Is $\mathbb Q(\sqrt{2},\sqrt{3},\sqrt{5})=\mathbb Q(\sqrt{2}+\sqrt{3}+\sqrt{5})$. [duplicate]

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

Solution 1:

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

Solution 2:

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.