Finite subunitals of the Hermitian unitals

Every finite subunital of any generalized hermitian unital is itself a hermitian unital; the embedding is given by an embedding of quadratic field extensions. In particular, a generalized hermitian unital with a finite subunital is a hermitian one (i.e., it originates from a separable field extension).

A hermitian unital in a pappian projective plane consists of the absolute points of a unitary polarity of that plane, with blocks induced by secant lines (see Sect. 1). The finite hermitian unitals of order q are the classical examples of 2-(q 3 + 1, q + 1, 1)-designs, cp. [6, II.8, pp. 57-63 and p. 246], or [2,Ch. 2]. We consider generalized hermitian unitals H(C|R) where C|R is any quadratic extension of fields; separable extensions C|R yield the hermitian unitals, inseparable extensions give certain projections of quadrics.
If one has an embedding of field extensions from E|F into C|R (in the sense of 1.5) then clearly the unital H(E|F ) is embedded into H(C|R). Our Main Theorem asserts, in the converse direction, that every embedding of a finite unital into one of those generalized hermitian unitals comes from an embedding of field extensions. This also implies that no finite subunitals exist in generalized hermitian unitals that are not hermitian unitals.
If U is a unital of order q embedded in the hermitian unital H(F r 2 |F r ) of order r then our result says that U is the hermitian unital of order q, and there is an odd integer e such that r = q e .
We remark that an analogous result holds for generalized polygons (where the Moufang property singles out the classical examples), see [12, 5.2.2]: any (thick) subpolygon of a Moufang polygon is Moufang, as well.

Generalized hermitian unitals
Let C|R be any quadratic (possibly inseparable) extension of fields; so C = R + εR, with ε ∈ C R. There exist t, d ∈ R such that ε 2 − tε + d = 0, since ε 2 ∈ R + εR. The mapping is a field automorphism which generates Aut R C: if C|R is separable, then σ has order 2; if C|R is inseparable, then σ is the identity. Now we introduce our geometric objects. We consider the pappian projective plane PG(2, C) arising from the 3-dimensional vector space C 3 over C, and we use homogeneous coordinates [X, Y, Z] := (X, Y, Z)C for the points of PG(2, C). Remark 1.2. The name "generalized hermitian unital" is motivated by the following observations.
If C|R is separable, then H(C|R) is the hermitian unital arising from the skew-hermitian form h : see [4, 2.2]. So the elements of H are the absolute points of a polarity of PG(2, C), in this case.
If C|R is inseparable, then H(C|R) is the projection of an ordinary quadric Q in some projective space of dimension at least 3 from a subspace of codimension 1 in the nucleus of Q, see [4, 2.2]. In this case, there is no polarity of PG(2, C) such that the elements of H are absolute points.
In any case, we have for each point p of H(C|R) a unique line of PG(2, C) passing through p and containing no other point of H(C|R), see [4, 2.3]. We will refer to this line as the tangent line to H(C|R) at p in PG(2, C).
The unital H(C|R) admits all conceivable translations (i.e., automorphisms that fix a point and every block through it), and these are induced by elations of PG(2, C), see [4, 2.13].
By an O'Nan configuration, we mean four blocks intersecting in six points of the unital (i.e., a (6 2 4 3 ) configuration). Naming this configuration in honor of O'Nan [8] is customary in the context of unitals, see [2, p. 87]; the configuration is named after Veblen and Young in the axiomatics of projective spaces, or after Pasch in the context of ordered (Euclidean) geometry.
We will use the following properties of generalized hermitian unitals:

Main result
Lemma 2.1. Let (P, L) be a linear space containing no O'Nan configuration, let (U, B) be a unital of order q, and let (α, β) be an embedding of (U, B) in (P, L). If two blocks B 1 , B 2 of (U, B) have no point of U in common, then B β 1 , B β 2 are disjoint blocks of (P, L).
Proof. Aiming at a contradiction, suppose that two blocks The absence of O'Nan configurations in (P, L) implies that two arbitrary blocks of (U, B) both intersecting B 1 ∪ B 2 in exactly two points have no points off B 1 ∪ B 2 in common. Hence the number of points in U lying on a block intersecting B 1 ∪ B 2 in exactly two points is greater than (q + 1) 2 (q − 1) ≥ q 3 + 1; this is a contradiction. Proof. Let q denote the order of (U, B). In this proof, we suppose q > 2; the case q = 2 is treated separately in 3.1 below.
Let p ∈ U be arbitrary, and let B ∈ B be such that p / ∈ B. Let B 0 , B 1 , . . . , B q be the blocks of (U, B) containing p and intersecting B nontrivially, say in x 0 , x 1 , . . . , x q , respectively. Let x be an arbitrary point on B 0 {p, x 0 }. We claim that at least one block of (U, B) contains x and intersects B 1 ∪B 2 ∪· · ·∪B q in at least two points (different from p). Indeed, if not, then there are q 2 blocks through x different from B 0 , which is a contradiction. So let B x be a block of (U, B) containing at least three points (including x) of B 0 ∪ B 1 ∪ · · · ∪ B q . We note that B x and B are disjoint by (NON). For the same reason (or by 2.1) their extensions to H(C|R) are also disjoint. It then follows from (TRA) that B x intersects B i for each i ∈ {0, 1, . . . , q}, and 2.1 yields that these intersection points belong to U . Hence we have shown that (TRA) holds in the subunital (U, B). Now let τ be the translation of H(C|R) with center p mapping x 0 to x. Let y be any point of U not on B 0 . Since B was arbitrary, we may assume that y ∈ B, so without loss of generality y = x 1 . By the uniqueness in (TRA), τ maps x 1 to the intersection B x ∩B 1 . Since this intersection point belongs to U , it follows that τ preserves U . Hence (U, B) admits all translations and hence is hermitian by the main result of [3]. Now consider the (standard) embedding of H(C|R) in the projective plane PG(2, C). Then also (U, B) is embedded in PG(2, C) and so by the Main Theorem of [4] there is a subfield E ≤ C of order q 2 and a subplane π ∼ = PG(2, E) containing U . (Here we use q > 2, for q = 2 the Main Theorem of [4] does not apply.) Hence there is a polarity ρ π of π with absolute point set U . We now show that ρ π extends to a polarity ρ of PG(2, C) where the absolute points are the points of H(C|R).
Consider the lines extending blocks of (U, B) that meet at least three blocks through p. By (TAN), any two of those lines have an intersection point on the tangent line T to U at p in π. Varying the blocks to be extended, we obtain more than one point on T . The same description applies, mutatis mutandis, to points on the tangent to H(C|R) at p in PG (2, C). So that line extends the tangent to (U, B) in π. This already implies that not all tangent lines to H(C|R) contain the same point and so C|R is separable by [4, 2.3(2)]. Hence there is a polarity ρ of PG(2, C) associated to H(C|R). Since U contains a quadrangle, and points of U are mapped to lines of π under the action of ρ, we see that ρ preserves π. Since tangent lines to (U, B) and H(C|R) coincide in π, we see that ρ |π = ρ π . Hence the generator of the Galois group of C|R preserves E and induces x → x q in E.
This proves our main result completely for q = 2. For q = 2, see 3.1.
If C is finite of order r 2 then E is unique with given order q 2 and the field automorphism x → x r is not trivial on E, which means that E is not contained in the unique subfield R of order r; hence C is an extension of E of odd degree. Thus 2.2 gives the following.

Corollary 2.3.
If U is a unital of order q embedded in the hermitian unital H(F r 2 |F r ) of order r, then U is the hermitian unital H(F q 2 |F q ), there is an odd integer e such that r = q e , and the embedding of the unital is standard.
Up to projective equivalence, the embedded unital is obtained by using a hermitian equation over F q 2 |F q to define the large unital in PG(2, F r 2 ), and restricting coordinates to F q 2 to define the small unital.
Remark 2.4. Wilbrink [13] has characterized the finite hermitian unitals by three conditions. His condition (I) is our (NON) in 1.3. Under the assumption (NON), his condition (II) is equivalent to our (TRA). For unitals of even order, these two conditions alone suffice to characterize the hermitian unitals, see [7]. Wilbrink's condition (III) is too technical to state it here; compare [5, p. 299]. Proof. Each unital of order 2 is isomorphic to the hermitian unital H(F 4 |F 2 ), and is clearly embedded in H(C|R) if R has characteristic 2 and C ∼ = R[X]/ (X 2 + X + 1).

The smallest unital
The unital of order 2 is also isomorphic to the affine plane of order 3 (see [11, 10.16 Asū − u = 0, these equations give a = b and then 0 = 2a. So R has characteristic 2, and F 2 (u) ∼ = F 4 is contained in C but not in R. We obtain an embedding of the field extension F 4 |F 2 into C|R, and the embedding of the unital H(F 4 |F 2 ) of order 2 in H(C|R) is a standard embedding; see 1.5.
It remains to show that the unital of order 2 is not embedded in H(C|R) if C|R is inseparable. As above, we know from [10, 3.7] that any embedding of H(F 4 |F 2 ) into PG(2, C) is projectively equivalent to the one where the embedded points have coordinates in {0, 1, u, u 2 }, with u ∈ C and u 2 = u + 1; we use that C has characteristic 2 because C|R is inseparable. Now the set {0, 1, u, u 2 } is a subfield of order 4 in C, and it is contained in R because otherwise the extension C|R would be separable. Thus the embedded unital of order 2 is contained in a finite subplane π ∼ = PG(2, F 4 ) of PG(2, C).
Let B be a block of H(C|R) joining two points of the embedded unital of order 2. Then B ∩ π is a set of three points. We introduce coordinates for the line L of PG (2, C) containing B in such a way that 0, 1, and ∞ are the coordinates of the three points in B ∩ π. Then B consists of the points with coordinates in R ∪ {∞} because the blocks of H(C|R) are Baer sublines with respect to C|R, see [4, 2.8]. Thus B ∩ π contains all points of L ∩ π (i.e., those with coordinates 0, 1, u, u 2 , and ∞), contradicting the fact that |B ∩ π| = 3. This contradiction yields that there is no embedding of H(F 4 |F 2 ) in H(C|R) if C|R is inseparable.
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