# Basis for $\mathbb{Q}(\sqrt{2},\sqrt{3})$

How can I find a basis for the field extension $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$?

I can show that [$\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=4$ (so I am looking for 4 basis elements).

$\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2})(\sqrt{3})$ and a basis for $\mathbb{Q}(\sqrt{2})$ is $a+b\sqrt{2}$, I can also show that $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$ but I do not know how this information can help me. Thanks.

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Can you name four linearly independent elements of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$? –  Hurkyl Apr 8 '12 at 3:43

Dedekind's Product Theorem, which proves that $$[M:K] = [M:L][L:K]$$ for any tower of extensions $K\subseteq L\subseteq M$, is proven by showing that if $\{\ell_i\}_{i\in I}$ is a basis for $L$ over $K$, and $\{m_j\}_{j\in J}$ is a basis for $M$ over $L$, then $\{m_j\ell_i\}_{(i,j)\in I\times J}$ a basis for $M$ over $K$.
You know that $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{2})\subseteq \mathbb{Q}(\sqrt{2})(\sqrt{3})$. You've shown that $\{1,\sqrt{3}\}$ is a basis for $\mathbb{Q}(\sqrt{2})(\sqrt{3})$ over $\mathbb{Q}(\sqrt{2})$ (since the extension is of degree $2$), and you know that $\{1,\sqrt{2}\}$ is a basis for $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$. You have all the ingredients you need.
Note: You write: "a basis for $\mathbb{Q}(\sqrt{2})$ is $a+b\sqrt{2}$". This is incorrect as written: a general element of $\mathbb{Q}(\sqrt{2})$ is of the form $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$, but neither a single element, nor all these elements, form a basis for $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$.