# Query regarding the use of the KLM theorem for field extensions

In my book the theorem is stated as:

Let $$K$$, $$L$$ and $$M$$ be fields such that $$K ⊂ L ⊂ M$$ and let the degrees $$[M : L]$$ and $$[L : K]$$ be both finite. Then $$[M : L][L : K]=[M : K]$$.

First question: The theorem specifies that it must be the case that $$K ⊂ L ⊂ M$$. Does this mean it cannot be used if the subfields are not proper subfields, so $$K \subseteq L \subseteq M$$?

Second question: Could I reverse the theorem to deduce a $$K\not\subset L\subset M$$ or $$K\subset L\not\subset M$$?

Suppose $$K ⊂ L ⊂ M$$ and I know that $$[M:K]=3$$ and $$[L:K]=2$$. Putting these values into the theorem should give $$[M:L]=\frac{3}{2}$$, which is not a valid degree due to not being an integer. Could I deduce from this that $$K\subset L\not\subset M$$?

• I am pretty sure that the book simply uses the symbol $\subset$ (as opposed to $\subseteq$) for "subset of" (proper or not). Many authors do this. – darij grinberg Mar 23 '17 at 19:31

First question: The theorem is also true if $K=L$ or $L=M$ or even $K=L=M$. If e.g. $K=L$ you simply get $[L:K] = 1$ and $[M:K]= [M:L]$.
Second question: Yes, indeed the theorem is often used like that. One might ask for example, if $\mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\sqrt{5})$. It is easy to see that $[\mathbb{Q}(\sqrt{5}) : \mathbb{Q}] = 2$ and $[\mathbb{Q}(\sqrt{5}) : \mathbb{Q}] = 3$. The assumption $\mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\sqrt{5})$ would then lead to the contradiction you mentioned, showing that $\mathbb{Q}(\sqrt{5}) \not\subset \mathbb{Q}(\sqrt{5})$.