When are nonintersecting finite degree field extensions linearly disjoint?

Question

Let $F$ be a field, and let $K,L$ be finite degree field extensions of $F$ inside a common algebraic closure. Consider the following two properties:

(i) $K$ and $L$ are linearly disjoint over $F$: the natural map $K \otimes_F L \hookrightarrow KL$ is an injection.

(ii) $K \cap L = F$.

It is well known that (i) $\implies$ (ii): see e.g. $\S$ 13.1 of my field theory notes. This implication ought to be (and usually is) followed up with the comment that (ii) does not imply (i) without some additional hypothesis: for instance take $F = \mathbb{Q}$, $K = \mathbb{Q}(\sqrt[3]{2})$, $L = \mathbb{Q}(e^{\frac{2 \pi i}{3}}\sqrt[3]{2})$, or more generally, any two distinct, but conjugate, extensions of prime degree. Thus some normality hypothesis is necessary. What is the weakest such hypothesis?

The following is a standard result: see e.g. $\S$ 13.3, loc. cit.

Theorem: If $K/F$ and $L/F$ are both Galois, then (ii) $\implies$ (i).

I remember this point coming up in a course I took as a graduate student, and the instructor claimed in passing that it was enough for only one of $K$, $L$ to be Galois. None of the standard field theory texts I own contains a proof of this. But by online searching I found an algebra text of P.M. Cohn which proves something stronger.

Theorem: If at least one of $K,L$ is normal and at least one is separable [it is permissible for the same field to be both normal and separable], then (ii) $\implies$ (i).

I wasn't able to freely view the proof, so if someone can pass it along to me I'd be appreciative. Still, I think I know of no counterexamples to the following stronger

Claim: If at least one of $K,L$ is normal, then (ii) $\implies$ (i).

Is this in fact true? (I believe I have seen it claimed in certain research papers, e.g. one by Piatetski-Shapiro. But because terminology and running separability hypotheses are not so standardized, I don't take this as conclusive evidence.)

Answer 1

There are counterexamples to the Claim. Recall that an extension of a field $F$ of characteristic $p>0$ is separable if and only if it is linearly disjoint from $F^{1/p^\infty}$ over $F$. Note that $F^{1/p^\infty}/F$ is a normal extension. There is an inseparable algebraic extension $K/F$ which does not intersect $F^{1/p^\infty} \smallsetminus F$ (see Does an inseparable extension have a purely inseparable element?). This gives a counterexample.

Answer 2

This is also very late, but here is a self-contained answer for future reference of those tired of chasing references (I was). This is, by the way, Exercise 5.26 in Morandi's Fields and Galois Theory.

Theorem: Let $K, L$ be algebraic extensions of $F$ such that at least one of $K$ and $L$ is normal and at least one separable, then $K\cap L=F$ implies that $K$ and $L$ are linearly disjoint over $F$.

Proof: We may assume without loss of generality that both $K/F$ and $L/F$ are finite. Suppose that $K/F$ is normal.

If $K/F$ is also separable, this is the "Theorem on Natural Irrationalities", the proof of which runs as follows. Note that $K/F$ is the splitting field of some separable polynomials over $F$, and then $KL/L$ is the splitting field of the same separable polynomials over $L$ and is hence Galois. The restriction map $\operatorname{Gal}(KL/L)\to \operatorname{Gal}(K/F)$ is clearly injective, and by finite Galois theory has image $\operatorname{Gal}(K/M)$ for the fixed field $M$ of the image, which is easily seen to be $K\cap L=F$. Therefore, $\operatorname{Gal}(KL/L)\to \operatorname{Gal}(K/F)$ is an isomorphism, so $[KL:L]=[K:F]$, whence $K$ and $L$ are linearly disjoint over $F$.

Now suppose that $L/F$ is separable instead, and let $S$ (resp. $I$) denote the separable (resp. purely inseparable) closure of $F$ in $K$. Then $S/F$ is Galois, so by the case already shown we have $[SL:L]=[S:F]$. Now $SL/F$ is separable and $I/F$ purely inseparable, so that $SL$ and $I$ are linearly disjoint over $F$; in particular, we have $[KL:SL]=[SLI:SL]=[I:F]$. Then the result follows from the computation $$\begin{aligned} [KL:L][SLI:L]&=[SLI:SL]\cdot [SL:L]\\&= [I:F]\cdot [S:F]\\&= [I:F]\cdot [K:I] = [K:F],\end{aligned}$$ where in the first equality on the last line we are using that $I=K^{\operatorname{Aut}(K/F)}$, whence $K/I$ is Galois with Galois group $\operatorname{Aut}(K/F)\cong \operatorname{Gal}(S/F)$ and hence, in particular, $[S:F]=[K:I]$. Q.E.D.