I want to prove this statement.
$$\mathbb{Q}(\sqrt 2, \sqrt 3, ..... , \sqrt n ) = \mathbb{Q}(\sqrt 2 + \sqrt 3 + .... + \sqrt n )$$ for any $n >1$.
It looks like a very hard problem.
How can I approach this one?
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2$\begingroup$ What about How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime? ? $\endgroup$– Martin RDec 14, 2015 at 18:58
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2$\begingroup$ An answer to your question is buried in the comments to this answer by yours truly. $\endgroup$– Jyrki LahtonenDec 14, 2015 at 18:59
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$\begingroup$ MartinR, Jyrki Thanks! $\endgroup$– nicksohnDec 14, 2015 at 19:45
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$\begingroup$ It does look hard, but some aspects are easy. Try to break it into an "easy" half and a "difficult" half, and then develop an approach to the difficult direction by considering the situation for small $n$. $\endgroup$– hardmathDec 15, 2015 at 0:50
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1 Answer
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Let us show that $$\mathbb{Q}(\sqrt[d_1]{a_1}, \ldots, \sqrt[d_n]{a_n}) = \mathbb{Q}( \sqrt[d_1]{a_1}+ \cdots + \sqrt[d_n]{a_n})$$ ( $a_l$, $d_l$ positive integers).
By Galois theory, if $K$ be a field, $L\supset K$ a Galois extension, $\alpha$, $\beta$ in $L$ so that every $\sigma \in \text{Gal}(L/K)$ that fixes $\alpha$ also fixes $\beta$, then $\beta \in K(\alpha)$.
Let now $\alpha = \sqrt[d_1]{a_1}+ \cdots + \sqrt[d_n]{a_n}$ and $\beta_l = \sqrt[d_l]{a_l}$, $1\le l \le n$. Consider a Galois extension $L$ containing all the $\beta_l$ (and so $\alpha$, too). Let $\sigma \in \text{Gal}(L/\mathbb{Q})$. We have $\sigma(\beta_l) = \omega_l \beta_l$ where $\omega_l$ is a $d_l$-th root of $1$. Assume that $\sigma(\alpha) = \alpha$, that is $$\sum \omega_l \beta_l = \sum \beta_l.$$ Now the $\beta_l$'s are real and positive and we have $$| \sum \omega_l \beta_l| \le \sum | \omega_l \beta_l | = \sum \beta_l$$ and if we have equality then all the modulus $1$ numbers $\omega_l$ have to be equal, and from the above we conclude $\omega_l=1$ for all $l$, and so $\sigma(\beta_l)= \beta_l$ for all $l$.
We conclude $\sqrt[d_l]{a_l} \in \mathbb{Q}( \sqrt[d_1]{a_1}+ \cdots + \sqrt[d_n]{a_n})$ for all $l$.
answered Dec 14, 2015 at 19:27
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1$\begingroup$ Great argument which is simple and yet powerful +1 $\endgroup$– Paramanand Singh ♦Dec 13, 2020 at 10:56
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1$\begingroup$ I had to think about "if $K$ is a field, $L\supseteq K$ a Galois extension, $\alpha, \beta \in L$ so that every $\sigma \in $Gal$(L/K)$ that fixes $\alpha$ also fixes $\beta$, then $\beta \in K(\alpha)$." So I addeded details. We have $H=\{\sigma \in $Gal$(L/K) \mid \sigma \alpha = \alpha \} = $Gal$(L/K(\alpha))$ and therefore $L^H=K(\alpha)$ (fixed field), since Gal$(L/K(\alpha))$ is a galois extension. Similarly, for $U:=\{\sigma \in $Gal$(L/K) \mid \sigma \beta = \beta \} $ we see that $L^U = K(\beta)$. Now, since $H \subseteq U$, we conclude that $K(\beta)=L^U \subseteq L^H=K(\alpha)$. $\endgroup$ Aug 6, 2022 at 13:13
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1$\begingroup$ @Fabio: Yes, indeed. But once you are used to the argument, the story is that short. :-) $\endgroup$ Aug 6, 2022 at 15:22