Example of finite extension which is not finitely generated extension I just read the theorem Finitely Generated Algebraic Extension is Finite. So a field being finitely generated and algebraic is a sufficient condition for it being finite. Is it also a necessary condition? In particular, can you give an example of:
A finite extension $K/F$ which is not finitely generated (by $F$ or its subfield)? 
or does this always hold?
 A: If $K/F$ is finite, say $n = [K:F] < \infty$, then (by definition) $K = \text{Span}_F\{k_1,\ldots,k_n\}$ where $k_i \in K$ for each $i$. Hence every element of $K$ can be written as a finite linear combination of the $k_i$ over $F$, and so we know $K \subseteq F[k_1,\ldots,k_n]$. At the same time $F[k_1,\ldots,k_n] \subseteq K$ is clear since $K$ is closed under addition and multiplication. So $K = F[k_1,\ldots,k_n] = F(k_1,\ldots,k_n)$ is finitely generated.
A: You seem to misunderstand what finite extension means: $K/F$ is a finite extension when $K$ is a finite-dimensional vector space of $F$. Clearly, this implies that $K$ is finitely generated (as an algebra) over $F$, since a basis is a generating set.
So every finite extension is finitely generated.
The converse does not hold because $F[X]$ is finitely generated as an algebra over $F$ but has infinite dimension.
However, every finitely generated algebraic extension is a finite extension, that is, has finite dimension. That's the theorem you've cited.
A: Suppose $K/F$ is not finitely generated. Let $k_{i}$ be elements satisfying $k_{i}\not\in K(k_{1}\cdots k_{i-1})$, such that
$$
K=F(k_{i}),i\in I, |I|\ge |\mathbb{N}|
$$
Then let $K_{1}=F(k_{1})$, $K_{2}=F(k_{1},k_{2})$, etc by picking a subset of $I$ with cardinality $\mathbb{N}$. We see $|K_{i}:F|=|K_{i}:K_{i-1}||K_{i-1}:F|$ must increase strictly as $i\rightarrow \infty$. But we know $K\supset K_{i}$ for any $i$ and $|K:F|<N$ for some $N$. This gives a contradiction. 
