When adding several $\sqrt[n]{p}$ to the rational numbers, what is the degree of field extension?

Question

When adding several $\sqrt[n]{p}$ to the rational numbers, what is the degree of field extension?

For example, does $[\mathbb{Q}(\sqrt[n]{2},\sqrt[m]{3}):\mathbb{Q}]=mn$ hold true? Are there more general statements?

Answer 1

Let $K = \mathbf Q(\sqrt[n_1]{p_1},\ldots,\sqrt[n_k]{p_k})$ where $p_1, \ldots, p_k$ are distinct primes and $n_1,\ldots,n_k \geq 1$. Then $[K:\mathbf Q] \leq n_1\ldots n_k$. It is intuitively plausible that this inequality is equality.

To show that intuition is right, let $n$ be a multiple of $n_1,\ldots,n_k$ and $L = \mathbf Q(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k})$, so $[L:\mathbf Q] \leq n^k$.

The root extractions made above are all positive roots, so $K \subset L$ and $\sqrt[n]{p_i}^{n/n_i} = \sqrt[n_i]{p_i}$, which tells us that $\sqrt[n]{p_i}$ has degree at most $n/n_i$ over $K$. Thus $[L:K] \leq (n/n_1)\cdots (n/n_k) = n^k/(n_1\cdots n_k)$.

If $[K:\mathbf Q] < n_1\cdots n_k$ then $[L:\mathbf Q] = [L:K][K:\mathbf Q] < (n_1\cdots n_k)(n^k/(n_1\cdots n_k) = n^k$. Therefore if $[L:\mathbf Q] = n^k$ we must have $[K:\mathbf Q] = n_1\cdots n_k$.

The only property I have used about primes here is that they are positive integers, so the above reasoning shows the next result, where all roots are positive real numbers.

Theorem. Let $a_1,\ldots, a_k$ be positive integers. If $[\mathbf Q(\sqrt[n]{a_1},\ldots,\sqrt[n]{a_k}):\mathbf Q] = n^k$ for all positive integers $n$, then $[\mathbf Q(\sqrt[n_1]{a_1},\ldots,\sqrt[n_k]{a_k}):\mathbf Q] = n_1\cdots n_k$ for all positive integers $n_1,\ldots, n_k$.

Thus, when you want to show $[\mathbf Q(\sqrt[n_1]{a_1},\ldots,\sqrt[n_k]{a_k}):\mathbf Q] = n_1\cdots n_k$ for all positive integers $n_1,\ldots, n_k$, it is enough to prove the special case when $n_1, \ldots, n_k$ are all equal. That is a convenient notational simplification!

Now we would like to prove $[\mathbf Q(\sqrt[n]{a_1},\ldots,\sqrt[n]{a_k}):\mathbf Q] = n^k$ for all $n \geq 1$, when $a_1,\ldots,a_k$ are suitable positive integers. It is enough to prove this when $n$ runs through a sequence of positive integers such that each positive integer is a divisor of some integer in the sequence, by the same reasoning that I made at the very start to reduce the case of exponents $n_1,\ldots,n_k$ to the case of a common exponent $n$ that is some multiple of $n_1,\ldots, n_k$. A proof when $a_1,\ldots,a_k$ are distinct primes, which is the setting of your question, can be found on pp. 234-237 in Lisl Gaal's book Classical Galois Theory with Examples (3rd ed.), where the argument is attributed to J. I. Richards and assumes $n$ is even (no loss in making that assumption on $n$). Richards published his result in "An Application of Galois Theory to Elementary Arithmetic", Adv. Math. 13 (1974), 268-273, a copy of which has been scanned online here.

Your question was asked 11 years ago on MSE here, and the answer there by Georges Elencwajg cites a place to see a proof only covering the case when $n$ is odd: Steven Roman's book Field Theory (2nd ed.), Theorem 14.3.2 on page 305.

Answer 2

Yes, this statement, as well as obvious extensions for more primes, are true. For $m,n$ relatively prime, this follows as an easy application of the Tower Law. For $n=m=2$, more generally for an arbitrary number of square roots, there is also a short and completely elementary argument. For an extension of the method to arbitrary roots, see Besicovitch, A. S, On the linear independence of fractional powers of integers. J. London Math. Soc.15(1940), 3–6. More general results can be found for example in Zhou, Jian Ping, On the degree of extensions generated by finitely many algebraic numbers, J. Number Theory 34 (1990), no. 2, 133–141, and no doubt elsewhere.