Let $k \subset \mathbb{F}$. $k, \mathbb{F}$ are fields. Vector space over $\mathbb{F}$ isn't trivial ($V_\mathbb{F} \neq 0$). $\dim_k V < |k|$. Is $\mathbb{F}$ an algebraic extension of $k$?

So I should prove that $\mathbb{F}$ got no transcendent elements over $k$. I don't know where to start.


If $F$ had a transcendental element $x$ then it has these $k$-linearly independent elements: $1/(x-a)$, $a \in k$, so ${\rm dim}_k(V) \geq {\rm dim}_k(F) \geq |k|$, contradicting the assumptions. Thus $F$ must be algebraic over $k$.

  • $\begingroup$ As I understood to prove linearly independency I must prove: Let $n=|k|$. $\alpha_1(x-a_2)(x-a_3)...(x-a_{n}) + ... + \alpha_{n}(x-a_1)...(x-a_{n-1})=0 \iff \alpha_1=...=\alpha_n = 0$. x isn't a root of this polynomial over k. And that's why all alphas must be zero. $\endgroup$ – Khan Mar 13 '17 at 6:20
  • $\begingroup$ That would be equivalent, but it's not necessary to clear the denominators of the functions $1/(x-a)$. A rational function with a pole at say $x=a_1$ can not possibly be a linear combination of rational functions that don't have a pole at $x=a_1$. Hence the functions $1/(x-a)$ (for all $a \in k$) are $k$-linearly independent. Another comment: the cardinality $|k|$ need not be finite, the same argument still works. $\endgroup$ – Mark Mar 13 '17 at 14:23

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