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Is $\mathbf Q(\sqrt 2,\sqrt 3,\sqrt 5)=\mathbf Q(\sqrt 2+\sqrt 3+\sqrt 5)$?

Say $L=\mathbf Q(\sqrt 2,\sqrt 3,\sqrt 5)$ and $K=\mathbf Q(\sqrt 2+\sqrt 3+\sqrt 5)$.

It is easy to show that $\mathbf Q(\sqrt 2,\sqrt 3)=\mathbf Q(\sqrt 2+\sqrt 3)$. Also, $[L:\mathbf Q]=8$.

If we assume that $L\neq K$, then we will have $[K:\mathbf Q]=4$. The reason for this is that $K(\sqrt 5)=L$.

Now since $K$ is a superfield of $\mathbf Q$, and $[\mathbf Q(\sqrt 5):\mathbf Q]=2$, we have $[K(\sqrt 5):K]\leq 2$.

Since we have assume that $K\neq L$, we must have $[K(\sqrt 5):K]=2$. By the tower law, $[K(\sqrt 5):K][K:\mathbf Q]=[K(\sqrt 5):\mathbf Q]=[L:\mathbf Q]=8$, giving $[K:\mathbf Q]=4$.

I am not able to make any further progress.

Generalization:

Let $p_1,\ldots,p_n$ be pairwise distinct primes. Is $\mathbf Q(\sqrt p_1+\cdots+\sqrt p_n)=\mathbf Q(\sqrt p_1,\ldots,\sqrt p_n)$?

Just something I think might be useful in solving the above: It is known that if $p_1,\ldots,p_n$ are pairwise distinct primes, then $[\mathbf Q(\sqrt p_1,\ldots,\sqrt p_n):\mathbf Q]=2^n$.

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  • $\begingroup$ Regarding the second question, do you know any Galois theory? Up to what? $\endgroup$ – Git Gud Jan 8 '15 at 18:50
  • $\begingroup$ I am doing a course on Galois Theory. The course has just started. $\endgroup$ – caffeinemachine Jan 8 '15 at 18:51
  • $\begingroup$ For the fact about the degree, please see this.. That is enough to solve your problem, for $\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_n})$ has degree $\le 2^n$ over $\mathbb{Q}$. $\endgroup$ – André Nicolas Jan 8 '15 at 19:04
  • $\begingroup$ You can easily prove the converse inclusion by proving that the conjugates of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ are $\pm\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}$. $\endgroup$ – Jack D'Aurizio Jan 8 '15 at 19:05
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    $\begingroup$ You only need $\le 2^n$. For clearly $\mathbb{Q}(\sqrt{p_1}+\cdots+\sqrt{p_n})$ is a subfield of $\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_n})$. $\endgroup$ – André Nicolas Jan 8 '15 at 19:16
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There's a trick for the first question.

Let $\alpha=\sqrt{2}+\sqrt{3}+\sqrt{5}$ and note that

$$\begin{align} \alpha ^3\mathop =^{(1)}&\,26\sqrt 2+24\sqrt 3+20\sqrt 5+6\sqrt {30},\\ \alpha ^5\mathop =^{(2)}&\,784\sqrt 2+664\sqrt 3+520\sqrt 5+200\sqrt {30},\\ \alpha ^7\mathop =^{(3)}&\,23024\sqrt 2+18976\sqrt 3+14720\sqrt 5+5936\sqrt {30}, \end{align}$$

(1), (2), (3).

Rewrite the four equalities above as $$\begin{bmatrix}1 & 1 & 1 & 0\\ 26 & 24 & 20 & 6\\ 784 & 664 & 520 & 200\\ 23024 & 18976 & 14720 & 5936 \end{bmatrix}\begin{bmatrix}\sqrt 2\\ \sqrt 3\\ \sqrt 5\\ \sqrt{30} \end{bmatrix}=\begin{bmatrix} \alpha\\ \alpha^3\\\alpha ^5\\ \alpha^7\end{bmatrix}_.$$

The matrix $M$ on the left is invertible and $M^{-1}\in\mathcal M_{4\times 4}( \mathbb Q)$. To see that $M$ is invertible in a quick way, it suffices to prove that its determinant is not a multiple of $5$ (because $0$ is a multiple $5$). Consider the matrix modulo $5$ and check that the determinant is not a multiple of $5$ (WA link).

It follows that $$\begin{bmatrix}\sqrt 2\\ \sqrt 3\\ \sqrt 5\\ \sqrt{30} \end{bmatrix}=M^{-1}\begin{bmatrix} \alpha\\ \alpha^3\\\alpha ^5\\ \alpha^7\end{bmatrix}$$ and each entry of the matrix on the RHS is in $\mathbb Q(\alpha)$.

This proves the hard inequality $\mathbb Q(\sqrt 2, \sqrt 3, \sqrt 5)\subseteq \mathbb Q(\alpha)$.

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  • $\begingroup$ Oops! I did not know this! Thanks! $\endgroup$ – Bombyx mori Jan 8 '15 at 19:18
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    $\begingroup$ This is nice. May be this argument can be generalized. The main difficulty would be to show that $M$ will be invertible in the general case. $\endgroup$ – caffeinemachine Jan 8 '15 at 19:18
  • $\begingroup$ @Bombyxmori I learned it from a book. It's very nice. I added another useful trick. $\endgroup$ – Git Gud Jan 8 '15 at 19:22
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    $\begingroup$ I just learned like 4 tricks I never knew about in this answer :O $\endgroup$ – Emily Jan 8 '15 at 19:55
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    $\begingroup$ Proposition $$\operatorname{det}A = 0 \iff \operatorname{det}A = 0 \operatorname{mod}p,\quad\text{for ever $p$ prime.} $$ I just realized this is actually true, so it is sufficient to find a prime for which the determinant is non-zero. I won't erase the comment for it may be helpful to someone else. $\endgroup$ – hjhjhj57 Jan 9 '15 at 7:51
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This is a duplicate of 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?. So you can take a look at the other post.

Without using Galois theory, I think some manipulation along the lines $$ \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}} $$ might be helpful, as similar manipulations should give you a few other equations. Then via some cancellation you can get $\sqrt{2},\sqrt{3},\sqrt{5}$ individually. But I do not have a proof based on this.

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  • $\begingroup$ Thank you. It is good to know that the claim is true. I need to learn more Galois theory before I can understand the answer in the link you provided. $\endgroup$ – caffeinemachine Jan 8 '15 at 19:15
  • $\begingroup$ @caffeinemachine: I wish I have a proof of it based on elementary means, too. Thanks! $\endgroup$ – Bombyx mori Jan 8 '15 at 19:17
  • $\begingroup$ The Primitive Element Theorem guarantees that there exists a single element $\alpha$ in $\mathbb Q(\sqrt p_1,\ldots,\sqrt p_n)$ such that $\mathbb Q(\alpha)=\mathbb Q(\sqrt p_1,\ldots,\sqrt p_n)$. $\endgroup$ – caffeinemachine Jan 8 '15 at 19:24

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