TITLE: Inverting the elements of a basis of a finite field extension
QUESTION [5 upvotes]: Let $K/k$ be a finite field extension of degree $d$. Suppose that $\{a_1, \dots, a_d\}$ is a basis of $K$ as a $k$-vector space. Is it true that $\{a_1^{-1}, \dots, a_d^{-1}\}$ is a basis of $K$ as a $k$-vector space as well? I'm primarily interested in the characteristic zero case.
In the characteristic zero setting, I think that a good start would be replacing $K$ by its normal closure. This allows us to assume that $K/k$ is a Galois extension. Let $A$ be a $d \times d$ matrix with coefficients in $k$ such that $(a_1^{-1}, \dots, a_d^{-1})^T = A \cdot (a_1, \dots, a_d)^T$. My question is equivalent to whether or not $A$ is invertible.

REPLY [4 votes]: No. Take $\alpha=\sqrt[3]2$ and $K/k = \Bbb Q(\alpha)/\Bbb Q$. Then
$$
\bigg\{ 1, \frac{\alpha^2}2, \frac{\alpha^2-\alpha+1}3 \bigg\}
$$
is a basis for $\Bbb Q(\alpha)$ over $\Bbb Q$. But
$$
\bigg\{ 1^{-1}, \bigg( \frac{\alpha^2}2 \bigg)^{-1}, \bigg( \frac{\alpha^2-\alpha+1}3 \bigg)^{-1} \bigg\} = \{ 1,\alpha,\alpha+1\},
$$
which is not a basis.
Of course, to construct this I started with $\{ 1,\alpha,\alpha+1\}$; there's no reason to think that their inverses would be linearly dependent (randomly chosen elements aren't). I'm sure similar examples can be found even in Galois extensions.