Inconsistency in standard definition of abstract vector spaces

abstract-algebralinear algebravector-spaces

A vector space $V$ over field $F$ is a nonempty set with two binary operations, "addition" and "multiplication." Addition is a function $V \times V \rightarrow: (v_1, v_2) \rightarrow v_1 + v_2$ where $v_1, v_2 \in V$. Scalar multiplication is a function $F \times V \rightarrow: (a,v) \rightarrow av$ where $a\in F, v\in V$.

The following axioms hold:

  1. $V$ is an abelian group under $+$.
  2. $a(v_1+v_2)=av_1+av_2$ for all $a\in F$ and $v\in V$
  3. $(a+b)v=av+bv$ for all $a,b\in F$ and $v \in V$
  4. $a(bv)=(ab)v$ for all $a,b\in F$ and $v \in V$
  5. $1_F v= v $ for all $v\in V$

source: paraphrased from https://sites.math.washington.edu/~greenber/VectorSpaces

Such a definition allows for $V$ to be an infinite abelian group over some finite field $F$, which appears to lead to an inconsistency.

For example, consider the vector space $\mathbb{R}^3$ over $GF(2)$. Let $v$ be any vector $(x_1,x_2,x_3)$ other than the zero vector. Then $1(x_1,x_2,x_3)+1(x_1,x_2,x_3)=(1+1)(x_1,x_2,x_3)=0(x_1,x_2,x_3)=0_V$. So, $(x_1,x_2,x_3) = -(x_1,x_2,x_3)$ implying that $(x_1,x_2,x_3)$ is the zero vector. Contradiction.

What am I missing here? Shouldn't an appropriate definition of a vector spaces avoid such issues? Perhaps by requiring that the order of $F$ be greater than or equal to the order of $V$?

Best Answer

Your most recent comment gets to the point:

I started this post with the assumption that any abelian group paired with any field can be used to construct a vector space.

That isn't the case - for there to be a way to make $V$ a vector space over $F$, we need $V$ and $F$ to be "not-too-different" in a couple ways. The coarsest of these is a relationship between the characteristic of the field $F$ and order considerations of the group $V$:

For $F$ a field of characteristic $p$, if $V$ can be made into an $F$-vector space then we must have the order of each element of $V$ divide $p$.

(Note that the above works even if $p=0$ - everything divides $0$.)

For example, the group $\mathbb{R}$ - in which every element has infinite order - cannot be a vector space over any field of positive characteristic. And the point where things will break down in any attempt is exactly point $(3)$ - for exactly the reason you describe.

As a coda, note that this isn't the only obstacle to the existence of a vector space structure. For example, cardinality considerations play a role: for $V$ a nontrivial abelian group, in order for $V$ to be an $F$-vector space we must have an injection from $F$ into $V$ (think about multiplying some nonzero element of $V$ by elements of $F$). So e.g. $\mathbb{Q}$ can't be an $\mathbb{R}$-vector space.

Related Question