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Let $\alpha$ and $\beta$ be two linearly independent algebraic numbers. Is it always true that $\mathbb{Q}(\alpha,\beta) \cong \mathbb{Q}(\alpha + \beta)$?

Certainly $\mathbb{Q}(\alpha + \beta) \subset\mathbb{Q}(\alpha, \beta)$. The tricky thing is showing the reverse inclusion.

Let $\deg(\alpha) = m$ and $\deg(\beta) = n$ then $\mathbb{Q}(\alpha,\beta)$ has dimension $mn$ over $\mathbb{Q}$. My thought here is that you can take successive powers of $\alpha + \beta$ and generate a system of equations. For some sufficient power, the matrix of coefficients will be solvable. With that, you can isolate $\alpha$ and $\beta$ which proves the reverse inclusion.

Does anyone have any insights?

Bo Rel
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  • I like the thought about taking powers of $\alpha + \beta$. My concern is that a generic term will look like $\begin{pmatrix}n \ k\end{pmatrix} \alpha^{n - k}\beta^k$, which a priori has no reason to belong to $\mathbb{Q}(\alpha + \beta)$, so with each equation you're expanding the list of variables you'll need to solve for. – Charles Hudgins Oct 01 '20 at 04:34
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    No. Consider $\alpha=\root3\of2$ and $\beta=\omega\alpha$. We have $\alpha+\beta=-\omega^2\alpha$. Here $[\Bbb{Q}(\alpha+\beta):\Bbb{Q}]=3$ and $[\Bbb{Q}(\alpha,\beta):\Bbb{Q}]=6$. Note also that here $m=n=3$ and $6<mn=9$. – Jyrki Lahtonen Oct 01 '20 at 04:37
  • @JyrkiLahtonen, I was trying to avoid those cases. Linearly independent wasn't the right condition. What if $m$ and $n$ are different? – Bo Rel Oct 01 '20 at 05:32
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    It seems to me that the concept you want is linearly disjoint. $[\Bbb{Q}(\alpha,\beta):\Bbb{Q}]=mn$ is equivalent to $\Bbb{Q}(\alpha)$ and $\Bbb{Q}(\beta)$ being linearly disjoint extensions. For this to happen it is obviously necessary (but as the above example shows not sufficient) that $\Bbb{Q}(\beta)\cap\Bbb{Q}(\alpha)=\Bbb{Q}$. (assuming both are seen as subfields of a common umbrella field like $\Bbb{C}$ so that it makes sense to talk about their intersection). – Jyrki Lahtonen Oct 01 '20 at 13:48
  • (cont'd) If both $\Bbb{Q}(\alpha)$ and $\Bbb{Q}(\beta)$ are Galois over $\Bbb{Q}$, then the above necessary condition is also sufficient. See this post by Pete L. Clark for more, not the least for a link to his nice set of lecture notes. – Jyrki Lahtonen Oct 01 '20 at 13:51
  • (cont'd) If both extensions are Galois and intersect trivially, then it easily follows that $\alpha+\beta$ generates the composition (and that the composition has the predicted degree). Otherwise it depends. The assumption of $m$ and $n$ being different leaves open possibilities like $\alpha=\root4\of2$, $\beta=\sqrt2$. In that case actually $\Bbb{Q}(\alpha,\beta)=\Bbb{Q}(\alpha+\beta)$, but also clearly $\Bbb{Q}(\beta)\subset\Bbb{Q}(\alpha)$. – Jyrki Lahtonen Oct 01 '20 at 13:58
  • Thank you @JyrkiLahtonen. I'm also a big fan of Dr. Clark. – Bo Rel Oct 02 '20 at 02:45

1 Answers1

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Let $\alpha=\sqrt{3}-\sqrt{2}$ and $\beta=\sqrt{2}$. Then $$\mathbb{Q}(\alpha+\beta)=\mathbb{Q}(\sqrt{3})\neq \mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\alpha,\beta).$$

Of course the primitive element theorem tells you that for all but finitely many $c\in \mathbb{Q}$, $$\mathbb{Q}(\alpha+c\beta)=\mathbb{Q}(\alpha,\beta).$$

tkf
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  • I was trying to avoid these situations, where you luck out with a cancellation. Maybe having them be linearly independent over $\mathbb{Q(\alpha,\beta)}$ would make it more interesting. But thank you for the theorem, I will be looking into that. – Bo Rel Oct 01 '20 at 05:34
  • $\alpha,\beta$ will always be linearly dependant over $\mathbb{Q}(a,b)$ as they lie in a $1$--dimensional vector space. – tkf Oct 01 '20 at 11:13
  • Touche. Do you see what I'm after though? – Bo Rel Oct 02 '20 at 02:47
  • Yes - from @JyrkiLahtonen's comments, it looks like the linear independence that you want in your question is for the ${\alpha^i\beta^j|,,i=0,1,2,\cdots,n-1, j= 0,1,2,\cdots,m-1}$ to be linearly independent over $\mathbb{Q}$. Here $n,m$ are the degrees of the minimal polynomials of $\alpha,\beta$ respectively. Perhaps you should ask if this linear independence guarantees that $\mathbb{Q}(\alpha+\beta)=\mathbb{Q}(\alpha,\beta)$, in a new question, and link it to this one. – tkf Oct 02 '20 at 21:57