# How to find the basis of an extension field

Sorry for asking a simple question but why is it obvious that $$\{1,\sqrt{3}+\sqrt{5}\}$$ is a basis of $$\mathbb{Q}(\sqrt{3}+\sqrt{5})$$ over $$\mathbb{Q}(\sqrt{15})$$?

I know that $$[\mathbb{Q}(\sqrt{3}+\sqrt{5}) : \mathbb{Q}(\sqrt{15})] = 2$$ and that $$\{1,\sqrt{3}+\sqrt{5}\}$$ is a linearly independent set. Have trouble understanding why it spans. Am I correct in saying any element of $$\mathbb{Q}(\sqrt{3}+\sqrt{5})$$ over $$\mathbb{Q}(\sqrt{15})$$ is of the form $$a+b(\sqrt{3}+\sqrt{5})$$, where $$a,b\in \mathbb{Q}(\sqrt{15})$$ and thus we are able to express any element of $$\mathbb{Q}(\sqrt{3}+\sqrt{5})$$ over $$\mathbb{Q}(\sqrt{15})$$ using $$\{1,\sqrt{3}+\sqrt{5}\}$$ and thus it is the basis? What would be the basis of $$\mathbb{Q}(\sqrt{3}+\sqrt{5})$$ over $$\mathbb{Q}$$ (I know that its cardinality = 4)? Thanks.

• Vector spaces have many different bases, so you should refer to sets as being "a basis" rather than "the basis." – Robert Shore Apr 16 at 23:25

There is a very important theorem in linear algebra that states that a linearly independent set of $$n$$ vectors (when $$n$$ is finite) in $$n$$-dimensional vector space is a basis. Also, a set of $$n$$ vectors that span an $$n$$-dimensional vector space is a basis. Of course this is not necessary true if the dimension is infinite.