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We have $n$ distinct prime numbers $p_1,\cdots ,p_n$ and I am asked to show that $[\mathbb{Q}(\sqrt[r]{p_1},\cdots ,\sqrt[r]{p_n}):\mathbb{Q}]=r^n$ where $r\in \mathbb{N}$.

I tried to solve it by induction. The case $n=1$ is trivial. If we let

$F=\mathbb{Q}(\sqrt[r]{p_1},\cdots ,\sqrt[r]{p_n})$

$E=\mathbb{Q}(\sqrt[r]{p_1},\cdots ,\sqrt[r]{p_n},\sqrt[r]{p_{n+1}})$

$L=\mathbb{Q}(\sqrt[r]{p_2},\cdots ,\sqrt[r]{p_n})$

then by the inductive hypothesis $[F:\mathbb{Q}]=r^n$, $[L:\mathbb{Q}]=r^{n-1}$, $[F:L]=r$. We want to show that $[E:F]=r$. We know that $[E:F]\leq r$. If $[E:F]<r$ then the minimal polynomial of $\sqrt[r]{p_{n+1}}$ over $F$ is

$\prod_{i\in I\subset \left \{0,\cdots ,r-1\right \}}X-\sqrt[r]{p_{n+1}}\zeta_r ^i$ where $\left |I\right |=m<r$. Looking at the last coefficient of this polynomial we obtain $\sqrt[r]{p_{n+1}^m}\zeta_r ^N\in \mathbb{R}$ for some $N\in \mathbb{N}$, whence $\zeta_r ^N=\pm 1$ and $\sqrt[r]{p_{n+1}^m}\in F$.

Therefore $\sqrt[r]{p_{n+1}^m}=\sum_{i=0}^{r-1}a_i\left (\sqrt[r]{p_1}\right )^i$ where $a_i\in L$.

I don't know how to continue, I don't even know if I am doing the right thing.

Is it true that the trace over $F/L$ of $\sqrt[r]{p_{1}^i}$ equals $0$ for every $i\neq 0$? Because in that case taking trace over $F/L$ we would obtain $0=ra_0$, thereby $a_0=0$ and I think that could be helpful.

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    $\begingroup$ See this answer for on way. $\endgroup$ – Bill Dubuque Nov 14 '16 at 21:21
  • $\begingroup$ But that answer is for the case $r=2$... $\endgroup$ – solomeo paredes Nov 14 '16 at 21:26
  • $\begingroup$ It gives citations for higher degree generalizations. $\endgroup$ – Bill Dubuque Nov 14 '16 at 21:29
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This question has been asked many times for the particular case $r=2$. The central technical difficulty consists in translating a condition of additive linear independence into a multiplicative one. The most convenient way is to use Kummer theory under certain hypotheses which render the problem easier than in the general references given by @Bill Dubuque. For example:

Fix $r\ge2$. Let $K$ be the cyclotomic field $\mathbf Q(\zeta_r)$ and let us prove by induction the property $(P_m)$ that $[K(\sqrt [r] {p_1} , ... , \sqrt [r]{p_m}): K]=r^m$ for all $m$'s and all primes $p_j$'s not dividing $r$, i.e. unramified in $K$. For $m=1$, the Eisenstein criterion shows that $[\mathbf Q(\sqrt[r] {p_1}):\mathbf Q]= r$, hence, by linear disjointness (which follows from the non ramification condition), $(P_1)$ holds. Suppose that $(P_{n})$ holds. Then, by Kummer theory, the subgroup $B_n$ of $K^{*}/K^{*r}$ generated by the classes of $p_1,...,p_n$ mod $K^{*r}$ is isomorphic to $(\mathbf Z/r)^{n}$ in additive notation. Since $(P_{n+1})$ is equivalent to the linear disjointness of $K(\sqrt [r] {p_1} , ... , \sqrt[r]{p_n})$ and $ K(\sqrt[r] p_{n+1})$, it amounts to showing that for any $a_{n+1} < r$, the class of $p_{n+1}^{a_{n+1}}$ cannot belong to $B_n$. Suppose the contrary, i.e. that that we have an equation $p_{n+1}^{a_{n+1}} = p_{1}^{a_1}...p_{n}^{a_n}x^r$, where $a_j < r$ and $x\in K^{*}$, or, after clearing denominators, an equality of ideals $(p_{n+1}^{a_{n+1}}).I^r = (p_{1}^{a_1}...p_{n}^{a_n}).J^r$ (¤), where $I$ and $J$ are two coprime integral ideals. To simplify notations, write $p$ for any of the $p_j$'s. Since $p$ is unramified, the principal ideal $(p)$ is prime or splits as a product $P_1...P_s$ of $s$ distinct prime ideals. The participation of $(p)$ or of any such $P_i$ to the equation (¤) is of the form $(p)^a$ or $P_i^{a}$, with $a<r$. Impossible,by the uniqueness of the prime decomposition of ideals in a Dedekind ring. Thus $(P_m)$ holds for $K$, hence, by linear disjointness, the analogous property holds for $\mathbf Q$.

Question: The non ramification hypothesis is not needed when $r=2$. When $r>2$ what happens for the divisors of $r$ ?

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