Linked Questions

6
votes
1answer
503 views

Elementary proof for $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ where $p_i$ are different prime numbers. [duplicate]

Take $p_1, p_2, \ldots, p_n, p_{n+1}$ be $n+1$ prime numbers in $\mathbb{P} \subseteq \mathbb{N}$. $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ seems to be quite ...
0
votes
2answers
230 views

Prove that $\left[\mathbb{Q}\left(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}}\right):\mathbb{Q}\right]=2^{n}$ [duplicate]

Let $p_{1},p_{2},\ldots,p_{n}$ be $n$ primes,$\left(p_{i},p_{j}\right)=1$ if $i\neq j$ . Prove that $\left[\mathbb{Q}\left(\sqrt{p_{1}},\sqrt{p_{2}},\ldots,\sqrt{p_{n}}\right):\mathbb{Q}\right]=...
4
votes
1answer
113 views

Finding degree of the extension [duplicate]

Is it true that the degree of extension $\mathbb Q(\sqrt {2},\sqrt {3},\sqrt {5},\dotsc,\sqrt {p_n}) / \mathbb Q$ is $2^n$ where $p_n$ is the $n$th prime number. If so, how to prove this? My idea is ...
1
vote
2answers
75 views

Dimension of the vector space $\mathbb{Q}(\sqrt{3}, \sqrt{5}, \sqrt{11})$ over $\mathbb{Q}$ [duplicate]

Let $\mathbb{Q}(\sqrt{3}, \sqrt{5}, \sqrt{11})$ be the smallest field that contains all rational numbers, $\sqrt{3}, \sqrt{5}$ and $\sqrt{11}$. Consider this field to be a vector space over $\...
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 ...
0
votes
1answer
48 views

Dimensions of Field Extensions [duplicate]

I have a Galois theory exercise to prove that $$[\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5):\mathbb Q]=8$$ I understand the proof up until the bit where If I prove $$\sqrt5 \notin \mathbb{Q}(\sqrt2,\sqrt3)$$ ...
0
votes
1answer
99 views

An infinite extension of $\mathbb Q$ [duplicate]

Let $S=\{\sqrt p \in \mathbb R | p $ is a primer number$\}$. How can I show that $\mathbb Q(S)|\mathbb Q$ is an infinite field extension?
0
votes
0answers
21 views

compute the grades over $\mathbb{Q}$ [duplicate]

Let $p_{1}$ $\neq$ $p_{2}$ $\neq$ $p_{3}$ prime numbers. Compute the grades over $\mathbb{Q}$ of the extension fields $\mathbb{Q} ( \sqrt{p_{1}}, \sqrt{p_{2}})$ and $\mathbb{Q} ( \sqrt{p_{2}}, \sqrt{...
129
votes
3answers
15k views

The square roots of different primes are linearly independent over the field of rationals

I need to find a way of proving that the square roots of a finite set of different primes are linearly independent over the field of rationals. I've tried to solve the problem using elementary ...
63
votes
3answers
6k views

Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$

Let $f(x) \in \mathbb{Z}[x]$. If we reduce the coefficents of $f(x)$ modulo $p$, where $p$ is prime, we get a polynomial $f^*(x) \in \mathbb{F}_p[x]$. Then if $f^*(x)$ is irreducible and has the same ...
60
votes
1answer
3k views

Is $\sqrt1+\sqrt2+\dots+\sqrt n$ ever an integer?

Related: Can a sum of square roots be an integer? Except for the obvious cases $n=0,1$, are there any values of $n$ such that $\sum_{k=1}^n\sqrt k$ is an integer? How does one even approach such a ...
13
votes
7answers
1k views

Proof that $\sqrt6 - \sqrt2 - \sqrt3$ is irrational. [duplicate]

I want to prove that: $$\sqrt6 - \sqrt2 - \sqrt3$$ is irrational. I have tried using squares, the $p/q$ definition of rationality and the facts that 1)rational$\times$ irrational=irrational (unless ...
17
votes
2answers
4k views

Is $\mathbb Q(\sqrt{2},\sqrt{3},\sqrt{5})=\mathbb Q(\sqrt{2}+\sqrt{3}+\sqrt{5})$. [duplicate]

Is $\mathbf Q(\sqrt 2,\sqrt 3,\sqrt 5)=\mathbf Q(\sqrt 2+\sqrt 3+\sqrt 5)$? Say $L=\mathbf Q(\sqrt 2,\sqrt 3,\sqrt 5)$ and $K=\mathbf Q(\sqrt 2+\sqrt 3+\sqrt 5)$. It is easy to show that $\mathbf Q(\...
7
votes
5answers
496 views

$p,q,r$ primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational.

I want to prove that for $p,q,r$ different primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational. Is the following proof correct? If $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is rational, then $(\sqrt{p}+\sqrt{q}+...
10
votes
5answers
231 views

Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.

Question: Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational. Following from the question, I tried: ...

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