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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 $K\subset L\not\subset M$?

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    $\begingroup$ 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. $\endgroup$ – darij grinberg Mar 23 '17 at 19:31
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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 $\mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\sqrt[3]{5})$. It is easy to see that $[\mathbb{Q}(\sqrt{5}) : \mathbb{Q}] = 2$ and $[\mathbb{Q}(\sqrt[3]{5}) : \mathbb{Q}] = 3$. The assumption $\mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\sqrt[3]{5})$ would then lead to the contradiction you mentioned, showing that $\mathbb{Q}(\sqrt{5}) \not\subset \mathbb{Q}(\sqrt[3]{5})$.

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  • $\begingroup$ Exactly what I needed, thanks :) $\endgroup$ – Mike A Mar 23 '17 at 19:23

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