$\operatorname{dim} \mathbb{Q}[2^{1/n},3^{1/m}] / \mathbb{Q} = nm$ I'm trying to prove that $d := \operatorname{dim} \mathbb{Q}[2^{1/n},3^{1/m}] / \mathbb{Q} = nm, n \geq 1, m \geq 1$. My attempts:

*

*It is obvious when $\operatorname{gcd}(n,m) = 1$ (because $n | d, m | d$).

*When $m \neq 0 \mod 2$  it is possible (by Hensel's lemma) to embed $\mathbb{Q}[3^{1/m}]$ into field of $2$-adic numbers $Q_2$, and then we can apply Eisenstein criteria to show that $x^n - 2$ is irreducible over $Q_2$, and, hence, irreducible over $\mathbb{Q}[3^{1/m}]$.

*Computer experiments show that $\mathbb{Q}[2^{1/n},3^{1/m}] = \mathbb{Q}[2^{1/n} + 3^{1/m}]$ and for every concrete pair of numbers $n,m$ it is possible to compute minimal polynomial of $2^{1/n} + 3^{1/m}$ and show that it have degree $nm$, but I don't see how to write minimal polynomial in general (one closed formula for any $n,m \geq 1$).

Hope for your help!
 A: Since $\sqrt[n]{2}$ has degree $n$ over $\mathbf Q$ and $\sqrt[m]{3}$ has degree $m$ over $\mathbf Q$, the field $\mathbf Q(\sqrt[n]{2},\sqrt[m]{3})$ has degree at most $mn$ over $\mathbf Q$. You want to show "at most" is "equal to".
First, we reduce to the case that $m$ and $n$ are the same by the following argument.  When $m$ and $n$ are arbitrary, let $d = mn$, $K = \mathbf Q(\sqrt[n]{2},\sqrt[m]{3})$, and $L = \mathbf Q(\sqrt[d]{2},\sqrt[d]{3})$, where the $d$th roots of $2$ and $3$ are built from the chosen $n$th root of $2$ and $m$th root of $3$ by extracting further roots from them (that is, $\sqrt[d]{2}^{d/n} = \sqrt[n]{2}$ and $\sqrt[d]{3}^{d/m} = \sqrt[m]{3}$), so $\sqrt[d]{2}$ has degree at most $d/n$ over $K$ and $\sqrt[d]{3}$ has degree at most $d/m$ over $K$.  This makes
$$
\mathbf Q \subset K = \mathbf Q(\sqrt[n]{2},\sqrt[m]{3}) \subset L = \mathbf Q(\sqrt[d]{2},\sqrt[d]{3})
$$
where $[L:K] \leq (d/n)(d/m) = d^2/(mn)$. Since $[K:\mathbf Q] \leq mn$, we have $[L:\mathbf Q] \leq d^2$. Therefore if we can show $[L:\mathbf Q] = d^2$, all the field degree inequalities we have obtained up to this point have to be equalities. Proving $[L:\mathbf Q] = d^2$ is an instance of what you want when using the same amount of root extractions on $2$ and $3$.
Thus if we can prove the result when $m = n$ we will have proved the result in general.
Note that there is nothing here so far that is special about $2$ and $3$. If we have an arbitrary base field $F$ in place of $\mathbf Q$ and pick nonzero $a$ and $b$ in $F$ in place of $2$ and $3$ in $\mathbf Q$, the conditions
(i) $[F(\sqrt[m]{a},\sqrt[n]{b}):F] = mn$ for all positive integers $m$ and $n$
and
(ii) $[F(\sqrt[d]{a},\sqrt[d]{b}):F] = d^2$ for all positive integers $d$
are equivalent: (ii) is a special case of (i), and the argument above shows how (i) can be deduced from (ii) using $d = mn$. Of course determining whether or not such conditions are actually true is another matter entirely from their equivalence, but at least this lets us focus on the case where we are taking the same level of root extraction on both numbers.
In fact, it is not even important that we are adjoining roots of just two elements of the base field.  For $a_1, \ldots, a_k$ in $F^\times$, the conditions
(i) $[F(\sqrt[n_1]{a_1},\ldots,\sqrt[n_k]{a_k}):F] = n_1\cdots n_k$ for all positive integers $n_1, \ldots, n_k$
and
(ii) $[F(\sqrt[d]{a_1},\ldots,\sqrt[d]{a_k}):F] = d^k$ for all positive integers $d$
are equivalent.  Again, this equivalence merely lets us reduce the proof of (i) to the case that $n_1, \ldots, n_k$ are all equal.  There is still a lot of work ahead in actually proving (ii) in some example, such as your question with $F = \mathbf Q$, $k=2$, $a_1 = 2$, and $a_2 = 3$.
As Serge Lang once wrote, there is a feeling in most people that the degree
$[F(\sqrt[d]{a_1},\ldots,\sqrt[d]{a_k}):F]$ should be affected by multiplicative relations among $a_1, \ldots, a_k$, or more precisely relations among them in $F^\times/(F^\times)^d$. For example, $\mathbf Q(\sqrt[3]{2})$ and $\mathbf Q(\sqrt[3]{4})$ each have degree $3$ over $\mathbf Q$ but $[\mathbf Q(\sqrt[3]{2},\sqrt[3]{4}):\mathbf Q]$ is $3$ rather than $9$ because $\mathbf Q(\sqrt[3]{2}) = \mathbf Q(\sqrt[3]{4})$ due to $\sqrt[3]{2}\sqrt[3]{4} = 2$. Even that is subtle: I had in mind real cube roots in both cases, but if $\sqrt[3]{2}$ were a real cube root of $2$ and $\sqrt[3]{4}$ were a complex (i.e., non-real) cube root of $4$, then that cube root of $4$ would be quadratic over $\mathbf Q(\sqrt[3]{2})$ (since $x^3 - 4$ would split over that field into a linear factor involving the real cube root of $4$ and a quadratic factor involving the non-real cube roots of $4$). So $[\mathbf Q(\sqrt[3]{2},\sqrt[3]{4}):\mathbf Q]$ would then be $6$ rather than $3$: the field degree depends not just on multiplicative relations among the numbers whose $d$th roots are being extracted, but also on the interaction between the choice of $d$th roots and $d$th roots of unity.
If the base field $F$ contains a full set of $d$ different $d$th roots of unity, then $F(\sqrt[d]{a_1}, \ldots,\sqrt[d]{a_k})/F$ is Galois and Kummer theory implies $[F(\sqrt[d]{a_1}, \ldots,\sqrt[d]{a_k}):F]$ equals the size of the group $\langle a_1, \ldots, a_k\rangle$ inside $F^\times/(F^\times)^d$, so having the field degree be $d^k$ (the maximal possible value) amounts to saying $a_1, \ldots, a_k$ are multiplicatively independent modulo $d$th powers (the only way a product of powers of them can be a $d$th power in $F^\times$ is for all the exponents to be multiples of $d$).
Your question is about base field $\mathbf Q$, which only has roots of unity $\pm 1$. That means the above paragraph doesn't immediately help us when $d > 2$. Still, the idea that when $a_1, \ldots, a_k$ are multiplicatively independent in $\mathbf Q^\times$ we should have $[\mathbf Q(\sqrt[d]{a_1}, \ldots,\sqrt[d]{a_k}):\mathbf Q] = d^k$ should feel reasonable. Since you are asking about the case where $a_i$'s are prime, I will now simply refer to a paper of Ian Richards that confirms the result in this case: for distinct primes $p_1, \ldots, p_k$, $[\mathbf Q(\sqrt[d]{p_1}, \ldots,\sqrt[d]{p_k}):\mathbf Q] = d^k$ for all positive integers $d$, where we use positive real $d$th roots each time.
The paper is
I. Richards, "An application of Galois theory to elementary arithmetic,"
Advances in Math. 13 (1974), 268-273.
A link to the paper is here.
