# Does existence of nonzero linear functional depends on axiom of choice? [duplicate]

Given an arbitrary nonzero vector space $V$, is there a nonzero linear functional on $V$, without assuming axiom of choice? I know that by assuming existence of a basis for $V$, we can consider the dual basis for a subspace of $V^*$, which justifies the existence of nonzero linear functional, but this argument fails without axiom of choice. I guess there may not always be a nonzero linear functional, but my knowledge on axiom of choice and infinite-dimensional vector spaces is lacking. A quick search on Google fails to give an answer.

Note that I am not talking about normed spaces or continuous linear functionals, just plain vector spaces with no additional structure.

## marked as duplicate by Asaf Karagila♦ axiom-of-choice StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 9 '17 at 12:16

The answer is negative. In other words, there are models os set theory (without the axiom of choice) for which there are vector spaces $V\neq\{0\}$ such that $V^*=\{0\}$. See here, for instance.