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# Tag Info

### Properties of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$-orbits

First question: giving a $\gamma\in\Omega(\alpha)$ is equivalent to giving an embedding $\mathbb Q(\alpha)\hookrightarrow\overline{\mathbb Q}$, by sending $\alpha$ to $\gamma$. Now standard arguments ...

### Proving the tower law through an inductive proof

We shall proof by mathematical induction True for n=1 [K1:k0] = [K1:k0] Suppose true for n=t [Kt:k0]=[kt:kt-1][kt-1:kt-2]...[k1:k0] (1) We have to proof for n =t+1 That is [Kt+1:k0]=[kt+1:kt][kt:...
1 vote
Accepted

### Is the Galois action on $k$-algebra homomorphisms transitive?

A $k$-map $\operatorname{Spec} F \to X$ is equivalent to a choice of point $x\in X$ and a map of residue fields $k(x)\to F$ (ref), so we get a composition of extensions $k\subset k(x) \subset F$. So ...

### Clarifying the meaning of insolvability of quintic

The trouble is, how do you define "can be expressed with"? The Galois-theoretic theorem and proof states the problem precisely as: For a given field $F$ and a given polynomial $f\in F[x]$, ...

### Galois group of $(x^3-2)(x^2-2)$ over $\mathbb{Q}$

Let $L_3$ denote the splitting field of $x^3 - 2$ and $L_2$ denote the splitting field of $x^2-2.$ Now, we compute the Galois groups of the two polynomials. The cubic has Galois group $S_3$ since we ...
Accepted

Accepted

### How large is the gap in Ruffini's 1813 proof that there is no general quintic formula?

The result by Abel (also called theorem on natural irrationalities) is very important in the proof of insolvability of a general quintic. It ensures that at any stage of forming radicals over field of ...
1 vote

### Number field, $K_i\neq K_{i+1}$

I personally found this question fascinating, and am surprised it didn't get more upvotes. Here is a solution. Let $K$ be a number field, and define $K_n$ as in the question. We can characterize ...

### How to rationalize the denominator $\frac{1}{1 + \sqrt{5} - \sqrt{25}}$

Yet an other solution, that covers the general case: $$\bbox[lightblue]{\qquad \xi=\xi(a) :=\frac 1{1+\sqrta-\sqrt{a^2}} =\frac 1{1+A-A^2} \ , \qquad A:=\sqrta\ .\qquad }$$ The problem ...

Accepted

### How does $x^5-5$ factor over $\mathbb{F}_p$ for different values of $p$ mod 5?

There is really no Galois theory involved here. Consider the map $$\varphi:\ \Bbb{F}_p^{\times}\ \longrightarrow\ \Bbb{F}_p^{\times}:\ x\ \longmapsto\ x^5.$$ This is a group homomorphism of cyclic ...

Hint: You can try to look at the union of the fields $\mathbb{F}_{l^n}$ inside $\overline{\mathbb{F}}_l$, for $n$ powers of $p$. $\bf{Added:}$ for a more sophisticated example, see Iwasawa theory. $\... 3 votes Accepted ### Fixed field theorem You are mis-reading the proof.$\gamma$is a primitive element, so it is not fixed by any$\sigma$, unless$\sigma$fixes every element of$K$. But then$\sigma$is the identity. This is exactly what ... 4 votes Accepted ### If$K \subseteq \mathbb{C}$is a Galois extension of$\mathbb{Q}$and$\mathrm{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/4\mathbb{Z}$, then$i \notin K$? Suppose$\mathrm{Gal}(K/\mathbb{Q})=\mathbb{Z}/4\mathbb{Z}$and$i\in K$. Suppose the group$\{\mathrm{id}, \sigma\}$fix the field$\mathbb{Q}(i).$Let$\tau$be the complex conjugation.$\sigma, \...
Every automorphism $Φ$ of the rational numbers $\Bbb Q$ under addition to itself has the form $Φ(x)=xΦ(1)$. Proof: Let $Φ:(Q,+)→(Q,+)$ be an automorphism. Let \$x∈Q ⇒ x=m/n \text{ where } m,n\in\Bbb Z \...