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Let $K,L$ be two number fields and let $KL$ denote the composite field (the smallest subfield of $\mathbb{C}$ containing both $K$ and $L$). Denote respectively by $R,S$ and $T$ the ring of algebraic integers of $K,L$ and $KL$. Let $m,n$ be the degrees of $K$ and $L$ over the rationals and assume that $[KL:\mathbb{Q}]=mn$. Now, let $\alpha_1,\ldots,\alpha_n$ be a $\mathbb{Z}$-basis for $R$, and $\beta_1,\ldots,\beta_m$ a $\mathbb{Z}$-basis for $S$. Thus the $mn$ products $\alpha_i\beta_j$ form a basis for $RS$ over $\mathbb{Z}$.

The book I'm reading (Marcus, Number fields pag. 34) says that they also form a basis for $KL$ over $\mathbb{Q}$. How can be proven this?

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  • $\begingroup$ Think of all the fields as vector spaces over $\mathbb{Q}$. Since the algebraic integers are known to span the field, the result follows. $\endgroup$ – Mikhail Katz Aug 8 '13 at 15:21
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Well, $\alpha_1, \dots, \alpha_n$ form a basis for $K$ over $\mathbf Q$, and $\beta_1, \dots, \beta_n$ form a basis for $L$ over $\mathbf Q$. Therefore the products $\alpha\beta$ form a basis for $KL$ over $\mathbf Q$.

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  • $\begingroup$ Would you please provide details ? I am a beginner, I can't understand that at first sight. $\endgroup$ – 王李远 Jan 10 '18 at 2:59

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