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Under what conditions is $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha+\beta)$? Or $\mathbb{Q}(\alpha\beta)$?

Working: Firstly note that for any $\alpha,\beta$ we can write $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\gamma)$ for some element $\gamma$. We can easily show that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$. Initially I suspected this would be the case for all pairs $(\alpha,\beta)$, but the following provides a counterexample:

$\mathbb{Q}(1+\sqrt{2},1-\sqrt{2})$ - to see why this is a counterexample, note that the sum of the elements is $2$, which is already in $\mathbb{Q}$, yet the element $1+\sqrt{2}\not\in\mathbb{Q}$.

So it would be reasonable to suggest that $\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha+\beta)$ in the case where $\alpha,\beta$ are not conjugates of one another. How would such a statement be proven?

  • $\alpha$ and $\alpha$ are conjugates of one another, yet $\Bbb Q(\alpha,\alpha)=\Bbb Q(\alpha+\alpha)$. I'm sure one can find less trivial examples, as long as one goes above degree 2. – Arthur Oct 28 '22 at 14:59
  • It is easy to cook up examples, where $[\Bbb{Q}(\alpha,\beta):\Bbb{Q}(\alpha+\beta)]$ becomes arbitrarily large. Consider $\alpha=\sqrt2+\sqrt3+\sqrt5+\sqrt7$, $\beta=-\sqrt3-\sqrt5-\sqrt7+\sqrt{11}$, when $\Bbb{Q}(\alpha,\beta)$ is a degree $2^5$ extension of the rationals, but $\Bbb{Q}(\alpha+\beta)$ is a quartic. Add/subtract more square roots of distinct primes if desired. – Jyrki Lahtonen Oct 28 '22 at 15:38
  • See 1, 2 for the non-obvious facts leading to my previous comment. – Jyrki Lahtonen Oct 28 '22 at 15:46

2 Answers2

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$\sqrt{2}+i$ and $\sqrt{3}-i$ are not conjugates of each other, but $\Bbb Q(\sqrt{2}+i,\sqrt{3}-i) \neq \Bbb Q(\sqrt{2}+\sqrt{3})$

However, something close is true and this is follows from one of the standard proofs of the primitive element theorem. We have $\Bbb Q(\alpha+\lambda \beta)=\Bbb Q(\alpha,\beta)$ for all but finitely many $\lambda \in \Bbb Q$. More precisely, for all but at most $[\Bbb Q(\alpha):\Bbb Q][\Bbb Q(\beta):\Bbb Q)]$ values of $\lambda$. (Regardless of whether $\alpha$ and $\beta$ are conjugate.) So in some sense, it is true that $\Bbb Q(\alpha,\beta)=\Bbb Q(\alpha+\beta)$ "most of the time", because for a counterexample, one has to choose $\alpha$ and $\beta$ so that $\lambda=1$ is one of the finitely many exceptions.

Lukas Heger
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This kind of question was asked on MathOverflow 12 years ago: look here.

KCd
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