Relation between the Axiom of Choice and a the existence of a hyperplane not containing a vector In a lot of problems in linear algebra one uses the existence, for each $E$ vector space over a field $k$, and each $x\in E$, of a Hyperplane $H$ such that $E=k\cdot x \oplus H$ (Let us denote $\mathcal{P}$ this property). With Zorn's Lemma, the existence of a such $H$ is trivial. However, as this seems weaker than the existence of a basis of $E$, maybe $\mathcal{P}$ does not imply the Axiom of choice. Thus my question is: is $\mathcal{P}$ equivalent to the Axiom of Choice and if not, is there a weaker form of the Axiom of Choice equivalent to $\mathcal{P}$?
 A: It is not hard to see that this statement is equivalent to "In every vector space, for every vector $v$ there is a functional $f$ such that $f(v)=1$". 
If $\cal P$ holds, then the projection onto $k\cdot x$ is a linear functional which is nontrivial; if there is a nontrivial functional then choose $x$ which is mapped to $1_k$, and consider $H$ as the kernel of the functional. Some $\cal P$ is equivalent to "If $V$ is nontrivial, then $V^*$ is nontrivial".
As of October 2018, it is still unknown whether or not this is equivalent to the axiom of choice in full. But there are some intermediate results, for example it is consistent that there is a vector space over a field $k$ whose dual is trivial, for any fixed $k$. Moreover, this is consistent with $\sf DC_\kappa$ for any prescribed $\kappa$.
One should remark that restricting to Banach spaces and continuous functionals, the nontriviality of the [topological] dual is equivalent to the Hahn–Banach theorem.
Let me also add that Marianne Morillon showed that in $\sf ZFA$, $\cal P(\Bbb Q)$, namely restricting our attention to vector spaces over $\Bbb Q$, is not enough to deduce that every vector space over $\Bbb Q$ admits a basis, and that a weaker form of choice called "Axiom of Multiple Choice" implies $\cal P$ for fields of characteristics $0$.

Morillon, Marianne, Linear forms and axioms of choice., Commentat. Math. Univ. Carol. 50, No. 3, 421-431 (2009). ZBL1212.03034.

These two facts are not telling us a whole lot about what happens in $\sf ZF$, since in $\sf ZF$ the axiom of multiple choice implies full choice.
