Linked Questions

2
votes
1answer
94 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 $...
2
votes
2answers
97 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
310 views

$\sqrt{p_1}$ is not in $Q[\sqrt{p_2},…,\sqrt{p_n}]$ [duplicate]

How to show $\sqrt{p_1}$ is not in $Q[\sqrt{p_2},...,\sqrt{p_n}]$ if $p_1,...,p_n$ are distinct primes? Intuitively, this is pretty clear, but it makes me very uncomfortable to just believe. Any idea ...
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})(...
2
votes
1answer
146 views

Exercise about field extensions [duplicate]

Consider $a_1,\ldots,a_n\in \mathbb Z$. i) Suppose $a_1,\ldots, a_n$ are pairwise relatively prime. I have to see by induction on n that $[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_n):\mathbb Q]=2^n$ Once ...
1
vote
1answer
235 views

How to get this equation solved?

I came across this equation $$\sqrt a-\sqrt b=\sqrt 7-\sqrt 5$$ And you have to find the value of '$a$' and '$b$' when both of them are primes. The solution was $a=7, b=5$. Now, my question is, ...
1
vote
2answers
709 views

An element not in a field extension [duplicate]

Possible Duplicate: Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were ...
1
vote
5answers
123 views

A question about dense subsets of Euclidean spaces and Hilbert space

Let $S$ be the Euclidean plane and let $p(S)$ be a fixed point of $S$. Does there exist a dense subset $D(S)$ of $S$, such that no pair of distinct points of $D(S)$ are at the same distance from $p(S)$...
1
vote
1answer
146 views

How to convert $\Bbb Q(\sqrt 2,\sqrt 3)$ to $\Bbb Q(\alpha)?$

I have a basic question about algebraic field extensions: How can I convert a multiple extension like $\mathbb{Q}(\sqrt{2},\sqrt{3})$ to a single (elementary) field extension (like $\mathbb{Q}(\sqrt{...
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 ...
1
vote
0answers
34 views

Is $\{\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\dots\}$ l.i. over $\mathbb{Q}$? [duplicate]

In order to prove that $[\mathbb{R}:\mathbb{Q}] = \infty$, I was trying to construct an infinite linear independent subset of $\mathbb{R}$. Before noticing that $\{1,\pi,\pi^2,\dots\}$ does the job, I ...
1
vote
0answers
52 views

If $k_1, \ldots, k_n$ are non-square, pairwise coprime, then $\sqrt {k_n} \not \in \mathbf{Q}(\sqrt {k_1}, \ldots, \sqrt {k_{n-1}})$ [duplicate]

Seems intuitive. Like the fact that $\sqrt 3 \not \in \mathbf{Q}(\sqrt 2)$. But how to approach actually proving it? The proof of this fact doesn't seem to generalize well.
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 ...
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{...
1
vote
0answers
89 views

Linear independence of square root of square free numbers [duplicate]

Possible Duplicate: The square roots of the primes are linearly independent over the field of rationals I am reading a research article in which there is a theorem regarding square roots of ...

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