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The question is:

Consider a finite vector space V over a field F where |V| > 1. Is F always finite?

My proof: Suppose for a contradiction that F was infinite. Since V is closed under scalar multiplication any element from our field, say c, multiplied by 1 will give us back c, which is in V. Since F has infinitely many elements this implies V will also have infinitely many elements. This is a contradiction since V is finite. Hence F must also be finite.

Is this proof sound?

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    $\begingroup$ $c$ is in the field? Then why is $1$ necessarily in $V$? Note that $1$ is not an element of, for instance, $\Bbb R^2$. Since $1$ is not necessarily in $V$, how can we word it to not make mention of the "element $1$"? $\endgroup$
    – JMoravitz
    Sep 14, 2020 at 23:22
  • $\begingroup$ Sorry I was being lazy, by 1 I meant the multiplicative identity, 1v = v for all v ∈ V. $\endgroup$
    – Chad McMath
    Sep 15, 2020 at 1:28
  • $\begingroup$ And yes, c is any scalar from the field $\endgroup$
    – Chad McMath
    Sep 15, 2020 at 1:36
  • $\begingroup$ Then c times 1 is a scalar times a scalar, is not in V but rather it's in F $\endgroup$
    – JMoravitz
    Sep 15, 2020 at 2:29
  • $\begingroup$ "For closure under multiplication, we demand that if u ∈ V, a ∈ F, then aF∈ V". My understanding is that any vector multiplied by a scalar must be in the set. Since the multiplicative identity is in V, every scalar must also be in V. Is this incorrect reasoning? $\endgroup$
    – Chad McMath
    Sep 15, 2020 at 3:41

1 Answer 1

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Suppose that $F$ is not finite. Let $v\neq 0, v\in V$, consider the map $f:F\rightarrow V$ defined by $f(x)=xv$, there exists $x,y$ not zero, $x\neq y$ such that $f(x)=f(y)$ this implies that $(x-y)v=0$, by multiplying by the inverse of $(x-y)$ we deduce that $v=0$ contradiction.

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