Do there exist two finite extensions of $\mathbb{Q}$, $L=\mathbb{Q}(\alpha)$ and $K=\mathbb{Q}(\beta)$ such that $L\cap K=\mathbb{Q}$, the minimal polynomial $\mu_{\alpha}(X)\in \mathbb{Q}[X]$ of $\alpha$ over $\mathbb{Q}$ is not irreducible over $K$, but the factorization of $\mu_\alpha(X)$ over $K$ has no linear factors ?
(see also here)
Answer
I think this works. Let $f$ be an $A_4$-quartic, let $\alpha$ and $\gamma$ be two of its zeros in a splitting field $M$, let $\beta=\alpha+\gamma$. Note that $L$ has degree 4 over the rationals, and the Galois group of $M/L$ is cyclic of order 3, while $K$ has degree 6 over the rationals (as $\beta$ has 6 distinct conjugates over the rationals), and the group of $M/K$ is cyclic of order 2. If $L\cap K$ had any nonrational elements, they would have to be of degree 2 over the rationals, but $M$ has no subfield of degree 2 over the rationals, since $A_4$ has no subgroup of index 2. Also, $\mu_{\alpha}$ can't be irreducible over $K$, since it has degree 4, but $M/K(\alpha)$ only has degree 2. And $\mu_{\alpha}$ can't have any zeros in $K$, since $K$ would then be a field of degree 6 containing a subfield of degree 4.
I hope someone will check this.
EDIT: A simpler example. Let $\alpha=\root4\of2, \beta=(1+i)\root4\of2$. Then $\mu_{\alpha}(x)=x^4-2$, both $K$ and $L$ are subfields of the splitting field $M$ of $\mu_{\alpha}$, both have degree 4 over the rationals and index 2 in $M$. $L\cap K$ is the rationals, and $\mu_{\alpha}$ factors over $K$ as $$(x^2-(1+i)\root4\of2x+i\sqrt2)(x^2+(1+i)\root4\of2x+i\sqrt2)$$
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