Can the integers be made into a vector space over any Finite Field?

Question

Given a Finite Field $F$, can the the abelian group $\mathbb Z$ be made into a vector space over $F$ without changing the additive structure of $\mathbb Z$?

This seems like it shouldn't be complicated, but having trouble on this one, any hints would be appreciated.

I already know that $F\cong \mathbb F_{p^n}$ for some prime $p$ and some $n\in \mathbb N$.

I have been trying to make a map $F\times \mathbb Z\rightarrow \mathbb Z$ by $(\overline n,m)\mapsto n\cdot m$ but this really seems like the wrong approach since the axioms aren't really met, maybe its not even possible.

REPLY: Thanks so much for your replies, I am now fairly convinced that there is no such Vector Space. Thank you again, I am very appreciative of your assistance ^_^

Answer 1

If your field $F$ has characteristic $p$, any element $v$ in a vector space over $F$ satisfies $\underbrace{v + \cdots + v}_{p\text{ times}} = 0$. This is not the case for $\mathbb Z$.

Answer 2

A vector space over a field is always isomorphic as an abelian group to a direct sum of copies of the additive group of the field. Over a finite field, therefore, there will always be nonzero elements of finite order, hence $\mathbb{Z}$ cannot be isomorphic to the additive group of a vector space over a finite field.

Answer 3

$\mathbb Z$ is not a vector space over any field $F$.

Proof. Assume, that $\mathbb Z$ is a vector space over field $F$ with unit $e$. Let $1<n\in\mathbb N$ be such that if $p:={\rm char}\ F>0$, then $p\nmid n$. Then $ne$ - invertible element of $F$, hence equation $nx=a\Leftrightarrow (ne)x=a$ has solution in space $\mathbb Z$ for any $a\in\mathbb Z$. But in $\mathbb{Z}$ equation $nx=1$ has no solutions - contradiction. $\Box$

Answer 4

Just for fun another argument, one that does not use the characteristic but only the fact that $F$ is finite. In$~\Bbb Z$, and two nonzero elements $n,m$ have a common nonzero scalar multiple, for instance $nm$. In a vector space that would mean they are linearly dependent, and a vector space in which any two nonzero vectors are linearly dependent has dimension${}\leq1$. But a $1$-dimensional vector space over a finite field is finite, which $\Bbb Z$ is not.

Answer 5

Since this matter has caused some interest, I will give a general description of the rings $R$, over which $\mathbb Z$ can be a unital module (i.e. $1_Ra=a$ for any $a\in\mathbb Z$). All modules below are right and unital, rings have a units and homomorphisms transfer units to units, rings may be noncommutative.

Theorem. Abelian group $\mathbb Z^+$ by addition can be made a module over the ring $R$ iff $R$ has an ideal $I$ such, that $R/I\cong \mathbb Z$.

Proof. Group $\mathbb Z^+$ can be made a module over the ring $R$ iff there exists ring homomorphism $\phi:R\to {\rm End}(\mathbb Z)$. But ${\rm End}(\mathbb Z)\cong\mathbb Z$. In such a way, $\mathbb Z^+$ can be made a module over the ring $R$ iff there exists ring homomorphism $\phi:R\to \mathbb{Z}$. Any such morphism must be epimorphic, since $\mathbb Z$ generated by $1$ and $1\in\phi(R)$. Thus, existence of such morphism is equivalent to the presence of the ideal $I$ of $R$ such, that $R/I\cong\mathbb Z$. $\Box$

Hence, in particular, it follows that $\mathbb Z^+$ can not be a module over any field and over any finite ring.