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?

• This seems like the sort of question that can be answered immediately by anyone who understands all of the terms in the question. So I recommend familiarizing yourself with the relevant definitions and make sure you understand them. – anon Feb 12 '14 at 0:48

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.

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.

• How can we find a "basis" for the extension field? I mean, how can we find some element $\alpha \in K$ where $K=F[\alpha]$? – Ninja Apr 28 '17 at 15:48
• @Ninja, see Primitive element theorem. – lhf Apr 28 '17 at 16:17

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.

• As you have defined $K_i$, one has $[K_i:F] = \aleph_0$ for all $i$. So the degrees are not strictly increasing. – Pete L. Clark Feb 12 '14 at 0:42
• No, each $k_{i}\in K$ so $|K_{i}:F|\le |K:F|<N$. – Bombyx mori Feb 12 '14 at 0:53
• I see I did not state it precisely. I hope this statement is more precise. – Bombyx mori Feb 12 '14 at 1:07
• In your earlier version you wrote that the $k_i$'s were algebraically independent elements over $F$ and $K_i = F(k_1,\ldots,k_i)$. In this setting one has $[K_i:F] = \aleph_0$ for all $i$ so the degrees are not strictly increasing, as you had claimed. I don't see a precision problem. Anyway, your current version is correct. (In my opinion it is unnecessarily complicated: the degree is finite means the extension is finitely generated as a module, which means that it is finitely generated as an algebra, which means that it is finitely generated as a field extension). – Pete L. Clark Feb 12 '14 at 1:40
• Yes. Thanks for the advice. – Bombyx mori Feb 12 '14 at 1:47