TITLE: Prove or disprove that if V is a finite set over F with a dimension 1, then F is a finite field.
QUESTION [0 upvotes]: I think it is a True statement, and my proof is as follows:
let $\theta $ be the additive identity in F
Given any element v in V, since v + v = $\theta$ v + $\theta$ v = 2$\theta$ v, then 2$\theta$ $\in$ F, and $\frac{\theta}{2} \in$ F, hence $\frac{\theta}{2}v \in$ V, by this argument we can construct infinite $\{2v, \frac{\theta}{2}v , 3v,  \frac{\theta}{3}v...\}$
Hence V cannot be finite, and we are done.

REPLY [1 votes]: A basic fact is that for a finite dimensional vector space $V $ over a field
$\mathbb F $ ,  we have $|V|=|\mathbb F|^{\rm {dim}V} $.
This forces $|\mathbb F|\le|V|$.

For if $\lambda\cdot v=0$ and $\lambda\ne0$, then $v=0$.  Proof: $0=\lambda^{-1}\cdot0=\lambda^{-1}\cdot(\lambda\cdot v)=(\lambda^{-1}\cdot\lambda)v=1\cdot v=v$.
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