# A generalization of the theorem on natural irrationalities?

Here is the statement of problem 26 of chapter 5 from Morandi's Field and Galois Theory:

Let $$K/F$$ be a normal extension and let $$L/F$$ be an algebraic extension. If either $$K/F$$ or $$L/F$$ is separable, show that $$[KL:L] = [K:K \cap L]$$ Give an example to show that this can be false without the separability hypothesis.

Let's assume that the extensions are finite. If $$K/F$$ is separable, then it is Galois, and the theorem on natural irrationalities applies (the theorem states that $$\mathrm{Gal}(KL/L) \cong \mathrm{Gal}(K/K\cap L)$$. If we only assume $$L/F$$ to be separable, then I am stuck. I tried using the primitive element theorem, so that $$L = F(a)$$ for some $$a \in L$$ and then show that $$[K(a):F(a)] = [K:K\cap F(a)]$$, but I can't get anywhere with this.

I'm also a bit surprised by this problem; the theorem on natural irrationalities is fairly well-known, but everywhere I have seen it is stated with the assumption that $$K/F$$ is Galois. I was hoping that someone could provide some insight on how to do this, or a source for a proof that doesn't use the assumption that $$K/F$$ is Galois.

Edit: A tiny bit of progress. Let $$L/F$$ be separable. If $$\min(F,a)$$ has a root in $$K$$, then it splits over $$K$$ by normality of $$K/F$$. Therefore, $$a \in K$$, and $$K(a) = K$$ and $$K \cap F(a) = F(a)$$. The degree formula follows trivially in this case. We can then assume that the none of the distinct roots of $$\min(F,a)$$ lie in $$K$$.

• In A Course in Galois Theory by Garling, the theorem on natural irrationalities does not require that $K:F$ is separable. The fixed field $L_0 = \phi(\Gamma(KL:L)) \supseteq L$ is introduced, and the theorem then states that $\Gamma(KL:L_0) \cong \Gamma(K:K \cap L_0)$. The book also mentions that the theorem becomes a little simpler if $K:F$ is separable, in which case $L = L_0$. Commented Nov 6, 2020 at 1:36

Assume that $$L/F$$ is separable, and note that$$[KL:L]=[K:K\cap L]\Leftrightarrow [KL:K]=[L:K\cap L]$$.

Let $$L'$$ be the normal closure of $$L$$. Because $$L/F$$ is separable, then $$L'/F$$ is separable, therefore is Galois. By natural irrationalities, $$[KL':K]=[L':K\cap L']$$

and \begin{alignat*}{1} [KL':K]&=&[L':K\cap L']\\ \Leftrightarrow [KL:K] &=& \frac{[L':L][L:K\cap L]}{[KL':KL][K\cap L':K\cap L]}\tag*{(1)} \end{alignat*}

By natural irrationalities \begin{alignat*}{1} [KL':KL]&=&[L':KL\cap L']\tag{2}\\ [KL\cap L':L]&=&[(K\cap L')L:L]\\ &=&[K\cap L':K\cap L'\cap L]\\ &=&[K\cap L':K\cap L]\tag{3} \end{alignat*}

Then we have

\begin{alignat*}{1} [KL':KL][K\cap L':K\cap L]&=[L':KL\cap L'][KL\cap L':L]\tag*{by (2)(3)}\\ &=[L':L]\\ \Rightarrow 1&=\frac{[L':L]}{[KL':KL][K\cap L':K\cap L]}\\ \Rightarrow [KL:K] &= [L:K\cap L]\tag*{by (1)}\\ \Leftrightarrow [KL:L]&=[K:K\cap L]\tag*{QED} \end{alignat*}

• Wow, nice! The theorem on natural irrationalities had to get used three times. That wasn't easy! When $K/F$ and $L/F$ are both not separable, do you have a good example with $[KL:L] \neq [K:K \cap L]$? Commented Dec 1, 2020 at 7:59