**Proposition 11.2.1**: *Let F = F(t ^{a}, t, t^{b}) be a monomial semi-field type flock in PG(3,q) with q = p^{n} for some prime p, a = p^{i}, b = p^{j} with 0 < i < j < n. Let d = gcd(i,j,n). Then, F is not a star flock, and there are exactly p^{d} planes of the flock which pass through each primary baseline (including the carrier plane). Dually, all primary baselines contain exactly p^{d} points of D_{F}.*

*Pf*: Under the assumption that 0 < i < j < n, it follows that the functions t^{a}, t, and t^{b} are linearly independent over GF(q). Thus, by Theorem 7.1 , F is not a star flock.

In the dual setting, the points of D_{F} have coordinates (t^{a}, t, t^{b},1) for t in GF(q). Consider the line joining (0,0,0,1) and (r^{a}, r, r^{b},1) for some fixed r in GF(q)\{0}. The points of this line, other than (0,0,0,1) have coordinates of the form (r^{a}, r, r^{b}, 1 +) as ranges over GF(q). When = -1, the point of this line is in W' = 0 and so, can not be in D_{F}. We may thus assume that -1. For a point (s^{a}, s, s^{b},1) of D_{F} to be on this line, we must have s = r/1+, s^{a} = r^{a}/1+ and s^{b} = r^{b}/1+. These equations are satisfied if, and only if, (1+)^{a-1} = 1 and (1+)^{b-1} = 1. Therefore, the order of the element 1+ in the multiplicative group GF(q)\{0} must divide the gcd(a-1, q-1) = gcd(p^{i}-1, p^{n}-1) = p^{gcd(i,n)}-1, and gcd(b-1,q-1) = p^{gcd(j,n)}-1. Hence, the order of this element must divide gcd(p^{gcd(i,n)}-1, p^{gcd(j,n)}-1) = p^{d} - 1, and so, 1+ is an element of the subfield GF(p^{d}) of GF(q). Conversely, any element of this subfield will satisfy the conditions. Thus, there are p^{d} -1 points of D_{F} on this line corresponding to non-zero values of 1+, and together with (0,0,0,1) we get p^{d} points on every primary baseline.