Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know that : $\mathbf{Q}(\sqrt{2}) = \mathbf{Q}+ \sqrt{2} \mathbf{Q}$ , but then what is $\mathbf{Q}(\sqrt{2},\sqrt{3})$?

share|improve this question

1 Answer 1

up vote 5 down vote accepted

$\mathbf{Q}(\sqrt{2},\sqrt{3})$ means $\mathbf{Q}+\sqrt{2}\mathbf{Q}+\sqrt{3}\mathbf{Q}+\sqrt{6}\mathbf{Q}$, or in other words

$$\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\mid a,b,c,d\in\mathbf{Q}\}.$$


Be careful though. For example, $\mathbf{Q}(\sqrt{2},\sqrt[4]{2})=\mathbf{Q}(\sqrt[4]{2})=\mathbf{Q}+\sqrt[4]{2}\mathbf{Q}+(\sqrt[4]{2})^2\mathbf{Q}+(\sqrt[4]{2})^3\mathbf{Q}$, because adding in the $\sqrt{2}$ is redundant: we already have $\sqrt{2}=(\sqrt[4]{2})^2$ inside $\mathbf{Q}(\sqrt[4]{2})$.

In general, the field $\mathbf{Q}(a_1,\ldots,a_n)$ is the smallest field containing $\mathbf{Q}$ and the elements $a_1,\ldots,a_n$.

share|improve this answer
    
Thank you, Zev. –  Tashi Dec 22 '11 at 13:24
    
@Zev And what about $\mathbb Q(x)\:$ ? –  Bill Dubuque Dec 22 '11 at 16:58
    
@Bill I think it is a correct intuition to say $\mathbb{Q}(x)$ is the smallest field containing $\mathbb{Q}$ and $x$ even when $x$ is an indeterminate, even if we must explicitly define $\mathbb{Q}(x)$ to avoid being circular. –  Zev Chonoles Dec 23 '11 at 12:47
    
@Zev My point was merely that this case too is worthy of mention. –  Bill Dubuque Dec 25 '11 at 15:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.