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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.

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

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