# Does a finite Vector space imply a finite field

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?

• $c$ 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 $1$"? Sep 14, 2020 at 23:22
• Sorry I was being lazy, by 1 I meant the multiplicative identity, 1v = v for all v ∈ V. Sep 15, 2020 at 1:28
• And yes, c is any scalar from the field Sep 15, 2020 at 1:36
• Then c times 1 is a scalar times a scalar, is not in V but rather it's in F Sep 15, 2020 at 2:29
• "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? Sep 15, 2020 at 3:41

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.