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.
 A: 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. 
A: I think you're looking at the problem the wrong way. Let $F$ be a field and let $\alpha$ be algebraic of degree $n$ over $F$. I claim that $F[\alpha]=F(\alpha)$ and that $1,\alpha,\dots,\alpha^{n-1}$ is a basis of $F(\alpha)$ over $F$. If $1,\dots,\alpha^{n-1}$ were linearly dependent over $F$ then $\alpha$ would satisfy a polynomial of degree less than $n$, so they are linearly independent. Let $p(t)$ be the minimal polynomial of $\alpha$ over $F$. Any element of $F[\alpha]$ is of the form $f(\alpha)$ for some polynomial $f \in F[t]$. We may divide $f(t)$ by $p(t)$ so that
$$f(t)=p(t)q(t)+r(t)$$
where $\deg r(t) <n$ evaluating at $\alpha$ we deduce that
$$f(\alpha)=p(\alpha)q(\alpha)+r(\alpha)=r(\alpha).$$
Since $r(t)$ is of degree less than $n$ it follows that $r(\alpha)$ can be expressed as a linear combination of $1,\alpha,\dots,\alpha^{n-1}$. Thereby $1,\dots,\alpha^{n-1}$ form a basis for $F[\alpha]$. Now to see that $F[\alpha]$ is indeed a field we have a map $\varphi: F[t] \rightarrow F[\alpha]$ generated by sending $\varphi(t)=\alpha$. By definition of the minimal polynomial $\ker \phi = \langle p(t) \rangle$. Since $\varphi$ is surjective we have by the first isomorphism theorem that $F[\alpha] \cong F[t]/\langle p(t) \rangle$. Finally $p(t)$ is irreducible so it generates a maximal ideal in $F[t]$ and we conclude that $F[t]/\langle p(t) \rangle$ is a field. 
