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Consider the vector space axiom $$ r(sv)=(rs)v$$

It is compatibility of scalar multiplication with field multiplication. Here $v$ is a vector in a finite dimensional vector space over a field $F$ and $r,s$ are elements in $F$. Sometimes the definition of a vector space is stated without this axiom. My question is: is it a consequence of the other axioms? Or is a definition without it a mistake?

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  • $\begingroup$ I guess that a definition without this axiom is wrong, because I don't see how it could be deduced from the others. But I have no explicit example either. $\endgroup$
    – Giuseppe Negro
    Aug 29, 2013 at 11:47

1 Answer 1

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It is most probably an error. A field action implies some things about the group addition in the vector space and some axioms can be removed, but not this one.

For a complete discussion of vector spaces axioms, see this paper

Independent Axioms for Vector Spaces, J. F. Rigby and James Wiegold. The Mathematical Gazette Vol. 57, No. 399 (Feb., 1973), pp. 56-62.

The first page is freely available and contains all you need, except the proofs. The main point is that the axioms below suffice and are independent, and so form a minimal set:

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  • $\begingroup$ Those axioms do not suffice, since they don't rule out $\: V = \{\hspace{-0.02 in}\} \;$. $\;\;\;\;$ On the other hand, I suspect that $\hspace{.19 in}$ replacing #5 with "$\hspace{.02 in}\mathbf{0} \: = \: 0\hspace{-0.03 in}\cdot \hspace{-0.03 in}a \;\;$ for all $a$ in $V\hspace{.03 in}$" would suffice. $\;\;\;\;\;\;\;\;$ $\endgroup$
    – user57159
    May 15, 2014 at 16:48
  • $\begingroup$ @RickyDemer, I think they implicitly assume that $V$ is not empty. $\endgroup$
    – lhf
    May 15, 2014 at 17:54

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