Let $k$ be a number field, and $\{\alpha_i\}_{i=1}^r$ a $\mathbb Q$-basis of $k$. Why is $k=\mathbb Q(\alpha_1,\dots,\alpha_r)$?

Question

When we say $\{\alpha_i\}_{i=1}^r$ a $\mathbb Q$-basis of $k$, doesn't this mean that $k=\operatorname{span}_{\mathbb Q}\{\alpha_1,\dots,\alpha_r\}$, and therefore we can't get things like $\frac1{\alpha_i}$ which are in $\mathbb Q(\alpha_1,\dots,\alpha_r)$?

Or can we somehow get things like $\frac{f(\alpha_1,\dots,\alpha_r)}{g(\alpha_1,\dots,\alpha_r)}$ ($f,g\in\mathbb Q[X_1,\dots,X_r]$) by linear combinations of the basis? Using minimal polynomials or something?

I appreciate any assistance.

Answer 1

Well, if $k$ is an extension field of $\Bbb Q$, then $k$ is also a $\Bbb Q$-vector space.

In your example, $\alpha_1,\ldots,\alpha_n$ form a $\Bbb Q$-basis of $k$ and so each element of $k$ can be written as $$\sum_i c_i\alpha_i$$ where the $c_i$'s are rational numbers.

This includes that the inverses of the $\alpha_i$'s belong to $k$ (as $k$ is a field) and can also be represented this way.

Think about $\Bbb Q(\sqrt 2) = \{a+b\sqrt 2\mid a,b\in\Bbb Q\}$ with basis $\{1,\sqrt 2\}$. Here $$(a+b\sqrt 2)^{-1} = \frac{a}{a^2-2b^2} - \frac{b}{a^2-2b^2}\sqrt 2.$$

Answer 2

In the case of $k = \mathbb Q (\alpha) $ , if $f =x^n + a_{n-1} x^{n-1} +...+ a_0 $ is the minimal polynomial of $\alpha$ then ${\alpha}^{-1} = -{a_0}^{-1} ( {\alpha}^{n -1} + a_{n-1} {\alpha}^{n-1} .. a_1) $ since $ f(\alpha) =0 $. Similarly we can get for other finite extensions also. (Here $a_0 \neq 0$ as it is the constant of minimal polynomial.