Intuition for the fact that, in a vector space V over a field F, av = 0 $\implies$ a = 0 or v = 0. (a $\in$ F, v $\in$ V). I have no trouble proving this:
Let av = 0.  If a = 0 then then we are done.  Otherwise, there exists $a^{-1} \in F$ such that $a{^-1} a = 1$.  Multiplying both sides of the equation by $a^{-1}$ gives v = $a^{-1}$ 0 = 0.
But on a gut level I don't really understand it.  I can't draw a picture of it.  Why can't a vector space have a periodic structure?  I've constructed "counter-examples" to the theorem above, even though I know it's true because I know my proof is correct.  If someone could point out why my couterexamples are fallacious (maybe by explaining that the "vector spaces" I've constructed are not in fact vector spaces, and why, I'd very much appreciate it).
My main counter-example is this: Let V be the set of rational numbers mod 2 (i.e., V = $\mathbb{Q}/2\mathbb{Z}$).  Let F = $\mathbb{R}$ and define scalar multiplication as you would in a modular ring (i.e, av = (av)%2, the remainder of av when divided by 2).  Then av is still a vector in V for every a, v.  Isn't it true that 7(1/7) = 0, with 7 $\in \mathbb{R}$ and (1/7), 0 $\in$ V?
Some more context: I'm encountering this problem while studying the difference between vector spaces and modules.  This naturally led me to think the following:
"OK, I know that a module can have zero-divisors; as an example, if I let V be the symmetry group of a circle (quite similar to my example above), and think of this as a module over $\mathbb{Z}$. (I understand that an abelian group can be considered a module over $\mathbb{Z}$, and I understand why). Then 3v = 0 has non-zero solutions: the rotations 2/3 and 1/3 around the circle.  But then what if I think of $\mathbb{Z}$ as a subset of $\mathbb{R}$?  Then there are zero-divisors in $\mathbb{R}$, which I showed right at the beginning is not possible.  So there must be some reason why letting the scalars be non-integer real numbers messes up the structure of the module/vector space.  But what is it?"  I've tested out all the components of the definition of a vector space, and they seem to all hold true when I extend this structure over $\mathbb{R}$.
I think there is something basic and important about the structure of rings, fields, and vector spaces that I'm missing.  Can you help me figure out what it is?  Thanks in advance.
 A: Generic modules can be so ill-behaved compared to vector spaces that I never recommend thinking about them as the same kind of beast. Spaces are free, divisible, torsionfree, semisimple, and the only simple module (the atoms of the semisimple universe) over a field (up to iso) is the field itself.
I cannot for the life of me figure out what you think your counterexample is. First off, since you're doing a quotient of additive groups, you might as well divide by $\Bbb Z$. Say we take the additive coset $\frac{1}{2}+\Bbb Z\in\Bbb Q/\Bbb Z$ and multiply by $\sqrt{2}$; what element of $\Bbb Q/\Bbb Z$ do you expect to get? But let's say you really wanted to ask why this wasn't a vector space over $\Bbb Q$, not $\Bbb R$. If scalar multiplication is naively defined as multiplying a representative by the scalar, the operation is not well-defined; in point of fact $0+\Bbb Z$ and $1+\Bbb Z$ are taken to $0+\Bbb Z$ and $\frac{1}{2}+\Bbb Z$ respectively, but $0+\Bbb Z=1+\Bbb Z$! Suppose you try to fix this by stipulating the representative be in $[0,1)\cap\Bbb Q$. Finally we end up butting heads with your original proof: multiply $\frac{1}{2}+\Bbb Z$ by $2$ first and then by $\frac{1}{2}$ and you will get $0+\Bbb Z$, however the vector space axioms tell us that $2^{-1}\cdot2\cdot(\frac{1}{2}+\Bbb Z)=1\cdot(\frac{1}{2}+\Bbb Z)\ne0+\Bbb Z$!
It turns out, you knew all along why you can't have "periodicity" (torsion) in a vector space; since multiplication in the scalar field corresponds to functional composition of the scalar multiplication maps on the vector space, the fact that nonzero elements of the field have inverses implies the scalar multiplication maps are invertible, so in particular they cannot annihilate anything nonzero.
In general modules over rings, the elements of the scalar ring need not generally be invertible, in which case the corresponding scalar multiplication maps needn't be invertible either, so they can end up annihilate things nontrivially.
