# Linked Questions

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 $\mathbb{Q}[\sqrt{2},\sqrt{3}]$? ...
3answers
1k views

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?
4answers
125 views

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

How can I show this field extension equality $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?
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...
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!
1answer
153 views

3answers
4k views

This is Exercise 18.14 from Algebra, Isaacs. $p_{1}\ ,\ p_{2}\ ,\ ... p_{n}$ are different prime numbers. How to show that $$\mathbb{Q}[\sqrt{p_{1}}, \sqrt{p_{2}}, \ldots, \sqrt{p_{n}} ] = \mathbb{Q}[... 3answers 533 views ### Is \sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}] true? Motivated by the positive answer to the following question: Is \mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})? I'm curious about whether {\Bbb Z}[\sqrt{2}+\sqrt{3}]=\Bbb{Z}[\... 4answers 2k views ### Minimal polynomial of \sqrt2+1 in \mathbb{Q}[\sqrt{2}+\sqrt{3}] I'm trying to find the minimal polynomial of \sqrt2+1 over \mathbb{Q}[\sqrt{2}+\sqrt{3}]. The minimal polynomial of \sqrt2+1 over \mathbb{Q} is$$ (X-1)^2-2. So I look at $\alpha = \sqrt2 ... 4answers 201 views ### I want know if my logic in showing that$\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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