If K is a number field, K/Q is a normal (Galois) extension, O_{K} its ring of algebraic integers, p is a prime number, and pO_{K} is the product of distinct prime ideals of norm p, then either all of these ideals are principal, or they are all nonprincipal. If they are nonprincipal, then there are no integers of norm p. In this case, let P be a prime ideal of norm p, and let c be its class in the class group. If there is a fractional ideal X of norm 1 whose class is c^{1}, then XP will be a principal ideal of norm p, but its principal generators will be noninteger fractions in K.
If Q is a prime ideal in the "numerator" of X, there has to be a corresponding Q' conjugate to Q in the "denominator" (this uses the assumption that K/Q is a normal extension).
Thus, any fractional ideal of norm 1 is a product of factors of the form Q/Q' where Q is a prime ideal in O_{K}.
I don't know in general how to determine whether such a product can be in any given ideal class, but I do know one case where it is not possible. If K has class number h = 2, then in the fractional ideal Q/Q', either Q and Q' are both principal, or they are both in the sole nonprincipal class, so that Q/Q' is principal. So in this case, all fractional ideals with norm 1 are a product of principal ideals, and are therefore principal.
So if K/Q is normal with class number h = 2, and p a prime number with pO_{K} the product of nonprincipal prime ideals of norm p, then there is no element of K having norm p.
The smallest cyclotomic example appears to be K = Q(\zeta_{39}) and p = 79.
