Linked Questions

0
votes
3answers
61 views

Prove that root of number is rational.

Consider $ x_1, x_2, ..., x_n \in \mathbb{R}$. We have to prove that each $\sqrt x $ is rational if the sum of $\sqrt x_1 + \ldots + \sqrt x_n $ is rational. I think that I could prove it using ...
0
votes
2answers
77 views

Why need open in the Baire Category Theorem

In the statement of the Baire Category Theorem, one needs to include openness of a set so that the theorem holds. Question: what is the example such that countable intersection of dense sets is not ...
2
votes
2answers
57 views

What are the applications of the result that $x$ is not a square in $F(x)$?

Let $F(x)$ denote the field of quotients of the ring $F[x]$. Prove that there is no element in $F(x)$ whose square is $x$. Solution:If possible let $\frac{f(x)}{g(x)} \in F(x),g(x)\neq 0$ such ...
1
vote
0answers
143 views

Characterizing $\operatorname{Gal}(\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_n})/\mathbb{Q})$ for $p_i$ primes?

For what $n$ is $$ \operatorname{Gal}(\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_n})/\mathbb{Q}) $$ known, where the $p_i$ are primes? By Kummer theory, I think that $$ \operatorname{Gal}(\mathbb{Q}(\sqrt{...
0
votes
0answers
140 views

Prove linear independence of roots of square-free numbers over $\mathbb{Q}$ using induction

I want to prove that this system $\{1,\sqrt{m_1},\sqrt{m_2},...,\sqrt{m_n}\}$, where all $m_i$ are different square-free natural numbers, is linear independent over $\mathbb{Q}$. My teacher has asked ...
2
votes
2answers
99 views

How to prove the set $\{\sqrt{n}:\textrm{$n$ is squarefree}\}$ to be a linearly independent set?

As the title goes, I am stuck on this problem. Prove that the set $\{\sqrt{n}:\textrm{$n$ is squarefree}\} =\{1,\sqrt{2}, \sqrt{3},\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{10},\ldots\}$ is a linearly ...
2
votes
1answer
96 views

For $n\ge 3, x_{1},…,x_{n} \in \mathbf{Q}^{\ast}$, $[\mathbf{Q}(\sqrt{x_{1}},…\sqrt{x_{n}}) : \mathbf{Q}] < 2^{n}$

For $n\ge 3, x_{1},...,x_{n} \in \mathbf{Q}^{\ast}$ and $[\mathbf{Q}(\sqrt{x_{1}},...\sqrt{x_{n}}) : \mathbf{Q}] < 2^{n}$ how can we conclude that there are non empty $I \subset \{1,...,n\}$ with $...
0
votes
1answer
69 views

How to prove that $\sqrt{p_n} \notin \mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{n-1}})$

I want to prove $\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{n-1}})=\mathbb{Q}(\sum\limits_{i =1}^n\sqrt{p_i})$, where $p_i$ are different prime integers. But I must solve this problem firstly: $\sqrt{p_n} ...
1
vote
2answers
49 views

Another (in)dependence over the nonzero rationals question

About one hour ago I asked a question which at first sight looked non-trivial to me but it is really trivial. Shame on me, whether I want it or not. Now I have, solely for fun, another question which ...
2
votes
1answer
69 views

Splitting field of $(x^2-2)(x^6-20)$ over $\mathbb{Q}$

I have to determine the splitting field $K$ of $f(x)=(x^2-2)(x^6-20)$ over $\mathbb{Q}$. My attempt of solution: $K=\mathbb{Q}(\sqrt2, \sqrt[6]{20}, i\sqrt3)$; $d_1:=[\mathbb{Q}(\sqrt2, \sqrt[6]{20})(...
0
votes
0answers
69 views

How to prove $a_{1}\sqrt[b_{1}]{c_{1}}+a_{2}\sqrt[b_{2}]{c_{2}}+…+a_{n}\sqrt[b_{n}]{c_{n}}$ is irrational?

Let's define the number $$A=a_{1}\sqrt[b_{1}]{c_{1}}+a_{2}\sqrt[b_{2}]{c_{2}}+.....+a_{n}\sqrt[b_{n}]{c_{n}}$$ where $a_{1}, a_{2}, ..., a_{n}$ are positive integers and $b_{1}, b_{2}, ..., b_{n}, ...
1
vote
1answer
54 views

$\sum_{i=1}^n {(a_i\sqrt{b_i})} \ne 0$

In a surd $a\sqrt{b}$   ($b \in \mathbb{Z^+}$)   the value of $b$ can assumed to be a square-free integer ($b = p_1p_2\dots p_k$, where $p_i$ are distinct primes), since otherwise a ...
0
votes
1answer
34 views

Identifying squares in field extension

I'm trying to solve problem 4.11 of http://homepages.warwick.ac.uk/~masda/MA3D5/Galois.pdf. The problem is, given $a,b\in k$ a field, and assuming that $\sqrt{a}\in k(\sqrt{b})$ prove that, either $\...
0
votes
0answers
20 views

Is $\mathbb Q(\sqrt 2) \times \mathbb Q(\sqrt 3)=\mathbb Q(\sqrt 2,\sqrt 3)$ if I prove $\sqrt 2,\sqrt 3$ are L.I. over $\mathbb Q$? [duplicate]

I proved that $\{1,\sqrt 2\}$ and $\{1,\sqrt 3\}$ are respective bases of $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$ over $\mathbb Q$. I want to show in some sense that since $\sqrt 2,\sqrt 3$ are ...

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