cdot 99$. If $H\le G$ is the Sylow $2$-subgroup of $G$, then its fixed field $K=\mathrm{Inv}(H)$ is a degree $99$ extension of the rationals with $Gal(K/\Bbb{Q})\simeq G/H$. I leave it as an exercise for the OP to show that $K=\Bbb{Q}(\cos(2\pi/19),\cos(2\pi/23))$ :-). – Jyrki Lahtonen Dec 24 '23 at 15:39
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    @JyrkiLahtonen: I can't remember right away but I have seen this technique elsewhere on this site. It appears that cyclotomic extensions can be used to find abelian Galois groups of a given order and hence provide explicit examples of abelian extensions of a given degree. – Paramanand Singh Feb 24 '24 at 01:03
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    @ParamanandSingh Yes. The Kronecker-Weber Theorem says every abelian extension of $\mathbb{Q}$ is a subfield of a cyclotomic extension. So not only can they be used to find examples, all examples are essentially of that type. – Arturo Magidin Feb 24 '24 at 01:46
  • OK so this is KWT in action. To find an abelian extension of degree $m$ we find an $n$ such that $m\mid \phi(n) $ and then work with field $\mathbb{Q} (\zeta _n) $ and its subfields. – Paramanand Singh Feb 24 '24 at 01:58
  • By the way I saw this technique used here: https://math.stackexchange.com/q/4865451/72031 – Paramanand Singh Feb 24 '24 at 02:07
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    @ParamanandSingh Not sure I would call it "Kronecker-Weber in action". The theorem tells you any example will be contained in a cyclotomic extension. It's the Galois correspondence that lets you go from a particular cyclotomic one to an abelian group of the order you want (plus the fact the abelian groups satisfy the converse of Lagrange, I guess). – Arturo Magidin Feb 24 '24 at 02:32
  • @ParamanandSingh Correct. The technique has been used on this site many times as the general mechanism of constructing (Galois) extensions of $\Bbb{Q}$ of a prescribed degree. As Arturo explained, you don't need Kronecker-Weber to understand what's going on. I'm afraid I cannot communicate the ideas underlying K-W myself, but I do know that it is a number-theoretic result rather than just Galois theory. Chiefly because it is specific to the field $\Bbb{Q}$ only! – Jyrki Lahtonen Feb 24 '24 at 06:15
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    A bit of searching gave this hit. A cursory reading revealed that I might be able to understand that (non-class-field-theoretic) proof, if I put in the effort. You will see that the proof relies on Galois theory to reduce the main claim to specific simpler claims handling a single prime at a time. Anyway, bits and pieces from a first (extensive) course in algebraic number theory are also needed. – Jyrki Lahtonen Feb 24 '24 at 06:20