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I'm wondering what things go horribly wrong by not requiring distributivity in the defitinion of an $R$-module, or $k$-vector space. A 'concrete' example for $\mathbb{R}$ would be quite satisfying.

EDIT: Arturo Magidin points out there are two distributivity laws, which completely slipped my mind. I had the law $\alpha(x+y)=\alpha x + \alpha y$ in mind when asking this.

He also gave a concrete answer, but I'd love to see other examples of this, and any references would be lovely, should they exist.

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1 Answer 1

up vote 4 down vote accepted

There are two "distributivity" laws in an $R$-module/vector space:

  1. For all $a\in R$, $x,y\in M$, $a(x+y) = ax+ay$;
  2. For all $a,b\in R$, $x\in M$, $(a+b)x = ax+bx$.

An example in which all axioms of a vector space except for (1) above holds is:

Take $V=\mathbb{C}^2$ with its usual addition; define scalar multiplication by: $$\alpha(x,y) = \left\{\begin{array}{ll} (\alpha x,\alpha y) &\text{if }x\neq 0;\\ (0,\overline{\alpha}y) &\text{if }x=0. \end{array}\right.$$

With an arbitrary ring/field, any nontrivial automorphism will do instead of complex conjugation.

If you don't mind other axioms failing, you can take $V=\mathbb{R}^n$ over $\mathbb{R}$, take $\alpha(x,y) = (0,0)$ if $\alpha\neq 1$, and $1(x,y)=(x,y)$, with $V=\mathbb{R}^2$. Note, however, that this does not satisfy associativity: if $\alpha\neq 0,1$ and $\beta=\frac{1}{\alpha}$, then $\alpha(\beta (x,y)) =(0,0)$, but $(\alpha\beta)(x,y) = (x,y)$.

An example in which all axioms of a vector space except for (2) above holds is: take $V=\mathbb{R}$ with its usual addition, and define scalar multiplication by $\alpha\cdot x = \alpha^2x$.

For a longer discussion of the independence of the sundry vector space axioms, see here.

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Oops! I forgot the second distributivity law, I'd the first in mind however. Thank you! –  John Stalfos Jul 26 '12 at 16:37

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