Algebraic dual $E^*$ separates $E$?

Question

Let $E$ be a vector space over a field $F$. Let $E^*:=\{f:E\rightarrow F : f \text{ is linear}\}$ be the algebraic dual of $E$. Suppose $E^*\ne \{0\}$ (i.e. take for granted the axiom of choice).

Is it always true that $E^*$ separates $E$ ? i.e. is it true that $$ \forall x\in E\backslash\{0\}, \; \exists f \in E^* \; s.t. \; f(x)\ne 0 \; \text{?} $$

Answer 1

Yes, because there exists a subspace $E'$ of $E$ such that $E=Fx\oplus E'$. Then $Fx\to F:ax\mapsto a$ and $0:E'\to F$ extends to a unique $F$-linear map $f:E\to F$ such that $f(x)=1\neq 0$.

Answer 2

With the axiom of choice: fixing $x$, find a basis $\{x\} \cup \{e_i\}_{i \in I}$ of $E$, so that each element of $E$ can be written uniquely as $\alpha x + \sum_{i \in I} \lambda_i e_i$, with only finitely many $\lambda_i$ unequal to 0. Now define $$ f\left(\alpha x + \sum_{i \in I} \lambda_i e_i \right) = \alpha. $$ Then $f \in E^*$ and $f(x) = 1 \neq 0$.