Linked Questions
11 questions linked to/from Why $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$?
10
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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 ...
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-2
votes
1
answer
72
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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 $\sqrt[3]{2}\notin \mathbb{Q}(\sqrt[3]{3})$. ...
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10
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3
answers
428
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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 $6$ ...
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9
votes
2
answers
798
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$\mathbb{Q}(\sqrt[3]{2})\cap\mathbb{Q}(\sqrt[3]{5})=\mathbb{Q}$?
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[...
- 19k
4
votes
3
answers
271
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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 ...
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8
votes
2
answers
149
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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{...
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4
votes
2
answers
334
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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 ...
- 14.5k
7
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1
answer
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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 $x^3+x+3$.
...
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2
votes
2
answers
386
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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 $Q$ ...
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2
votes
2
answers
120
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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 ...
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3
votes
0
answers
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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{...
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