# Does scalar multiplication have to be faithful on subspaces?

I was recently considering how one could motivate studying abelian groups from the study of linear algebra, and I presumed that the multiplicative group of the complex numbers would be a good a example of an abelian group that is not a vector space. However, upon closer inspection it seems that perhaps it is a real vector space, with appropriate scalar multiplication.

In particular, consider the scalar multiplication $$(\lambda, z) \mapsto z^{\lambda}$$. If our "vector addition" is actually complex multiplication, then this gives a two dimensional real vector space. However, this behaves quite a bit differently than ordinary finite-dimensional real vector spaces, in the sense that there exists nonzero $$\lambda$$ and $$z$$ with $$z^{\lambda} = 1$$ (which is the "zero" vector).

More generally, for many complex $$z$$ (e.g. roots of unity) there exists multiple distinct $$\lambda_{j}$$ with $$z^{\lambda_{j}} = z^{\lambda_{j'}}$$ (e.g. for $$n$$-th roots, $$\lambda_{j} = j \cdot n$$).

In other words, at least on one subspace (the unit circle) the action does not seem to be faithful (sort of - on closer reflection this is not strictly true). This comes as a shock and suggests to me that perhaps this isn't a vector space and that there is something I am missing. It is my understanding that all finite dimensional vector spaces are isomorphic to $$\mathbb{R}^{n}$$ for appropriate $$n$$, which does not seem to be the case here.

Thus my question: What I am missing here? Is there something that prevents a vector space from having this curious behavior, or is this actually a valid real vector space?

What I've looked at so far: it seems the only mention of this structure on MSE is this answer which says that, restricted to the positive reals under multiplication, it gives a vector space.

What you're looking for is the concept of a module over a ring $$R$$, which is an abelian group (written additively) together with compatible scalar multiplication by elements in $$R$$. You recover the notion of an abelian group by letting $$R=\mathbb Z$$: an abelian group is the same thing as a $$\mathbb Z$$-module. Vector spaces are modules over a field (or division ring).

Vector spaces are extremely special modules. The backbone theorems of linear algebra are mostly false for modules over more general rings. Most modules don't have a basis, in the sense that it's almost never true that there is a subset such that all elements of the module are unique linear combinations of elements of the subset. In a sense, that's the most "boring" case.

If a module does have a basis, then it's called a free module, and even then there's no need for a submodule of a free module to itself be free. This is true for abelian groups, but over more general rings it is not.

A free module need not have a concept of "dimension" either. The size of a basis of a free module is called the "rank" of the module. In general, a free module of rank $$n$$ is isomorphic to $$R^n$$. However, the rank need not be unique! There are rings where $$R^m\cong R^n$$ for all $$m, n\geq 1$$. Rings for which the rank is unique are said to have the invariant basis number property.

Already with the simplest non free abelian groups, $$\mathbb Z/n\mathbb Z$$, we have nonzero elements $$m$$ of the module and nonzero ring elements $$a$$ such that $$am=0$$. Elements for which such an $$a$$ exists that is not a zero divisor are called torsion elements, and your example has lots of them (specifically, the roots of unity) . Even if a module is not free, it need not have torsion elements. For example, $$\mathbb Q$$ is a torsion-free $$\mathbb Z$$-module, but it is about as far from being a free module as you can get: any two free submodules of rank $$1$$ intersect each other nontrivially.

Your group can be stripped to become a $$\mathbb Q$$-vector space, and indeed an $$\mathbb R$$-vector space. The group is the direct product $$\mathbb R^+\times S^1$$ when you decompose it into magnitude and direction. $$S^1$$, the unit circle, is what is making it fail to be a vector space, because the roots of unity are torsion elements and vector spaces can't have torsion. $$\mathbb R^+$$ under multiplication is actually isomorphic to $$\mathbb R$$ under addition via the exponential map $$t\mapsto e^t$$, and multiplication by a scalar becomes exponentiation, so it is a vector space in this way.

• Thanks for the answer! This particular construction doesn't seem like it forms a module, however, since $a \cdot (b \cdot v) = (ab) \cdot v$ doesn't always hold. But modules might actually be a good way to introduce abelian groups. Interesting thought! – Jacob Maibach Dec 23 '18 at 20:57
• @Jacob Being an abelian group, it's a module over $\mathbb Z$, which is integer exponentiation. Since it has integer torsion, it can't possibly be a module over $\mathbb Q$, no matter which roots you end up choosing. – Matt Samuel Dec 23 '18 at 20:59
• @Jacob However, if you take out imaginary numbers and negative numbers, you're left with the positive real numbers under multiplication. This is indeed a module with rational exponentiation. – Matt Samuel Dec 23 '18 at 21:02
• Could you add/emphasize that point about why it can't be extended to a $\mathbb{Q}$-module in your answer? I think that is exactly what I was looking for. – Jacob Maibach Dec 23 '18 at 21:11
• @Jacob See the edit. – Matt Samuel Dec 23 '18 at 21:17

Embarrassingly, the same question I linked has an answer explaining just this.

Namely, for $$z = i$$, note that $$(z^{4})^{1/2}$$ is not equal to $$z^{(4 \cdot 1/2)} = z^{2}$$. Why? Because $$i^{4} = 1$$ so $$(i^{4})^{1/2} = 1^{1/2} =1$$ which is quite distinct from $$i^{2} = -1$$. That is, the lack of uniqueness of square-roots (or roots more generally) in the complex numbers either prevents scalar multiplication from being well-defined, or it prevents it from satisfying the axiom that $$a \cdot (b \cdot v) = (ab) \cdot v$$ (for scalars $$a, b$$ and vector $$v$$).

Edit: Generalizing this gives a short proof that a vector space is torsion-free. Presume there exists a nonzero scalar $$a$$ and vector $$v$$ such that $$a \cdot v = 0$$. Then note that $$a^{-1} \cdot (a \cdot v) = (a^{-1}a) \cdot v = v,$$ yet also $$a^{-1} \cdot (a \cdot v) = a^{-1} \cdot 0 = 0.$$ Therefore $$v = 0$$. More generally, this shows that scaling must be bijective. That is, for distinct scalars $$a$$ and $$b$$, and vector $$v$$ with $$a \cdot v = b \cdot v$$, we have $$(a - b) \cdot v = 0$$ implying once again that $$v = 0$$.