0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $k$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.