VAN DER VRIES. — MULTIPLE POINTS OF TWISTED CURVES. 511 



m p — (m — 2) k + 1 ^ p (p + 2) — k 2 . 



If there are still points left at our disposal, we may use them in different 

 ways, and may cause the residual curve to break up in different ways. 

 The complete intersection of K 2 and S^ is of order 2p; and as C m is of 

 order m the residual is of order 2 p — m. As the complete intersection 

 has a point of multiplicity 2 k at A, the residual has a point of multi- 

 plicity 2 k — m -f 2 there. Every edge of K 2 meets S^ in k + 1 points 

 of which k are at A. If we can take p — k additional points on some of 

 these edges, these edges will contain p + 1 points in all and will lie com- 

 pletely on Sfj.. Therefore if there are enough points left at our disposal 

 to enable us to take p — k additional points on each of 2 p — m — 1 

 edges of K 2 , the residual will consist of these 2p — m — 1 edges of K 2 

 and (since the entire residual is of order 2 p — m) another straight line, 

 which can only he another edge of K 2 . The residual can then be made 

 to consist entirely of straight lines if 



(p — k) (2 p — m — I) ^ p (p + 2) — k* — mp + (m — 2) k — 1, 

 i. e. if - k - 1) (p - k — 2) ^ 1 ; 



which can only occur if k = p — 1, or k =■ p — 2. We may then always 

 take k = p — 1, that is take S^ to be a monoid. Thus p =■ k + 1 = 



11X 7)1 ~\- 1 



— or — - — according as m is even or odd. Therefore : — 



2 2 s 



A unicursal curve of order m that has a point of multiplicity m — 2 can 

 always be considered as the partial intersection of a quadric cone and a 



monoid of order p, where p = — ■ or — - — according as m is even or odd. 



Thus the quintic curve with a triple point can be obtained, as we have 

 seen before, as the partial intersection of a quadric cone and a cubic 

 monoid that have the same vertex. The cubic monoid is determined by 

 15 points, but 4 of these must be taken off the quadric in order not to 

 have the monoid contain the quadric as a component. We thus have 11 

 points at our disposal. In order "to make the cubic monoid contain a 

 quintic curve that has a triple point at the vertex of the monoid, we 

 must make it contain 3*5 — 2-3 + 1 or 10 additional points of the 

 curve. We can do this and still have one point left at our disposal. This 

 point can be taken on any generator of the cone, causing this generator 

 to be the residual intersection of the cone and the monoid. 



Similarly, a sextic curve with a quadruple point can be obtained as the 



