Is $\mathbb{Z}_2$ a vector space? If $\mathbb{Z}_2$ is defined in the usual way, is it a vector space?
It passes $\vec{0}\in \mathbb{Z}_2$ and all the addition properties, my problem is with the scalar multiplication properties.  Is $\alpha [x] \in\mathbb{Z}_2$ for $\alpha\in\mathbb{F}, [x]\in\mathbb{Z}_2$? I would assume so but I just wanted to make sure. (ie. $5[1] = 5 $? or $5[1]=1$)
Then $\mathbb{Z}_2$ would be a vector space with only two vectors, $\{[0],[1]\}$
 A: $\newcommand{\Integers}{\mathbf{Z}}$This is a nitpicky (yet naive, i.e., category-free) answer. The context of the question suggests that excessive pedantry may not completely misplaced, and may be useful to posterity.
The set $\Integers_{2}$ of residue classes mod $2$ is not a group, it's not a ring or a field, and it's not a vector space (over some field) until you "equip it with appropriate operations". Taking the question "Is $\Integers_{2}$ a vector space?" at face value, the answer is "No".
Instead, let's assume the question is "Can $\Integers_{2}$ be made into a vector space (for some field of scalars)?" As multiple commenters note, you have to choose the right field of scalars (see below), namely $(\Integers_{2}, +, \cdot)$, in which case the answer is "Yes".
The "usual" ways of making $\Integers_{2}$ into a group, a ring (in fact a field), or a vector space, are to let $+$ and $\cdot$ denote addition and multiplication mod $2$, viewed as binary operations on the set $\Integers_{2}$. Then


*

*$(\Integers_{2}, +)$ is a group with identity element $[0]$.

*$(\Integers_{2}, +, \cdot)$ is a field, with additive identity element $[0]$ and multiplicative identity $[1]$.

*$\Integers_{2}$ becomes a one-dimensional vector space over the field $(\Integers_{2}, +, \cdot)$ if vector addition is defined to be ordinary addition mod $2$, i.e., we identify "vectors under addition" with the Abelian group $(\Integers_{2}, +)$, and scalar multiplication is ordinary multiplication mod $2$, with the first operand interpreted as a scalar and the second operand interpreted as a vector. That is, in writing $c \cdot v$, you should think of $c$ as an element of the field $(\Integers_{2}, +, \cdot)$ and $v$ as an element of the Abelian group $(\Integers_{2}, +)$.

Why can't the field of scalars be anything but the field of two elements?
One reason is, the vector space axioms imply $c \cdot [0] = [0]$ for every scalar. Consequently, $c \cdot [1] = [1]$ for every non-zero scalar $c$; if $c \cdot [1] = [0]$ for some non-zero scalar $c$, you get into trouble multiplying the preceding equation by $1/c$:
$$
[0] = (1/c) \cdot [0] = (1/c) \cdot (c \cdot [1]) = \bigl((1/c) \cdot c\bigr) \cdot [1] = [1] \cdot [1] = [1].
$$
Now, if your field has more than two elements, there exist non-zero scalars $c$ and $c'$ (not necessarily distinct) whose sum is non-zero; just take $c' \neq -c$. This forces you to run afoul of the distributive law for scalars:
$$
[1] = (c + c') \cdot [1] = c \cdot [1] + c' \cdot [1] = [1] + [1] = [ 0].
$$
In summary, if $F$ is a field with more than two elements, there is no scalar multiplication map $\cdot:F \times \Integers_{2} \to \Integers_{2}$ that satisfies the vector space axioms.
