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2 votes

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 ...
Kenta S's user avatar
  • 12.4k
0 votes

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:...
Iqra Arshad's user avatar
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 ...
KReiser's user avatar
  • 58.3k
0 votes

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 $f\in F[x]$, ...
FShrike's user avatar
  • 26k
2 votes

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 ...
Caden Matthews's user avatar
3 votes
Accepted

Minimal Polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}(e^{\frac{2\pi i}{3}})$ and degree of field extension.

Actually there is a simpler argument. Look at the two towers : $\mathbf{Q} \subset \mathbf{Q}(\sqrt[3]{2}) \subset \mathbf{Q}(\sqrt[3]{2}, \zeta_3)$ and $\mathbf{Q} \subset \mathbf{Q}(\zeta_3) \subset ...
Anton Odina's user avatar
1 vote
Accepted

Proof of $Gal(K_1K_2/K_1)\cong Gal(K_2/(K_1\cap K_2))$

I will rephrase your question to a more general case. Let $K \subset \Omega$ be an arbitrary field extension and assume that there exist two intermediate fields $M$ and $L$, such that $K \subset L$ is ...
Anton Odina's user avatar
1 vote
Accepted

Issues understanding $|\text{Gal}(E/F)| = [E:F] \iff E/F$ is Galois.

I believe $L$ is an algebraic closure of $F$. Let's keep it that way. I believe for $(2)$ it is supposed to say $\tau_1$ becomes the identity on $F$, because that is your inclusion in $L$. I will ...
Anton Odina's user avatar
0 votes

Is a polynomial equation of degree $\ge 5$ not solvable by any way?

If it is a trinomial you can solve it analytically in terms of Lambert-Tsallis function. Look some of my previous answers.
K Z Nobrega's user avatar
1 vote

Field extension generated by coefficients of minimum polynomial

We have the extensions $K \subset E \subset K(\alpha)$, where $\alpha$ is algebraic over $K$. Therefore, $\alpha$ is algebraic over $E$. We write $f^{\alpha}_E := \sum^n_{k=0} a_k x^k \in E[x]$ and $L:...
Anton Odina's user avatar
6 votes
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 ...
Paramanand Singh's user avatar
  • 83.4k
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 ...
Pace Nielsen's user avatar
0 votes

How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

Yet an other solution, that covers the general case: $$ \bbox[lightblue]{\qquad \xi=\xi(a) :=\frac 1{1+\sqrt[3]a-\sqrt[3]{a^2}} =\frac 1{1+A-A^2} \ , \qquad A:=\sqrt[3]a\ .\qquad } $$ The problem ...
dan_fulea's user avatar
  • 29.4k
0 votes

How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

Let $x = \sqrt[3]{5}$ and let $y = \frac{1}{1 + x - x^2}$ (the value you want to solve for). Then: $$(1 + x - x^2)y = 1 \tag{1}$$ Multiplying by $x$ (and simplifying $x^3 = 5$) gives: $$(x + x^2 - 5)...
Dan's user avatar
  • 12.1k
0 votes

How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

By polynomial long division, $$\frac{x^3-5}{-x^2+x+1}=-x-1+\frac{2x-4}{-x^2+x+1}=0\tag1$$ $$x\frac{x^3-5}{-x^2+x+1}=-x^2-x-2+\frac{-2x+2}{-x^2+x+1}=0\tag2$$ and by adding $(1)$ and $(2)$ we have $-x^2-...
Bob Dobbs's user avatar
  • 6,262
0 votes

How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

Applying formulas for sum and difference of cubes, we can get immediately: $$\dfrac1{1+\sqrt[\large3]5-\sqrt[\large3]{25}} =\dfrac1{2-(1-\sqrt[\large3]5+\sqrt[\large3]{25})} =\dfrac{1+\sqrt[\large3]5}{...
Yuri Negometyanov's user avatar
4 votes
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 ...
Servaes's user avatar
  • 60.2k
2 votes

How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

Let $\,x = \sqrt[3]5,\,$ so $\,\color{#90f}{x^3 = 5},\,$ so $\,(x\!-\!a)(\overbrace{x^2\!+\!ax\!+\!a^2}^{\textstyle \color{#c50}{g_a}})=\!\color{#90f}{\overbrace{x^3}^{5}}\!\!-a^3 = \begin{cases}\ \ \ ...
Bill Dubuque's user avatar
3 votes
Accepted

How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

Now, since $x^3-5$ is the minimal polynomial of $\sqrt[3]{5}$ in $\mathbb Q[x]$, I think I want to find the inverse of $1+x-x^2$ in the quotient ring $\mathbb Q[x]/(x^3-5)$, but I'm not sure how. ...
dxiv's user avatar
  • 74.9k
8 votes

How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

We use $\color{red}{\text{Extended Euclidean Algorithm}}$ (for polynomials), which is a general method to deal with this type of rationalization problems, without introducing any tricks. Let $P=x^3-5, ...
MathFail's user avatar
  • 16.4k
3 votes

How can I prove this regarding Cyclotomic Cosets?

You are looking for the smallest integers in cyclotomic cosets modulo $N=2^m-1$. If $\mathcal{C}$ is such a coset and $N-x$ is the smallest element in it, then you cannot have $2x<N$ for otherwise $...
Jyrki Lahtonen's user avatar
5 votes

How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

Set $z=\sqrt[3]{5}$ and solve with the help of $z^3=5$ \begin{align*} (az^2+bz+c)\cdot (-z^2+z+1)&=1\\[6pt] -az^4+(a-b)z^3+(a+b-c)z^2+(b+c)z+c&=1\\[6pt] (a+b-c)z^2+(b+c-5a)z+(5a-b5+c)&=1\\[...
Marius S.L.'s user avatar
  • 1,529
4 votes

How to rationalize the denominator $\frac{1}{1 + \sqrt[3]{5} - \sqrt[3]{25}}$

My goal will be to simply derive expressions in the denominator in terms of $\thinspace x^{3n}\thinspace . $ Indeed, $\thinspace x=\sqrt [3]{5}\thinspace $ leads to the fraction $\thinspace \frac{1}{...
lone student's user avatar
  • 12.8k
0 votes

Example of a field on which every irreducible polynomial has degree a power of $p$

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. $\...
orangeskid's user avatar
  • 51.6k
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 ...
student91's user avatar
  • 2,615
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, \...
Acrobatic's user avatar
  • 794
-1 votes

Automorphism of the group of rational numbers under addition

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 \...
Mubeen Ansari's user avatar

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