91Ó°ÊÓ

First, we need to find the splitting field of the polynomial, \(X^{\prime}-a\) for any given prime number \(p\). Note that the polynomial factors as \((X^\prime - a) = X^p - a = (X - a^\frac{1}{p})(X^2 + aX^\frac{p-1}{p} + a^2X^\frac{p-2}{p} + \cdots + a^{p-1})\). Due to the factors' irreducibility, this factorization must hold in any field extension containing all of these roots. This means that the splitting field \(K_p\) must contain these roots of the polynomial. Therefore, \(K_p = \mathbf{Q}\left(a^\frac{1}{p}, \zeta_{p}, \zeta_{p}^2, \cdots , \zeta_{p}^{p-1}\right)\), where \(\zeta_{p}\) is a primitive \(p\)-th root of unity. Now, let's calculate the degree of this extension, \(\left[K_{p}: \mathbf{Q}\right]\).

Recall that the degree of a field extension is the dimension of its vector space over the base field. Since \(\mathbf{Q}\) is linearly independent over the roots in \(K_p\), we can find the degree of the extension by identifying the dimension of the smallest vector space containing these roots. Pairwise multiply the powers of each \(\zeta_{p}^k\) for \(0 \leq k \leq p-1\) with the basis \(\left\{a^\frac{1}{p}\right\}\), yielding a set of \(p(p-1)\) linearly independent elements in the extension field. Hence, we have \(\left[K_{p}: \mathbf{Q}\right]=p(p-1)\).

Now we define the compositum \(K_m\) as \(K_{m}=\prod_{p \mid m} K_{p}\). We want to find its degree \(d_m=[K_m : \mathbf{Q}]\).

If \(m\) is odd, we claim that \(d_m = \prod_{p \mid m} d_{p}\). Recall that for any two field extensions \(K\) and \(L\) over a field \(F\), if \(K \cap L = F\), then \([KL:F] = [K:F][L:F]\). In our case, for any distinct primes \(p\) and \(q\) dividing \(m\), we have \(K_p \cap K_q = \mathbf{Q}\), as they are both linearly independent due to the roots contained in each field. Since each \(K_p\) is a subfield of \(K_m\), the claim holds.

If \(m\) is even, we can write it as \(m = 2n\). Now, we need to show that \(d_{2n} = d_{n}\) or \(2d_{n}\), depending on whether \(\sqrt{a}\) is or is not in the field of \(m\)-th roots of unity \(\mathbf{Q}\left(\zeta_{m}\right)\). If \(\sqrt{a} \in \mathbf{Q}\left(\zeta_{m}\right)\), then the field \(\mathbf{Q}\left(\zeta_{m}\right)\) contains all the necessary primitive roots of unity to obtain extensions \(K_{2^n}\). Since \(\sqrt{a}\) is a square root of \(a\), the degree of the smallest such extension is equal to the degree of the smallest extension for the odd case, i.e., \(d_{2n} = d_{n}\). On the other hand, if \(\sqrt{a} \notin \mathbf{Q}\left(\zeta_{m}\right)\), then we need to consider the compositum of \(K_m\) and the smallest field extension containing \(\sqrt{a}\), i.e., \(K_{2n} = K_{m} \prod \mathbf{Q}\left(\sqrt{a}\right)\). Since \([\mathbf{Q}\left(\sqrt{a}\right) : \mathbf{Q}] = 2\) and the fields are linearly independent, we have \(d_{2n} = 2d_{n}\).