Why is necessary ask $F$ to be infinite in order to obtain: $ f(v)=0$ for all $ f\in V^* \implies v=0 $

Question

Let $V$ be a (possibly infinite dimensional) vector space over a field $F$ and $V^*$ its dual. Why is necessary ask $F$ to be infinite in order to obtain:

$$ f(v)=0 \text{ for all } f \in V^* \implies v=0. $$

(Probably not important, but to me $V$ is the Symetric Algebra)

Answer 1

Perhaps you're leaving something out. If $V^*$ is the space of all linear maps from $V$ to $F$ then the condition you ask about holds for an vector space over any field, finite or not:

Suppose $v\ne0$. Then $\{v\}$ is independent, so there is a basis $B$ for $V$ with $v\in B$. Let $f:B\to F$ be any function with $f(v)\ne0$. Then $f$ extends to a linear map from $V$ to $F$; that is, $f$ extends to an element of $V^*$.