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Given algebraic numbers $a$ and $b$, is it the case that all algebraic conjugates of $a+b$ take the form $a'+b'$ where $a'$ and $b'$ are algebraic conjugates of $a$ and $b$ respectively?

Alex Kindel
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  • Yes: this is true. Do you know a bit of Galois theory? – Crostul Feb 25 '18 at 21:45
  • @Crostul Vaguely. I know the basic definitions involved, but I don't know many results that relate the theory to more concrete things like my question. – Alex Kindel Feb 25 '18 at 21:52
  • @Crostul Maybe my question doesn't seem very concrete. The idea is that if I have a number that is a sum of surds, I can easily generate a set of numbers that I can be sure contains its conjugates. – Alex Kindel Feb 25 '18 at 21:54
  • The conjugates of $a + b$ is ${} \pm a \pm b$ such that the $\pm$ operations are independent of each other. – Mr Pie Feb 25 '18 at 22:37
  • @user477343 That's not true in general. If $a$ and $b$ are square roots, then it's true that all conjugates of $a+b$ are of that form, but not all numbers of that form are guaranteed to be conjugates. For a trivial example, take $a$ and $b$ both equal to $\sqrt{2}$. Then the conjugates of $a+b$ are only $2\sqrt{2}$ and $-2\sqrt{2}$ - the numbers formed by taking conjugates of $a$ and $b$ of opposite signs aren't included. Things also get more complicated with roots of indices larger than two. – Alex Kindel – Alex Kindel Feb 26 '18 at 02:33
  • @AlexKindel oh... well, thank you for teaching me :) – Mr Pie Feb 26 '18 at 02:40

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There is a characterization of conjugates by means of morphism of fields.

Let $K$ be a finite normal extension of $\Bbb Q$, and $a \in K$. Then, all conjugates of $a$ are $$\{ \sigma (a) : \sigma \in \mathrm{Gal} (K / \Bbb Q) \}$$

Now, if you have two algebraic numbers, say $a,b$, consider the normal closure of $\Bbb Q (a,b) / \Bbb Q$, and call it $K$. Then the conjugates of $a+b$ have the form $$\sigma (a+b)= \sigma (a) + \sigma (b)$$ where $\sigma \in \mathrm{Gal} (K / \Bbb Q)$. This means that the conjugates of $a+b$ are exactly sums of conjugates of $a$ and $b$.

Let's make an example. The conjugates of $\sqrt 2 + \sqrt 3$ are exactly $$\sqrt 2 + \sqrt 3 , \ \sqrt 2 - \sqrt 3 , \ -\sqrt 2 + \sqrt 3 , \ -\sqrt 2 - \sqrt 3$$

Crostul
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    What I tried and failed at before posting the question was to show this using more elementary means, because they're what I'm comfortable with, but this is a very direct consequence of the theory indeed. If I understand correctly, all one needs is "there exists a field homomorphism that sends conjugates to conjugates" together with the fact that field homomorphisms preserve addition. – Alex Kindel Feb 25 '18 at 23:29