Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Can you name four linearly independent elements of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$? – Hurkyl Apr 8 '12 at 3:43
up vote 8 down vote accepted

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 $\mathbb{Q}$.

share|cite|improve this answer

Your Answer


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.