4

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?

blue
  • 2,884
  • 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. – Giuseppe Negro Aug 29 '13 at 11:47

1 Answers1

10

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:

enter image description here

lhf
  • 216,483
  • 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. $;;;;;;;;$ –  May 15 '14 at 16:48
  • @RickyDemer, I think they implicitly assume that $V$ is not empty. – lhf May 15 '14 at 17:54