[akalin]

> If you try to define a vector space over a set without multiplicative closure, the vector space cannot be closed under scalar multiplication. Among other things, linear combinations stop being invertible (or even possible in general), and linear relations don't exist.

Mind clarifying this part? As someone else already pointed out, the integers are multiplicatively closed, but I suspect you're using "multiplicatively closed" to also mean "closed under multiplicative inverses". But I don't see how linear combinations stop being possible, e.g., "3x + 2y" is still a linear combination in a Z-module, or what it means for a linear combination to be invertible.

(Also not sure what exactly you mean by linear relations not existing if you have a module and not a vector space...)