Dimension of Example Quotient Space

Question

Let $V$ be the vector space across the set of all integers, with subspace $W$ as the vector space across all even numbers, both with the field as the integers. The quotient space $V/W=\{v+W|v\in{V}\}=\{W,X\}$ where $X$ is the set of all odd numbers, right? Since if $v$ is even, $v+W$ is all the evens, and if $v$ is odd, $v+W$ is all the odds. So, if $\dim{V/W}=\dim{V}-\dim{W}$, $\{W,X\}$ should have dimension 0, so why? Why does $\{W,X\}$ have dimension 0? If it did have a basis, wouldn't that basis just be $W$? Or is there no way to find the dimension of this set for some reason, defaulting it to 0?

Answer 1

It appears that you are confused between different algebraic subjects. The theory of vector spaces (called linear algebra) starts by assuming the choice of a field of numbers (called scalars in the context of linear algebra). That is a structure in which the four arithmetic operations between numbers are always defined, with the exception only of division by zero, and return a number. The most popular choice seems to be the field $\Bbb R$ of real numbers (probably because geometric intuition works best for that choice), but one can perfectly well choose to work with the rational numbers $\Bbb Q$ as scalars, or on the contrary allow all complex numbers $\Bbb C$ as scalars, and many more sophisticated choices are possible. But one cannot choose the integers $\Bbb Z$ as scalars, because many divisions of integers like $24\div 15$ cannot be performed with an integer as result. There is an important reason to insist that division by any nonzero scalar is always possible, though this reason is not immediately clear when one starts to set up linear algebra.

The definition of vector spaces require that the set of vectors be closed under addition, subtraction, and multiplication by scalars. If you apply that definition to the set of integers, you will find that (whatever field you chose to use) there are scalars that you cannot always multiply by; for the most modest choice of $\Bbb Q$ as scalars, you cannot always multiply by$~\frac23$, or more generally by any fraction that is not an integer (and if you choose $\Bbb R$ or $\Bbb C$ as field, there are even more problematic scalars). So $\Bbb Z$ is not a vector space, over any field.

It is possible to set up a theory that is not linear algebra, in which one does not insist that division by nonzero scalars is always possible. One then has a ring of scalars, and the notion of vector space is replaced with that of a so-called module over the ring. But dropping the division requirement does not simplify things; on the contrary. Rings that are not fields can be complicated structures that are hard to work with, and often one needs additional assumptions on the ring used to ensure that some complications will not happen; this leads to a great variety of kinds if rings with additional requirements weaker than that of being a field (like principal ideal domains, which are fairly nice rings and of which $\Bbb Z$ is an example). The theory of modules over rings is harder to develop than that of vector spaces, and one of the reasons for this is that there does not exists a neat notion like dimension. Therefore what you want to argue cannot be done in (the $\Bbb Z$-module of) the integers; the notion of dimension simply does not apply there. For what its worth, you can form the submodule $W$ of the even integers in the module $V$ of all integers, and form the quotient module $V/W$, which has two elements as you correctly argue. But that's all one can say; $V/W$ has no dimension (nor do $V$ and $W$).

One might wonder where in linear algebra theory the requirement of dividing by scalars really plays a role. The most basic place where this is crucially the case is the statement: if one has a linear combination of distinct vectors with nonzero scalar coefficients that gives the zero vector, then and one of those vectors can be expressed as linear combination of the others. For instance if $\alpha u+\beta v+\gamma w=\vec0$ (with the Greek letters scalars and the Romans vectors), then $u=-\frac\beta\alpha v-\frac\gamma\alpha w$. Working in a module over a ring that is not a field, one can easily have such linear relations for which it is not possible to express any of the elements $u,v,w$ in the relation in terms of the others. The possibility to use such expressions is vital to obtain the theoretical result that makes the notion of dimension possible (namely that all choices of a basis for a given vector space involve the same number of vectors).