$F$ is algebraically closed $\iff$ $\nexists$ $K$ s.t. $F \leq K$, $K \neq F$ and $[K:F] < \infty$

Question

$F$ is algebraically closed $\iff$ $\nexists$ $K$ s.t. $F \leq K$, $K \neq F$ and $[K:F] < \infty$.

Proof:

($\implies$) assume by way of contradiction that $F$ is an algebraically closed field and there does exist such a field $K$, say $[K:F] = n$.

Let $a_1,....a_n$ be a basis for $K$ over $F$.

Okay, now I'm lost, haha

$(\impliedby)$

I need help with this direction also.

Thanks all! You're the best!

Answer 1

The correct statement is :

$F$ is algebraically closed if and only if there does not exist a finite field extension $K$ of $F$, i.e. there does not exist $K$ such that $F\leq K$ and $[K:F]<\infty$.

Proof : $(\Rightarrow)$ Say $K$ is a finite field extension of $F$ of degree $n$. Let $\alpha\in K-F$. Then $\{1,\alpha,\alpha^2,\dots,\alpha^n\}$ is linearly dependent over $F$. Then there exists $\lambda_i\in F$ such that $\sum_i\lambda_i\alpha^i=0$, i.e. $\alpha$ satisfies a polynomial with coefficients in $F$, which contradicts the fact that $F$ is algebraically closed.

$(\Leftarrow)$ Say $F$ is not algebraically closed. Then there exists an irreducible polynomial $f\in F[x]$ such that $f$ has no root in $F$. Then $K=F[x]/(f)$ is a finite field extension of $F$, which is a contradiction. Hence $F$ is algebraically closed.

Answer 2

Depends on your definition for algebraically closed fields, but let me give it a try. Choose an element $a \in K$ which is not in $F$ and consider the corresponding minimal polynomial over $K$. Since $a \not\in F$, the minimal polynomial does not split over $F$ which is a contradiction to $F$ being algebraically closed.

For the other direction you can assume that $F$ is not algebraically closed and thus construct a finite extension by adjoining a root.