# $L/K$ field extension and $[M : K] ≤ C$ holds for all real intermediate fields $M$. Then $[L : K]$ is also finite

I want to check my solution for this exercise:

Let $$L/K$$ be a field extension and there exists a constant $$C ∈ \mathbb{N}$$ such that $$[M : K] ≤ C$$ holds for all proper intermediate fields $$M$$. Then $$[L : K]$$ is also finite.

To prove this can I not simply do this?:

$$L=\cup_{i=0}^n M_i$$ for $$K\subset M_i \subset L$$ $$\Rightarrow [L:K]\leq C^n\Rightarrow$$ $$[L:K]$$ is finite.

• No, this can’t work. The easiest counter-example is that of finite extensions with infinitely many intermediate subfields – so you’d need infinitely many $M_i$. Also, the $M_i$ cannot be disjoint since they all contain $K$, and you didn’t prove you could consider finitely many of them. Finally, you should clarify what you meant by “real intermediate”. Do you mean fields that are proper subfields of $L$ and contain $K$? Commented Feb 5 at 14:24
• Yes I mean proper subfields of L and contain K @Aphelli Commented Feb 5 at 14:30
• How can I prove that there are finitely many $M_i$? @Aphelli Commented Feb 5 at 14:35
• You can’t. There are finite extensions with infinitely many intermediate subfields. Commented Feb 5 at 14:40
• so how can I solve this problem @Aphelli ? Commented Feb 5 at 14:41

First note that $$L$$ is algebraic over $$K$$: were an element $$t \in L \setminus K$$ transcendental, then $$K\subsetneq K(t^2)$$ would be a proper infinite subextension.
The poset of proper intermediate fields (wrt inclusion) can be empty only when $$L/K$$ is finite, so we can consider the nonempty case. By problem statement the poset has finite height, so we can find a maximal element $$L/F/K$$. Take $$\alpha \in L \setminus F$$. Then the compositum $$F(\alpha)$$ is equal to $$L$$, and it has $$K$$-dimension at most $$[F:K] [K(\alpha):K] \le C^2$$.
Exercise: find a tower $$L/F/K$$ that attains the equality.