Linked Questions

3
votes
4answers
3k views

Difference between fields $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ and $\mathbb{Q}[\sqrt{2},\sqrt{3}]$? [duplicate]

Possible Duplicate: Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$? How would one describe elements from $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ and $\mathbb{Q}[\sqrt{2},\sqrt{3}]$? ...
2
votes
3answers
1k views

How to prove $\mathbb{Q}[\alpha,\beta]=\mathbb{Q}[\gamma]$? [duplicate]

Possible Duplicate: Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$? On Page 379, Algebra, Artin(1991) Let $\mathbb{Q}[\alpha,\beta]$ denote the smallest subring of $\...
1
vote
3answers
2k views

Is it obvious that $\mathbb Q(\sqrt 3+\sqrt 5)=\mathbb Q(\sqrt 3,\sqrt 5)$? [duplicate]

Is it obvious that $\mathbb Q(\sqrt 3+\sqrt 5)=\mathbb Q(\sqrt 3,\sqrt 5)$ ? If not how can I show it ?
1
vote
3answers
536 views

Prove $\mathbb Q (\sqrt2 + \sqrt3 ) = \mathbb Q (\sqrt2 , \sqrt3 )$ [duplicate]

So correct if I am wrong, but these sets are: $\mathbb Q (\sqrt2 + \sqrt3 )=\{a+b\sqrt2 + c\sqrt3 +d \sqrt6 : a,b,c,d \in \mathbb Q \}$ $\mathbb Q (\sqrt2 , \sqrt3 )=\{a+b\sqrt2 + c\sqrt3 : a,b,c \...
6
votes
1answer
563 views

$ \Bbb Q [ \sqrt{2} + \sqrt{3} ] = \Bbb Q [ \sqrt{2} , \sqrt{3} ] $ [duplicate]

Prove, that $ \Bbb Q [ \sqrt{2} + \sqrt{3} ] = \Bbb Q [ \sqrt{2} , \sqrt{3} ] $ I don't know the definition of $\Bbb Q [ \sqrt{2} , \sqrt{3} ]$, can anyone help me with this?
1
vote
4answers
125 views

How can I show this field extension equality? [duplicate]

How can I show this field extension equality $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?
0
votes
2answers
327 views

Primitive element of the extension $\mathbb Q(\sqrt{2},\sqrt{3})$ over $\mathbb Q$ [duplicate]

The title says it. I want to find an element $\alpha$ such that $\mathbb Q(\alpha)=\mathbb Q(\sqrt{2},\sqrt{3})$. I tried something like $\sqrt{2}+\sqrt{3}$ but that didn't help...
-2
votes
1answer
188 views

Why is $\mathbb Q(\sqrt 2+\sqrt 3)=\mathbb Q(\sqrt2,\sqrt 3)$? [duplicate]

Why is $\mathbb Q(\sqrt 2+\sqrt 3)=\mathbb Q(\sqrt2,\sqrt 3)$ ? I am Having problems understanding why this is true. Any input would be greatly appreciated!
0
votes
1answer
153 views

Find $\alpha$ such that the given field is $\mathbb{Q}(\alpha)$ [duplicate]

This question is in regards to separable field extensions. I am to show that this $\alpha$ is in the given field and verify by direct computation that the given generators for the extension of $\...
-1
votes
1answer
179 views

Showing $\mathbf Q(\sqrt2,\sqrt3)=\mathbf Q(\sqrt2+\sqrt3)$ [duplicate]

Showing $\mathbf Q(\sqrt2,\sqrt3)=\mathbf Q(\sqrt2+\sqrt3)$ BUT I want to show this using The Theorem of the Primitive Element, So I have to verify that $c$ cannot be $1$ and I need the $\mathbf Q$...
30
votes
5answers
3k views

Can $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m$ are perfect squares?

Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}...
32
votes
3answers
4k views

How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?

This is Exercise 18.14 from Algebra, Isaacs. $p_{1}\ ,\ p_{2}\ ,\ ... p_{n}$ are different prime numbers. How to show that $$\mathbb{Q}[\sqrt{p_{1}}, \sqrt{p_{2}}, \ldots, \sqrt{p_{n}} ] = \mathbb{Q}[...
11
votes
3answers
533 views

Is $\sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}]$ true?

Motivated by the positive answer to the following question: Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$? I'm curious about whether ${\Bbb Z}[\sqrt{2}+\sqrt{3}]=\Bbb{Z}[\...
4
votes
4answers
2k views

Minimal polynomial of $\sqrt2+1$ in $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$

I'm trying to find the minimal polynomial of $\sqrt2+1$ over $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$. The minimal polynomial of $\sqrt2+1$ over $\mathbb{Q}$ is $$ (X-1)^2-2.$$ So I look at $\alpha = \sqrt2 ...
3
votes
4answers
201 views

I want know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct

I want to know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct Now firstly, it seems that $\Bbb Q(\sqrt 2,\sqrt 3)=(\Bbb Q (\sqrt 2))(\sqrt3){\overset{{\...

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