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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 ^_^

Andrew TheDark
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5 Answers5

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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 $\mathbb Z$.

user26857
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Robert Israel
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    I think I went with this answer for its simplicity, although the other answers may be perfectly excellent! Sorry I forgot to thank you all for your contribution! – Andrew TheDark Jun 30 '16 at 22:47
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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.

Matt Samuel
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    This is kind of roundabout. Also, this argument makes heavy use of axiom of choice (in the form of the statement that vector spaces are free), while the fact that a vector space has the same characteristic as the base field is immediate from the definition. – tomasz May 12 '15 at 01:42
  • @tomasz: Non-trivial vector spaces have the same characteristic as their base field. – Martin Brandenburg May 13 '15 at 12:25
  • @MartinBrandenburg: True. And of course I should have said that the exponent of a (nontrivial) vector space is the same as the characteristic of the base field. – tomasz May 14 '15 at 18:55
  • @tomasz Can you please explain why exponent of vector space is same as $char(F)$ base field? – grayQuant Sep 19 '15 at 18:53
  • @grayQuant: Because $v+v+v+v+\ldots+v=(1+1+1+\ldots+1)v$. – tomasz Sep 19 '15 at 23:56
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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.

Marc van Leeuwen
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    You don't need linear dependence. Since $\mathbb{Z}$ is finitely generated as abelian group, it would be finite dimensional also as vector space. More generally, it can't be a module over a finite ring. – egreg May 12 '15 at 10:30
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$\mathbb Z$ 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. $\Box$

Alex W
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    Then $\mathbb Z$ has what? Do you mean $\mathbb Z$ is a direct sum of a family of subgroups, i.e. the decomposition over a basis? Either way this answer is not very clear. – Patrick Da Silva May 11 '15 at 22:54
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    @PatrickDaSilva some explanations. – Alex W May 11 '15 at 23:15
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    By the way, $\mathbb{Z}$ is a vector space (actually, a commutative algebra) over $\mathbb{F}_1$, the "field with one element". – Martin Brandenburg May 13 '15 at 10:30
  • @MartinBrandenburg I used the usual field's definition, which requires $1\neq 0$. Hence order of any field not less 2. – Alex W May 13 '15 at 10:44
  • @AlexW: You didn't understand my point. I don't mean the trivial ring (which has only trivial modules). I mean $\mathbb{F}_1$. See http://arxiv.org/pdf/0909.0069.pdf for an overview. And I wanted to mention this because actually the failure of $\mathbb{Z}$ being a vector space over some field is a big problem in arithmetic geometry. The usual reaction would be "ok, we have to deal with this", but a more "modern" reaction is "ok, so let's generalize the notion of a field so that it becomes true!". – Martin Brandenburg May 13 '15 at 12:20
  • @MartinBrandenburg Thank You! It is very interesting. – Alex W May 13 '15 at 16:55
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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.

Alex W
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