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 $\mathcal{P}$?

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    $\begingroup$ Just for the record, choice is definitely necessary to show this result: taking $E$ to be $\mathbb R$ as $\mathbb Q$-vector space, $H$ would have to be a subset of $\mathbb R$ isomorphic to $\mathbb R/\mathbb Q$, but it is consistent that this set can't be linearly ordered. $\endgroup$ – Wojowu Oct 22 '18 at 13:21

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.

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    $\begingroup$ I have literally no idea what any of that means. $\endgroup$ – Asaf Karagila Oct 22 '18 at 13:59
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    $\begingroup$ @მამუკაჯიბლაძე 1-dimensional vector space always has nonzero functionals - if $v$ is nonzero vector, then every vector is of the form $\alpha v$ and $\alpha v\mapsto\alpha$ is a nonzero functional. I think your confusion might arise from misunderstanding what "nonzero" here is supposed to mean - it just means any functional which is not zero everywhere. This is different from vector bundle terminology, where "nonzero" means that it's everywhere nonzero. $\endgroup$ – Wojowu Oct 22 '18 at 14:05
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    $\begingroup$ @მამუკაჯიბლაძე: Let me point out yet again, I have literally no idea what you're talking about. $\endgroup$ – Asaf Karagila Oct 22 '18 at 14:42
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    $\begingroup$ The statement in the question was "for each $x$ ... there is $H$ ...." So I think the equivalent statement should not be just that there is a non-zero linear functional but that, for each vector $x\neq 0$, there is a linear functional $f$ with $f(x)\neq0$. $\endgroup$ – Andreas Blass Oct 22 '18 at 15:56
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    $\begingroup$ Another equivalent formulation is "a system of linear equations over a field F has a solution in F if and only if every finite sub-system has a solution in F." The big advantage of this reformulation is that it's in the Howard-Rubin book "Consequences of the Axiom of Choice" as form 284, so one can look up what's known about it. Unfortunately, it turns out that essentially nothing is known. $\endgroup$ – Andreas Blass Oct 24 '18 at 0:04

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