I know that : $\mathbf{Q}(\sqrt{2}) = \mathbf{Q}+ \sqrt{2} \mathbf{Q}$ , but then what is $\mathbf{Q}(\sqrt{2},\sqrt{3})$?
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$\mathbf{Q}(\sqrt{2},\sqrt{3})$ 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 $a_1,\ldots,a_n$.
answered Dec 22, 2011 at 13:22
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$\begingroup$ @Zev And what about $\mathbb Q(x)\:$ ? $\endgroup$ Dec 22, 2011 at 16:58
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$\begingroup$ @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. $\endgroup$ Dec 23, 2011 at 12:47
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$\begingroup$ @Zev My point was merely that this case too is worthy of mention. $\endgroup$ Dec 25, 2011 at 15:49