Linked Questions

10 votes
5 answers
3k views

Prove $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[3]{2})$ [duplicate]

I've tried solving $\sqrt[3]{3} =a + b* \sqrt[3]{2}+c* \sqrt[3]{4}$, but there is no obvious contradiction, even when taking the norms/traces of both sides. I can't think of another approach. This is ...
Tahlor's user avatar
  • 225
-2 votes
1 answer
85 views

Show that the extension has degree 9 [duplicate]

Let $F=\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3})$. Show that over $\mathbb{Q}$ this is an extension of degree 9. I see that this is equivalent of showing that $\sqrt[3]{2}\notin \mathbb{Q}(\sqrt[3]{3})$. ...
questions's user avatar
10 votes
3 answers
534 views

Irreducibilty of polynomial $x^9-6x^6+282x^3-8$ over $\mathbb {Q} $

While trying to deal with the final parts of this answer I found that one needs to establish $$a=\sqrt[3]{1+\sqrt{11}}\notin\mathbb{Q} (\sqrt{11},\sqrt[3]{10})=F$$ Since both $a, F$ are of degree $6$ ...
Paramanand Singh's user avatar
  • 88.3k
9 votes
2 answers
837 views

$\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?

Is there an easy way to see that $$\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}?$$ I know that $\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})$ is a subfield of $\mathbb{Q}(\sqrt[...
Spenser's user avatar
  • 19.7k
4 votes
3 answers
302 views

Prove $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[4]{3})$

The question asks us to prove $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[4]{3})$ I realise this is similar to another question - Show that $\sqrt{2}\notin \mathbb{Q}(\sqrt[4]{3})$ - and in this question ...
user402253's user avatar
8 votes
2 answers
171 views

Are $a=3, b=7$ the only solutions to $\sqrt{3}+\sqrt[3]{7}=\sqrt{a}+\sqrt[3]{b}$ for $a,b \in \mathbb{Q}$?

So I had a lesson about calculating surds e.g. $\sqrt{5+2\sqrt{6}}$, then the teacher wrote the steps like that: For some $a,b\in\mathbb{Q}_{\ge 0}$ $$\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b} \\ 5+2\sqrt{...
MafPrivate's user avatar
  • 4,043
5 votes
2 answers
392 views

Why $[\mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q}):\mathbb{Q}] = nm$?

As title says, I want to prove that $\mathbb{Q}(\sqrt[n]{p}, \sqrt[m]{q})$ is degree $nm$ extension of $\mathbb{Q}$ when $p \neq q$ are distinct primes. By Eisenstein's criterion, $x^{n} - p$ is ...
Seewoo Lee's user avatar
  • 15.3k
2 votes
2 answers
537 views

Proving that $\sqrt 3\not\in Q(\sqrt[4]2)$ [duplicate]

I came across this problem while solving another one. I will show how far I could get on my own: Suppose that $\sqrt 3 \in Q(\sqrt[4]2)$. Since $Q(\sqrt3)$ is the smallest field containing both $Q$ ...
user2345678's user avatar
  • 2,895
7 votes
1 answer
227 views

Let $\alpha$ be a root of $x^3+x+1$ and $\beta$ be a root of $x^3+x+3$. Show that it is not possible that $\alpha\in\mathbb Q(\beta)$

Question : Let $\alpha$ be a root of $x^3+x+1$ and $\beta$ be a root of $x^3+x+3$. Show that it is not possible that $\alpha\in \mathbb Q(\beta)$. My proof : Given $\beta$ is a root of $x^3+x+3$. ...
Nick Diaz's user avatar
  • 403
1 vote
2 answers
210 views

Proving $\sqrt[3]{5} \notin \mathbb{Q}(\sqrt[3]{2})$ with solvability of system in $\mathbb{Q}$

We were asked to show that $\sqrt[3]{5} \notin \mathbb{Q}(\sqrt[3]{2})$ by reducing the problem to the solvability of system in $\mathbb{Q}$. There was also a hint with it that a system does not need ...
riescharlison's user avatar
2 votes
2 answers
127 views

Fundamental Question on how to prove $a \not\in K(b)$ where $a,b$ algebraic over $K$

I have a very fundamental question on how to prove something like $\sqrt{3} \not\in \mathbb{Q}(\sqrt{2})$. In all of the proofs trying to show something similar eg. here, or here it is shown that (for ...
G. Chiusole's user avatar
  • 5,456
3 votes
0 answers
91 views

A proof needed in minimal polynomial.

How do you prove this statement? If a, b are different prime numbers, the minimal polynomial of $\sqrt[n]{b}$ over the extension field $\mathbb{Q} (\sqrt[m]{a})$ of the rational number field $\mathbb{...
Richard's user avatar
  • 31