The integer q is a rational prime. When is (q) a prime ideal in a Kummer ring based on the rational prime p? This question appears on page 196 of "Elements of Abstract Algebra" by Allan Clark, Wadsworth (1971), Dover (1984). The subject is treated in a general way in the following articles:

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*Huber, C.M., "On the Prime Divisors of the Cyclotomic Functions", Trans. of the American Math. Society, Vol. 27, No. 1 (Jan. 1925) pp. 43-48.

*Hilbert, D., Collected Papers, Vol. 1, Number Theory, Chapter 7.21 (1932).

*Kummer, E.E., "Uber die Zerlegung der aus Wurzeln der Einheit Gebildeten complexen Zahlen in ihren Primfactoren", J. Reine u. Angew Math., 35, pp. 327-367 (1847).

*Ribenboim, P., "13 Lectures on Fermat's Last Theorem", Lecture V, "Kummer's Monument", Springer (1979), reprint (2010).

The answer to the question posed is that q is a primitive root (mod p). This is a special case in the literature cited above. What is a more direct proof of the question that would satisfy a layman or student in mathematics?
 A: Perhaps a more transparent phrasing of the question is "when does a rational prime $q$ remain prime in the cyclotomic extension $\mathbb Z[\zeta_p]$, where $\zeta_p$ is a primitive $p$\th root of unity, for prime $p$?"
Classical algebraic number theory very nicely resolves this (and the answer is in-principle well known): this is so exactly when the smallest power $q^f$ of $q$ such that $p$ divides $q^f-1$ is $f=p-1$ (which is the degree of the extension $\mathbb Q(\zeta_p)/\mathbb Q$). Let $k=\mathbb Q(\zeta_p)$.
The standard, central result that is decisive here is that the sum of the local degrees is the global degree. That is, for all primes $v$ lying over $q$ in that extension, $p-1=[k:\mathbb Q]=\sum_v [k_v:\mathbb Q_q]$. To say that $q$ is a primitive root mod $p$ is to say that there is no $p$th root of unity in $\mathbb F_{q^n}$ until $n=p-1$ (or larger). Thus, already the residue class field extension at a single place $v$ over $q$ is of degree $p-1$. Thus, the degree of the corresponding completion is $p-1$, and there's not room for any other prime lying over $q$. And conversely...
