Inverses of elements in field extensions and inverses of basis elements Say I create an extension $K$ over a field $F$ obtained by adjoining an element $\alpha$, i.e. $K = F[\alpha]$ ($\alpha$ does not necessarily have to be the root of a polynomial with coefficients in $F$). In this process I might need to adjoin some powers of $\alpha$. Suppose I have a basis $B$ for $K$ and want to show that every element in $K$ has an inverse - that $K$ is indeed a field. If every element of $B$ has an inverse, is that enough to guarantee that every element in $K$ have an inverse, and is there a nice way to compute it? If not, is it true if the extension is finite-dimensional i.e. $|B|$ is finite?
An example of what I'm talking about: take (letting $\alpha = \sqrt[3]{2}$) $K = \mathbb{Q}(\alpha)$. I have my basis $B = \left\{1, \alpha, \alpha^2\right\}$, and clearly $\alpha^{-1} = \frac{\alpha^2}{2}$ and $\alpha^{2^{-1}} = \frac{\alpha}{2}$. Any element in $K$ is of form $a + b\alpha + c\alpha^2$ and has some inverse of the same form. If I were asked to prove that adding the two elements $\alpha, \alpha^2$ is enough to make $K$ into a field, i.e. we do not need to add more powers, by showing that everything with just the adjunction of those two elements is multiplicatively closed and has inverses, would the existence of inverses of $\alpha, \alpha^2$ guarantee the existence of inverses of general elements?
This is in trying to prove properties about the extension. This is indirectly homework, but this exact question is not so I would appreciate a full answer or a link to one.

Consider the extension $K$ of the field $F$, adjoining a root $\alpha$ of $x^2 - 1$. Clearly $1,\alpha$ is a basis, and both elements are invertible, but $\alpha-1$ is a zero-divisor, and thus not invertible.
PS I am assuming you only want $F$ to be a field. 
