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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 $[\mathbf Q(\sqrt p_1,\ldots,\sqrt p_n):\mathbf Q]=2^n$.

caffeinemachine
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2 Answers2

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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 $\mathbb Q(\sqrt 2, \sqrt 3, \sqrt 5)\subseteq \mathbb Q(\alpha)$.

Git Gud
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  • Oops! I did not know this! Thanks! – Bombyx mori Jan 08 '15 at 19:18
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    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. – caffeinemachine Jan 08 '15 at 19:18
  • @Bombyxmori I learned it from a book. It's very nice. I added another useful trick. – Git Gud Jan 08 '15 at 19:22
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    I just learned like 4 tricks I never knew about in this answer :O – Emily Jan 08 '15 at 19:55
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    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. – hjhjhj57 Jan 09 '15 at 07:51
  • @GitGud Do you by chance still remember the book where you found this extraordinary trick? And what was the reason for choosing $\alpha^3, \alpha^5, \alpha^7$ and not other exponents? Thanks for such a great post! [+1] – Dr. Mathva Oct 24 '22 at 20:42
  • @Dr.Mathva The book, which is in Portuguese, was "Teoria de Galois" (I don't remember which edition, but it shouldn't matter) by Owen J. Brison. You can find a reference to it here, however I'm not sure how to purchase it. About the exponents, I just tried a few low numbers and used Wolfram Alpha to do the work for me, I have no insight there. – Git Gud Oct 30 '22 at 09:30
  • Got this answer by asking ai…I am looking for any trick to find out the system of equations, that is exactly what I want, great! – Tim Jun 21 '23 at 17:47
  • @Tim Interesting! Thank you for sharing. I never used these systems for mathematics. – Git Gud Jun 26 '23 at 00:19
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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.

Community
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Bombyx mori
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  • 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. – caffeinemachine Jan 08 '15 at 19:15
  • @caffeinemachine: I wish I have a proof of it based on elementary means, too. Thanks! – Bombyx mori Jan 08 '15 at 19:17
  • The Primitive Element Theorem guarantees that there exists a single element $\alpha$ in $\mathbb Q(\sqrt p_1,\ldots,\sqrt p_n)$ such that $\mathbb Q(\alpha)=\mathbb Q(\sqrt p_1,\ldots,\sqrt p_n)$. – caffeinemachine Jan 08 '15 at 19:24