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Let $\Bbb{F}_p$ be the finite field of integers modulo $p, p$ a prime, let $A$ be an abelian group. Precisely when can $A$ be made into a vector space over $\Bbb{F}_p$?

rakhil11
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  • How much linear algebra do you know? Like, for example, can there be two non-isomorphic vector spaces of dimension $2$ over any given field? Or is your focus on the infinite-dimensional case? – pjs36 Sep 21 '15 at 03:30
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    If dim(U) = dim(V), they're isomorphic, so no. I don't think this is referring to the infinite-dimensional case, although I guess that's not specified in the problem – rakhil11 Sep 21 '15 at 03:36
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    Can you find a necessary condition? What needs to be checked then to make it a sufficient condition as well? Can you do the finitely generated case? Go step by step. – guest Sep 21 '15 at 03:41
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    @rakhil11 Given that comment, I would just think (in the finite-dimensional case) about all the vector spaces $V$ over $\Bbb F_q$ you can think of, paying particular attention to the additive group $(V, +)$. Before trying to characterize everything, just characterize what you know. – pjs36 Sep 21 '15 at 03:53
  • Taking pjs36's hint one step further. In $\Bbb{F}_p$ we have $0=1+1+\cdots+1$ ($p$ summands on the right, all equal to $1$). What does this imply about the scalar multiplication (or addition) in $A$? – Jyrki Lahtonen Sep 21 '15 at 05:12
  • @JyrkiLahtonen I believe the only condition then is that $p\neq 2$. But I don't what we can say about scalar multiplication here? – grayQuant Sep 21 '15 at 19:24
  • @grayQuant: It is quite possible for $p$ to be equal to two. The key here is that multiplication by $0$ should give the same result as multiplication by $1+1+\cdots+1$. As it happens that is the only thing we need to require from $A$ to be able to turn it into a vector space over $\Bbb{F}_p$. – Jyrki Lahtonen Sep 21 '15 at 20:28

1 Answers1

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I'll try to give a full answer based on the comments

An abelian group $A$ is a $\mathbb{Z}-$module and $ann_{\mathbb{Z}}A=(d)$ for some $d\in\mathbb{Z}$.

I'll show that $A$ is a $\mathbb{F}_p$ vector space iff $(p)\subset (d)\Leftrightarrow d|p\Leftrightarrow p=d$

  • Let $ann_{\mathbb{Z}}=(p)$. Then we can define a $\dfrac{\mathbb{Z}}{p\mathbb{Z}}-$module structure as follows $$\dfrac{\mathbb{Z}}{p\mathbb{Z}}\times A\to A$$ with $(z+p\mathbb{Z},x)\mapsto zx$. This is well-defined : if $(z+p\mathbb{Z},x)=(z'+p\mathbb{Z},x')\Rightarrow x=x',z-z'\in p\mathbb{Z}\Rightarrow (z-z')x=0\Rightarrow zx=z'x'.$ This way $A$ becomes a $\mathbb{F}_p-$module (=$\mathbb{F}_p$-vector space)
  • Let $A$ an $\mathbb{F}_p-$vector space (=$\dfrac{\mathbb{Z}}{p\mathbb{Z}}$-module). Then consider the ring homomorphism $$f:\mathbb{Z}\to \dfrac{\mathbb{Z}}{p\mathbb{Z}}$$ with $f(z)=[z]_p$. Then we can see $A$ as a $\mathbb{Z}-$module with multiplication $$z\cdot a=f(z)a,\quad \forall a\in A,z\in \mathbb{Z}$$. Then $$z\cdot a=0\Leftrightarrow f(z)a=0\Leftrightarrow f(z)=[0]_p\Leftrightarrow p|z$$ hence $ann_{\mathbb{Z}}=(p)$
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