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I have some issues with this exercise. I have that F is a field and $d_1,...,d_k\in F\backslash\{0\}$

I have that $K:=F\left(\sqrt{d_1},...,\sqrt{d_k}\right)$

I have to show that $K=F\left(\sqrt{d_1}+...+\sqrt{d_k}\right)$. I have shown that $F\left(\sqrt{d_1}+...+\sqrt{d_k}\right) \subseteq F \left(\sqrt{d_1},...,\sqrt{d_k}\right)$ and now wants to show that $F\left(\sqrt{d_1},...,\sqrt{d_k}\right)\subseteq F\left(\sqrt{d_1}+...+\sqrt{d_k}\right)$. I am thinking of finding the conjugates of $\sqrt{d_1}+...+\sqrt{d_k}$ by finding the minimal polynomium in which $\sqrt {d_1}+...+\sqrt{d_k}$ is a root and then find the other roots of this minimal polynomium which will then be the conjugates. And then I am thinking of using these conjugates to say something about that $\sqrt{d_1},...,\sqrt{d_k} \in \left(\sqrt{d_1}+...+\sqrt{d_k}\right)$.

But I have problems finding the minimal polynomial in which $\sqrt{d_1}+...+\sqrt{d_k}$ is a root, so I am hoping that some of you can help me with that.

I have until now as well shown $[K:F]=2^k$. And I have shown that $K/F$ is Galois, and that for each $i=1,...,k$ $\sigma_i \in Gal(K/F)$ such that $\sigma_i\sqrt{d_j}=\sqrt{d_j}$ for $i\neq j$, but $\sigma_i \sqrt{d_i}=-\sqrt{d_i}$.

Sumi
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  • Have you tried induction on $k$? – Gerry Myerson Oct 01 '20 at 14:08
  • Hi Gerry! Yes I have. For k=1 it is obvious because $F(\sqrt{d_1}) \subseteq F(\sqrt{d_1})$. Then I assume it holds for k i.e.$F(\sqrt{d_1},...,\sqrt{d_k}) \subseteq F(\sqrt{d_1}+...+\sqrt{d_k})$ and want to show that it holds for k+1 i.e. $F(\sqrt{d_1},...,\sqrt{d_{k+1}}) \subseteq F(\sqrt{d_1}+...+\sqrt{d_{k+1}})$ which gives me some problems. – Sumi Oct 01 '20 at 14:36
  • You won't necessarily need to find the minimal polynomial of the sum of square roots. You can reuse the argument from this old answer of mine if you also read the comments under it. But, it is not necessarily true that the degree of the extension is $2^k$, so you need to exercise a bit of care. For example, if $F=\Bbb{Q}$, $d_1=6$, $d_2=10$ and $d_3=15$ then ... See this thread for more. Many good answers there. – Jyrki Lahtonen Oct 01 '20 at 21:57
  • Thank you for your help, but I have get a hint about considering the conjugates of $\sqrt{d_{1}}+...+\sqrt{d_{k}}$ and that's why I wanna find the conjugates, and that's what I am having a hard time doing – Sumi Oct 02 '20 at 08:17
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    Ok. The details really depend on the eventual dependencies such as $\sqrt{6}\sqrt{10}\sqrt{15}=30$ among the different square roots. The Galois theoretic idea is to show that the sum $S=\sqrt{d_1}+\cdots+\sqrt{d_n}$ satisfies $\sigma(S)\neq S$ for all the non-trivial elements $\sigma\in Gal(K/F)$. That is, after all, equivalent to $K=F(S)$. But there are a few pitfalls along the way. For example if $F$ has characteristic two, the extension is not Galois. – Jyrki Lahtonen Oct 02 '20 at 11:15
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    (cont'd)For a different kind of a problem: If $a_1+a_2+a_3=0$, then, by suitable choices of square roots, and $d_1=da_1^2, d_2=da_2^2, d_3=da_3^2$ we get $S=0$ even though $\sqrt{d}\in K\setminus F$. In other words, the result may be false, if we are not given some extra information. – Jyrki Lahtonen Oct 02 '20 at 11:16
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    The problems mentioned by @Jyrki are resolved by requiring that the radicals are multiplicatively independent over $F$, i..e. no product of distinct radicals lies in $F$, as in this proof, This generalizes to higher degree radicals - see the papers of Mordell and Siegel that I cite there. The algebraic essence of the matter is clarified when one studies Galois theory (of Kummer extensions). – Bill Dubuque Oct 14 '20 at 08:43

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