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I've always been curious about this, why do we use fields as the only algebraic structure to put vector spaces over? It seems a bit arbitrary to me, so I was wondering what would happen it we replaced the requirement with something less structured like a ring (with unity, we don't want to violate the axioms). Anyone have any idea what the consequences would be? Anyone have any ideas/reasons why we don't?

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    $\begingroup$ Here is a reference: en.wikipedia.org/wiki/Module_%28mathematics%29 $\endgroup$ – Jonas Meyer Jan 24 '13 at 18:16
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    $\begingroup$ We have such a thing over rings, it is called module. But when we impose the ring to be a field then we can talk about basis, dimension, etc. Of course some of these notions are still applicable for modules over some specific rings (like PID rings). $\endgroup$ – user56706 Jan 24 '13 at 18:18
  • $\begingroup$ Oh, wow, thanks. I'll look around at it thanks. $\endgroup$ – Pax Kivimae Jan 24 '13 at 18:27
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We can talk about rings acting on abelian groups nicely and we frequently do, we call them modules. The main issue is that many notions of linear algebra break down in this theory. For instance generating subsets are not at all what one would expect. An extreme example is to let $\mathbb Z$ act on $\mathbb Q$ in the natural manner, by multiplication. We would say consider $\mathbb Q$ as a $\mathbb Z$-module. Then there are no linearly independent subsets of size greater than $1$ but $\mathbb Q$ is not finitely generated as a $\mathbb Z$-module. Even worse things can happen for instance if we consider $\mathbb Z/(2)$ as a $\mathbb Z$ module then there are no non-trivial linearly independent subsets!

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    $\begingroup$ The good news is that a lot of linear algebra works in modules. The bad news is that a lot of linear algebra doesn't work in modules. $\endgroup$ – Gerry Myerson Jan 25 '13 at 3:50

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