Laurent powers of a transcendental element are linearly independent

For any universe-polymorphic types $K$ and $E$, equipped respectively with field structures and with an algebra structure of $E$ over $K$, and for any $z\in E$ that is transcendental over $K$, the family $(z^n)_{n\in\mathbb Z}$, using integer powers and hence both positive and negative powers, is linearly independent over $K$. Explicitly, every finitely supported family of coefficients $a_n\in K$ satisfying $\sum_{n\in\mathbb Z}a_n z^n=0$ in $E$ has every $a_n=0$. There is no finiteness or finite-dimensionality assumption on either field extension. No separate hypothesis $z\ne0$ is stated; the transcendence hypothesis excludes $z=0$, since $0$ is algebraic. For field extensions in which no transcendental element exists, there is no $z$ satisfying the hypothesis, so the theorem has no applicable instance.