Inverse of an algebraic element over finite extension I am trying to understand finite extensions of $\mathbb{Q}$ and I have to ask something little that I could not able to understand and may be trivial to you. I will write down few arguments and I would like to know whether they are correct or not and some $2$ related questions.
Now, let $K=\mathbb{Q}(\alpha_1,\dots,\alpha_n)$.Then,
i.$K$ is the smallest field containing both  $\mathbb{Q}$ and $\{\alpha_1,\dots,\alpha_n\}$ and it is defined as $K:=\{\cfrac{a}{b}: b\neq 0, a,b \in \mathbb{Q} \cup \{\alpha_1, \dots, \alpha_n\}\}$.
ii.$K$ can be considered as a vector space $V$ over $\mathbb{Q}$ which has a basis $\{1,\alpha_1,\dots\alpha_n\}$ so that any $x \in K$ can be written as $q_0+q_1\alpha_1 +q_2\alpha_2+\alpha_nq_n$ for some $q_1, \dots, q_n \in \mathbb{Q}$.
iii.$K$ as a vector space has to have dimension $n$ by definition of a finite extension but my basis has $n+1$ many elements, so what do I think wrong?
iv.How can I write the inverse of $\alpha_1$ with respect to my basis? 

$\mathbb{Q}(\alpha_1,\dots,\alpha_n)$ is the set of all rational functions of $\alpha_1,\dots,\alpha_n$, not just the fractions on them.
$\mathbb{Q}(\alpha_1,\dots,\alpha_n)$ does not in general have dimension $n$ over $\mathbb Q$ because $\alpha_1,\dots,\alpha_n$ is not in general a linear basis.
If $\alpha$ is algebraic with minimal polynomial $a_n X^n + a_{n-1} X^{n-1} + \cdots + a_1 X + a_0$, then $\alpha^{-1}=\dfrac{1}{-a_0}(a_n\alpha^{n-1}+a_{n-1}\alpha^{n-2}+ \cdots +a_1)$. Note that $a_0\ne0$ because of minimality.
