For a vector space $V$ over a field $F$ one of the axioms is that for any vector $v\in V$, we have $1v=v$. My intuition told me that this should follow from the other axioms, or more specifically the axiom $a(bv)=(ab)v$. The way I tried to show it was that $1(av)=(1a)v=av$ so that $1u=u$ for any vector $u$ that can be written as $av$ for some other vector $v$ and a scalar $a$. Of course $u$ can be written as $1u$ if we assume the axiom, but not necessarily otherwise.
I suppose this axiom for a vector space does not follow from the other axioms, because then it would be redundant. My question however is then, if there is an example of a "pseudo" vector space, where all the usual axioms hold, but not $1v=v$.