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Let $K$ be a field of characteristic $0$, $\alpha,\beta$ algebraic elements in an algebraic closure of $K$.

Let $K(\alpha,\beta)/K$ be the associated field extension. If both $K(\alpha)$ and $K(\beta)$ are both proper subfields of $K(\alpha,\beta)$, then do we have $K(\alpha,\beta)=K(\alpha+\beta)$?

I believe the answer is no, but I am struggling to find a counterexample, given that this is generically true.

By passing to a galois closure and using the fundamental theorem of galois theory, I believe this is equivalent to finding subgroups $H,K$, where $H\neq K$ of $G$ the galois group where $\alpha$ has stabiliser $H$, $\beta$ has stabiliser $K$, but the stabiliser of $\alpha+\beta$ is strictly larger than $H\cap K$.

Chris H
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    Have you tried, $\beta=-\alpha$? – Mohan Jun 13 '20 at 02:13
  • Then they aren't proper in $K(\alpha,\beta)$, that condition was to exclude the trivial cases, edited my equivalent statement, since it didn't exclude this. – Chris H Jun 13 '20 at 02:15
  • I think there is some trick like $\alpha=2+\sqrt2$ and $\beta=2-\sqrt2$. I’d have to check the details, but something like this definitely works. – Clayton Jun 13 '20 at 02:54

2 Answers2

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$\Bbb{Q}(2^{1/4}+ i 2^{1/4},2^{1/4}- i 2^{1/4})$ is larger than $\Bbb{Q}(2^{1/4}+ i 2^{1/4})\cong \Bbb{Q}((-8)^{1/4})$ and $\Bbb{Q}(2^{1/4})$

reuns
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The answer is positive when $L/K$ is a biquadratic extension, see e.g. https://math.stackexchange.com/a/3325514/300700. The property does not hold in general, but I guess it could hold for specific kummerian extensions "mimicking" the biquadratic case, e.g. $L=K(\sqrt [p] a, \sqrt [p] b)$, where $p$ is a prime s.t. $K$ contains a primitive $p$-th root of unity. Then $Gal(L/K)\cong C_p\times C_p$ is dual to the Kummer radical $R=<[a],[b]>$, an $\mathbf F_p$-plane generated (multiplicatively) by the classes of $a,b$ mod ${K^*}^p$. All the $(p+1) \mathbf F_p$-lines in that plane are generated by vectors of the form $[a^ib^j]$, and it remains only to pick up an adequate sub-extension of $L$ of the form $K(\sqrt [p] a^i + \sqrt [p] b^j)$.

NB This is not in contradiction with the example given by @reuns

nguyen quang do
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