# Tagged Questions

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### Prove that $K=k(\alpha)$

Prove: If K\k is a Galois extension and $\alpha \in K$ with $\sigma(\alpha)\neq \alpha$ for all $\sigma \in Gal(K,k)\backslash \lbrace id_K \rbrace$, than $K=k(\alpha)$.
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### Proving that a certain maximal subfield has countably infinite index

Let $K$ be a maximal subfield of $\mathbb C$ which does not contain $\sqrt{2}$. I've shown that $\mathbb C$ is an algebraic extension of $K$, and that the Galois group of any finite extension of $K$ ...
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### Criteria for a solvable septic equation

I have a certain 7x7 matrix whose elements are all symbolic, and I want to know the eigenvalues. I have to solve a septic equation, but it is generally impossible. However, I am only interested in the ...
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### Determining if a given equation is solvable given a set of ultra-radicals

So suppose someone is armed with the tools of standard arithmetic, exponents (and of course that comes along with roots) AS WELL AS a set of inverses for some polynomials which are not solvable using ...
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### Is the intersection of the conjugates of a subnormal subgroup of prime power index also a subgroup of prime power index?

I was wondering if it's really the case that, if $G$ is a group with subgroups $H$ and $N$ such that $H\unlhd N\unlhd G$ such that $G/N$ and $N/H$ is a $p$-group, then the intersection of all the ...
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### Is the order of $\operatorname{Gal}(\bar{\Bbb Q}/\Bbb Q)$ known?

Is the order of $\operatorname{Gal}(\bar{\Bbb Q}/\Bbb Q)$ known? And if so is there a description on how the order can be found? My initial thoughts is that because $\bar{\Bbb Q}$ is countable and ...
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### Characterisation of Galois Group with the action of $\sigma \in S_n$ on the roots

Let $f \in K[X]$ be irreducible and separable with roots $x_1,...,x_n$ in a splitting field $L$ of $f$ over $K$. We identify $\text{Gal}(L|K)$ with $\text{Gal}(L|K)\cong G\subset S_n$. How can I see ...
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### Galois Group of a cubic polynomial

Let $P=X^3+a_1X^2+a_2X+a_3 \in \mathbb{Q}[X]$ be irreducible, $x_1,x_2,x_3$ the roots of $P$ and $L:=\mathbb{Q}[x_1,x_2,x_3]$. The galois group of $P$ is isomorph to $S_3$. Now we define ...
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### How can i give this isomorphism between a Galois group and $S_3$?

I have shown that if $x$ is the nontrivial cubic root of the unity and that if $y$ is the real cubic root of $2$, then $Q(x,y)$ is a Galois extension whose Galois group has order $6$. I know that the ...
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### Problem of Galois Extension

$\Bbb K$ is a non-Galois extension of $\Bbb Q$ and $[K:\Bbb Q]=4$. If $\Bbb F$ is the Galois closure of $\Bbb K$ then show that $Gal(\Bbb F/\Bbb Q)$ is either $S_4, A_4$ or $D_8$ with order 8. ...
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### Galois Theory problem (primitive roots of unity)

If $e_1,e_2......,e_{p-1}$ denote primitive $p^{th}$ roots of unity. Here $p$ is prime. And set the sum of the $n^{th}$ powers of the $e_i$ as $p_n=e^n_1+e^n_2......+e^n_{p-1}$ Now I want to show that ...
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$Gal(\Bbb K/\Bbb F)\cong S$. Where $\Bbb K$ is extension of $\Bbb F$. And $S= G_1\times G_2\times\cdots\times G_K$ is a solvable group. Where each $G_i$ are groups of prime power order and $o(S)= n.$ ...
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### A question on solvable groups.

Let's suppose $S$ is a group and can be written as $S= G_1\times G_2\times\cdots\times G_K$; here each $G_i$ is a group of prime power order. And $o(S)= n.$ Is this $S$ solvable group? If yes how will ...
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### Galois groups of intermediate fields

Suppose $k\subset E$ is Galois and let $F$ and $F'$ be two intermediate fields. Let $FF'$ be the smallest intermediate field containing $F$ and $F'$. Also let $G$ denote $\text{Gal}(E/k)$. Let ...
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### Splitting field of resolvent equals that of $f$

Lemma: Let $\Psi \in k[X_1,...,X_n]=:B$ be s.t. $stab_{S_n}(\Psi)=H \subset S_n$, $S_n/H=\{ \Psi, t_2 \Psi,..., t_e\Psi \}$, $\Delta_\Psi$ the discriminant of $L_\Psi:=\prod_{i=1}^e (X-t_i \Psi)$, and ...
### Possibilities for $[KL:F]$ when $[K:F]=[L:F]$ is prime
Suppose $K/F$ and $L/F$ are extensions of $F$ (contained in some common field) of degree $p$, where $p$ is prime. Standard arguments show that $[KL:F]$ must be in $\{p,2p,\ldots,p^2\}$. But are all ...