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Arab. J. Math. https://doi.org/10.1007/s40065-023-00425-0 Arabian Journal of Mathematics G. Andablo-Reyes · B. L. De La Rosa-Navarro · M. Lahyane The effective monoids of some blow-ups of Hirzebruch surfaces Dedicated to Professor Brian Harbourne on the occasion of his 65th birthday Received: 16 February 2022 / Accepted: 27 February 2023 © The Author(s) 2023 Abstract We mainly give a numerical condition to ensure the ﬁnite generation of the effective monoids of some smooth projective rational surfaces. These surfaces are constructed from the blow-up of any ﬁxed Hirzebruch surface at some special conﬁgurations of ordinary points. Under this numerical condition, we determine explicitly the list of all (−1) and (−2)-curves. In particular, we complete a result obtained by Harbourne (Duke Math J 52(1):129–148, 1985) and another result obtained by the third author (C R Math 338(11):873–878, 2004). Moreover, the Cox rings of these surfaces are ﬁnitely generated. Our ground ﬁeld is assumed to be algebraically closed of any characteristic. Mathematics Subject Classiﬁcation 14J26 · 14C20 · 14C22 · 14C17 · 14Q20 1 Introduction We are interested in characterizing the smooth projective rational surfaces whose effective monoids are ﬁnitely generated. This is the reason why we study the surfaces whose minimal models are the Hirzebruch ones; see [6], and also [16,40]and [34]. For any smooth projective rational surface Z, the Néron-Severi group NS(Z) of Z is the quotient group of the group of divisors on Z modulo numerical equivalence, and it is a free ﬁnitely generated Z-module of ﬁnite rank ρ(Z ). A special subset of NS(Z) is the effective monoid Eff(Z) of Z,which is deﬁned as the set of elements γ of NS(Z), such that there exists an effective divisor D on Z with γ is the class of D modulo numerical equivalence [26]. It is well known that Eff(Z) has an algebraic structure of a monoid. The importance of studying the ﬁniteness of the effective monoid of some rational surface appears clearly in the characterization of ﬁnite generation of the Cox ring of such surface; see [5–7,13,15,18,37]and [16]. G. Andablo-Reyes Facultad de Ciencias Físico Matemáticas, Universidad Michoacana de San Nicolás de Hidalgo, Avenida Francisco J. Múgica S/N Ediﬁcio Alpha, Ciudad Universitaria, Colonia Felicitas Del Rio, C. P. 58040 Morelia, Estado de Michoacán de Ocampo, Mexico E-mail: andablo@umich.mx B. L. De La Rosa-Navarro Facultad de Ciencias, Universidad Autónoma de Baja California (UABC), Km. 103 Carretera Tijuana-Ensenada, C. P. 22860 Ensenada, Baja California, Mexico E-mail: brenda.delarosa@uabc.edu.mx M. Lahyane (B) Instituto de Física y Matemáticas, Universidad Michoacana de San Nicolás de Hidalgo, Avenida Francisco J. Múgica S/N, Edi- ﬁcio C-3, Ciudad Universitaria, Colonia Felicitas Del Rio, C. P. 58040 Morelia, Estado de Michoacán de Ocampo, Mexico E-mail: mustapha.lahyane@umich.mx 123 Arab. J. Math. Recall that the Cox ring, Cox(X ), of a projective variety X over an algebraically closed ﬁeld k is the k-algebra given by Cox(X) = H (X, L), L∈Pic(X) where Pic(X) is the Picard group of X,and H (X, L) is the ﬁnite-dimensional k−vector space of global sections of L; for more details, see [4,27], and [28]. In [7], we show the equivalence between the ﬁnite generation of the Cox ring of an anticanonical rational surface that satisﬁes the anticanonical orthogonal property, and the ﬁnite generation of its effective monoid. In [1,2,9,10,12,14,20–22,24,25,30–33,35,36,38,39]and [17], one may ﬁnd more results about the ﬁniteness of the effective monoids of some surfaces. Here, an anticanonical rational surface is a smooth projective rational surface whose complete anticanonical linear system is not empty [23], and we say that a surface has the anticanonical orthogonal property whenever every nef and effective divisor class (modulo numerical equivalence) on such surface that is orthogonal to an anticanonical class is the zero class. Thus, our surfaces are Harbourne–Hirschowitz ones; see [11]. In this article, we construct a family of smooth projective rational surfaces (see Sect. 2) whose sets of (−1)-curves and (−2)-curves are both ﬁnite under certain reasonable numerical condition; see Theorem 3.1. Consequently, we are able to infer the ﬁnite generation of the effective monoids of these surfaces; see Corollary 3.6. On the other hand, in Sect. 4, we may observe that under the same numerical condition, these surfaces satisfy the anticanonical orthogonal property (see Lemma 4.1) and, therefore, their Cox rings are ﬁnitely generated (see Theorem 4.2). 2 The construction of a family of smooth projective rational surfaces First, we remind some notion about Hirzebruch surfaces over an algebraically closed ﬁeld k of any characteristic. Fix a non-negative integer n. The Hirzebruch surface associated with n is the projectivization of the locally free sheaf O 1 ⊕ O 1 (−n) of rank two on the projective line P . It is well known that the set {C , F} is a P P k k k minimal set of generators of the Néron–Severi group NS( ) of as a Z-module, where C is the class of n n n a section C of (it is unique if n is positive, and in this case, that section is usually called the exceptional n n section), and F is the class of a ﬁbre f of . The intersection form on is given by the three equalities n n 2 2 (C ) =−n,(F) = 0, and C · F = 1; for more details, see for example [26], and [40]. Furthermore, if D n n is a prime divisor on some smooth projective surface Z , then Supp(D) denotes the support of the invertible sheaf O (D) associated with D [26]. Finally, for ﬁxed non-negative integers r and r ,let G = C + (n + 2) f where f is aﬁbreof ,and let 1 2 n n C C C G G G n n n f , f ,..., f , f , f ,..., f be r +r different ﬁbres of .Now,let P , P ,..., P , Q , Q ,..., Q r 1 2 n 1 2 r 1 2 r 1 2 r 1 2 2 1 2 be ordinary points of in general position, such that P ∈ Supp(G) ∩ Supp(f ) \Supp(C ) for every n i n i ∈{1, 2,..., r }, and Q ∈ Supp(C ) ∩ Supp(f ) \Supp(G) for every j ∈{1, 2,..., r };see Fig. 1.Next, 1 j n 2 we denote the blow-up of at the zero-dimensional subscheme ={P , P ,..., P , Q , Q ,..., Q } n 1 2 r 1 2 r 1 2 r ,r 1 2 by X ; see Fig. 2. r ,r 1 2 A minimal set of generators of NS(X ) as a Z−module is the set {C , F , −E , −E ,..., −E , −E , −E ,..., −E }, n P P P Q Q Q 1 2 r 1 2 r 1 2 Fig. 1 The conﬁguration of the points of 123 Arab. J. Math. Fig. 2 The blow-up of at the points of where C is the class of the total transform of C , F is the class of the total transform of a ﬁbre f of , E n n n P is the class of the exceptional divisor corresponding to the point P for every i ∈{1, 2,..., r }, and E is the i 1 Q class of the exceptional divisor corresponding to the point Q for every j ∈{1, 2,..., r }. The intersection j 2 r ,r 1 2 2 2 form on X is given by the following equalities: C =−n, F = 0, C · F = 1, C · E = 0, F · E = 0, n n ω ω E =−1for every ω ∈ , and E · E = 0for p, q ∈ , such that p = q. p q r ,r 1 2 3 The ﬁniteness of (−1)-curves and (−2)-curves of X r ,r 1 2 In this section, we prove that there are only a ﬁnite number of (−1)-curves and (−2)-curves of X under the assumption that (n + 2) + nr + 4r − nr − r r is positive. Here, a (−1)-curve, respectively, a (−2)-curve, 2 2 1 1 2 is a smooth rational curve of self-intersection −1, respectively −2. Using the notation of the last section, we present one of the our results: r ,r 2 1 2 Theorem 3.1 If (n +2) +nr +4r −nr −r r is positive, then X has only a ﬁnite number of (−1)-curves 2 2 1 1 2 n and (−2)-curves. Proof By the forthcoming Lemma 3.2, one can assume, without loss of generality, that r and r are positive. 1 2 r ,r r 1 2 2 Let D be a (−2)-curve on X , such that the class of D does not belong to {C − E , C + (n + 2)F − n n Q n j =1 j r r ,r r r 1 1 2 1 2 E }. One may write the class of D in NS(X ) as D = aC + bF − γ E − μ E , for P n n j P Q i j i =1 j =1 =1 r ,r some integers a, b,γ ,γ ,...,γ ,μ ,μ ,...,μ . We have the equalities D =−2and D · (−K ) = 0, 1 2 1 2 r 1 2 r 1 2 r r ,r 2 1 2 r ,r and −K = C − E + G is the class of an anticanonical divisor in NS(X ), where G is the 1 2 n Q n =1 r ,r 1 2 class of the strict transform of G in NS(X ); this implies the following equalities: r r 1 2 2 2 2 2ab − a n − γ − μ =−2, (1) j =1 =1 r r 2 1 b − an − μ = 0, and 2a + b − γ = 0, =1 j =1 2 2 since D =−2, D · C − E = 0and D · G = 0. n Q =1 Using the above equalities, and the fact that γ ≥ 0and μ ≥ 0for every j ∈{1, 2,..., r } and j 1 2a + b ∈{1, 2,..., r }, it is sufﬁcient to prove that a and b are bounded. Now, let γ¯ = γ − for all 2 j j b − an r j ∈{1, 2,..., r }, and let μ ¯ = μ − for each ∈{1, 2,..., r }. Therefore, γ¯ = 0, 1 2 j r j =1 μ ¯ = 0, and =1 r r r r 1 2 1 2 2 2 (2a + b) (b − an) 2 2 2 2 γ + μ = γ¯ + μ ¯ + + . (2) j j r r 1 2 j =1 =1 j =1 =1 123 Arab. J. Math. From Eqs. (1)and (2), we have r r 1 2 2 2 (2a + b) (b − an) 2 2 2 γ¯ + μ ¯ = 2ab − a n + 2 − − . r r 1 2 j =1 =1 Therefore, the following inequality is satisﬁed: 2 2 2 2abr r − a nr r + 2r r − (b + 2a) r − (b − an) r ≥ 0. 1 2 1 2 1 2 2 1 Then, we obtain after completing the square in a 2 2 2 2 (r r − 2r + nr )b (r r − 2r + nr ) b 2r r − b r − b r 1 2 2 1 1 2 2 1 1 2 2 1 a − − − ≤ 0. 2 2 2 2 n r + nr r + 4r (n r + nr r + 4r ) n r + nr r + 4r 1 1 2 2 1 1 2 2 1 1 2 2 This implies that 2 2 2 2 (r r − 2r + nr ) b 2r r − b r − b r 1 2 2 1 1 2 2 1 + ≥ 0. 2 2 2 (n r + nr r + 4r ) n r + nr r + 4r 1 1 2 2 1 1 2 2 Therefore 2(n r + nr r + 4r ) 1 1 2 2 b − ≤ 0. (n + 2) + nr + 4r − nr − r r 2 2 1 1 2 Thus, a and b are bounded. Indeed, by our hypothesis and the last inequality, b is bounded. To see that a is bounded, we use the fact that b is bounded and the inequality above 2 2 2 2 (r r − 2r + nr )b (r r − 2r + nr ) b 2r r − b r − b r 1 2 2 1 1 2 2 1 1 2 2 1 a − − − ≤ 0. 2 2 2 2 n r + nr r + 4r (n r + nr r + 4r ) n r + nr r + 4r 1 1 2 2 1 1 2 2 1 1 2 2 r ,r 1 2 Therefore, X contains a ﬁnite number of (−2)-curves. r ,r 1 2 For the ﬁniteness of the set of (−1)-curves, let N beaclass of a (−1)-curve on X , so N = aC + n n r r 1 2 bF − γ E − μ E , for some integers a, b,γ ,γ ,...,γ ,μ ,μ ,...,μ , such that N does j P Q 1 2 r 1 2 r j =1 j =1 1 2 r r 2 1 not belong to {C − E , C + (n + 2)F − E , F − E ,..., F − E , F − E ,..., F − n Q n P P P Q j i 1 r 1 j =1 i =1 r ,r E , E ,..., E , E ,..., E }. Then, N · (−K ) = 1and N =−1, so we have the following two 1 2 Q P P Q Q r 1 r 1 r 2 1 2 n cases to study: Case (1) N · (C − E ) = 1and N · G = 0, and n Q =1 Case (2) N · (C − E ) = 0, and N · G = 1. n Q =1 Assume that we are in Case (1), then r r r r 2 1 1 2 2 2 2 μ = b − an − 1, γ = b + 2a, and γ + μ = 2ab − a n + 1, =1 j =1 j =1 =1 2 2 since N · C − E = 1, N · G = 0and N =−1. n Q =1 2a + b b − an − 1 Now, let γ¯ = γ − for all j ∈{1, 2,..., r }, and let μ ¯ = μ − for each ∈ j j 1 r r 1 2 r r 1 2 {1, 2,..., r }. Therefore, γ¯ = 0, μ ¯ = 0, and 2 j j =1 =1 r r 1 2 2 2 (2a + b) (b − an − 1) 2 2 2 0 ≤ γ¯ + μ ¯ = 2ab − a n + 1 − − . r r 1 2 j =1 =1 This implies that 2 2 (2a + b) (b − an − 1) 2ab − a n + 1 − − ≥ 0. r r 1 2 123 Arab. J. Math. Then, we obtain that r r b − 2r b + nr b − nr 1 2 2 1 1 a − nr r + 4r + n r 1 2 2 1 is less than or equal to 2 2 2 r r − r b − r b + 2r b − r (r r b − 2r b + nr b − nr ) 1 2 2 1 1 1 1 2 2 1 1 + . 2 2 2 nr r + 4r + n r (nr r + 4r + n r ) 1 2 2 1 1 2 2 1 Therefore, we have the following inequality: 2 2 2 r r − r b − r b + 2r b − r (r r b − 2r b + nr b − nr ) 1 2 2 1 1 1 1 2 2 1 1 + ≥ 0, 2 2 2 nr r + 4r + n r (nr r + 4r + n r ) 1 2 2 1 1 2 2 1 and then 2(n + 4)b nr r + 4r + n r − nr − 4 1 2 2 1 1 b − − ≤ 0. 2 2 (n + 2) + nr + 4r − nr − r r (n + 2) + nr + 4r − nr − r r 2 2 1 1 2 2 2 1 1 2 Thus, completing the square in b and using our hypothesis, it follows that b is bounded. Consequently, a is r ,r 1 2 bounded too. Therefore, in Case (1), X contains a ﬁnite number of (−1)-curves. Now, assume that we are in Case (2),thatis, N · (C − E ) = 0, and N · G = 1, then n Q =1 r r r r 2 1 1 2 2 2 2 μ = b − an, γ = b + 2a − 1, and γ + μ = 2ab − a n + 1. =1 j =1 j =1 =1 2a + b − 1 b − an Let γ¯ = γ − for all j ∈{1, 2,..., r }, and let μ ¯ = μ − for each ∈{1, 2,..., r }. j j 1 2 r r 1 2 r r 1 2 Therefore, γ¯ = 0, μ ¯ = 0, and j =1 =1 r r 1 2 2 2 (2a + b − 1) (b − an) 2 2 2 0 ≤ γ¯ + μ ¯ = 2ab − a n + 1 − − . r r 1 2 j =1 =1 This implies that 2 2 (2a + b − 1) (b − an) 2ab − a n + 1 − − ≥ 0. r r 1 2 Then, after completing the square in a, we get that r r b − 2r b + nr b + 2r 1 2 2 1 2 a − nr r + 4r + n r 1 2 2 1 is less than or equal to 2 2 2 r r − r b + 2r b − r b − r (r r b + nr b + 2r − 2r b) 1 2 2 2 1 2 1 2 1 2 2 + . 2 2 2 nr r + 4r + n r (nr r + 4r + n r ) 1 2 2 1 1 2 2 1 Therefore, we have the following inequality: 2 2 2 2(nr + n + 2r + 2n)b nr r + 4r + n r − nr − n 2 2 1 2 2 1 2 b − + ≤ 0. 2 2 (n + 2) + nr + 4r − nr − r r (n + 2) + nr + 4r − nr − r r 2 2 1 1 2 2 2 1 1 2 r ,r 1 2 contains a ﬁnite number of (−1)-curves. Thus, a and b are bounded. Therefore, in case Case (2), X The following lemma is the special case of Theorem 3.1,when r and r are zero. 1 2 Lemma 3.2 With the above notation, the surface has ﬁnitely many (−1)-curves and (−2)-curves. 123 Arab. J. Math. Fig. 3 The conﬁguration of the ordinary points of a nodal cubic of P Proof Since K = 4, is an anticanonical rational surface (see [3, Lemma 2.1, p. 3]). Then, the result holds from [3]and [31]. n+4,r The following result gives the list of (−1)-curves and (−2)-curves on the surface X . n+4,r Corollary 3.3 With notation as above. The (−1)-curves and (−2)-curves on X are those given in Tables 1, 2,and 3. Proof It follows from the bounds given in the proof of the last theorem. n+4,r Remark 3.4 It is worth noting that all the (−1)-curves (which are not exceptional) and (−2)-curves on X come from smooth curves in for every non-negative integers n and r . n 2 4,10 Consequently, we show that the surface X has no (−2)-curves, as in the case of blowing up the projective plane P at points in general position. 4,10 Example 3.5 With the notation of Theorem 3.1, the surface X has not (−2)-curves. However, it has 556 (−1)-curves. r ,r 1 2 Now, we handle the ﬁnite generation of the effective monoid of X . Corollary 3.6 With notation as above, if (n + 2) + nr + 4r − nr − r r is a positive integer, then the 2 2 1 1 2 r ,r 1 2 effective monoid of the surface X is ﬁnitely generated. Proof It follows from Theorem 3.1 and [31]. Remark 3.7 Let R, P , P , P , P , P be ordinary points of a nodal cubic D on the projective plane P ,such 1 2 3 4 5 that R is the singular point, P , P and P are collinear, but P , P ,and P are not for every i = 1, 2, 3; see 1 2 3 i 4 5 Fig. 3. The surface obtained as the blow-up of P at these 6 points has 21 (−1)-curves and only one (−2)- curve, instead of 27 (−1)-curves and no (−2)-curves as in the case of six points in general position of P ; see [8, Table 1, p. 34] and also Table 2. It is worth nothing that this surface is the blow-up of at the points P , P , P , P , P . Moreover, allowing r > 0, our result completes a result obtained by Harbourne in [19] 1 2 3 4 2 and another result obtained by the third author in [29]. Also, one may observe that blow-ups of P at the node of an irreducible cubic r times do not affect the ﬁnite generation of the effective monoid. r ,r 1 2 4 The ﬁniteness of the Cox ring of X r ,r 1 2 In this section, we prove that the surface X satisﬁes the anticanonical orthogonal property, and we use r ,r 1 2 Theorem 3.1 to prove the ﬁnite generation of the Cox ring of X . r ,r 1 2 r ,r Lemma 4.1 With notation as above, let D be the class of a nef divisor in NS(X ), such that D · K = 0. 1 2 If (n + 2) + nr + 4r − nr − r r is a positive integer, then D = 0. 2 2 1 1 2 123 Arab. J. Math. 4,r Table 1 List of (−1) and (−2)-curves of X 4,1 4,1 (−1)-curves of X Cardinality of the set of (−1)-curves on X 0 0 F − E , i ∈{1, 2, 3, 4} 4 F − E 1 E , i ∈{1, 2, 3, 4} 4 E 1 C − E , i ∈{1, 2, 3, 4} 4 0 P C − E 1 0 Q C + F − E − E , i , i ∈{1, 2, 3, 4} 6 0 P Q 1 2 i 1 =1 C + F − E , i , i , i ∈{1, 2, 3, 4} 4 0 P 1 2 3 =1 i C + 2F − E − E 1 0 P Q i =1 i 1 2C + F − E − E 1 0 P Q i =1 i 1 4,1 4,1 (−2)-curves of X Cardinality of the set of (−2)-curves on X 0 0 4,1 There are no (−2)-curves on X Zero 4,r 4,r 2 2 (−1)-curves of X with r = 1 Cardinality of the set of (−1)-curves on X with r = 1 2 2 0 0 F − E , i ∈{1, 2, 3, 4} 4 F − E , j ∈{1, 2,..., r } r Q 2 2 E , i ∈{1, 2, 3, 4} 4 E , j ∈{1, 2,..., r } r Q 2 2 C − E , i ∈{1, 2, 3, 4} 4 0 P C + F − E − E , i , i ∈{1, 2, 3, 4}, j ∈{1, 2,..., r } 6r 0 P Q 1 2 2 2 =1 i j C + F − E , i , i , i ∈{1, 2, 3, 4} 4 0 P 1 2 3 =1 i 3 2 2 C + 2F − E − E , i , i , i ∈{1, 2, 3, 4}, j , j ∈{1, 2,..., r } 4 0 P Q 1 2 3 1 2 2 i j =1 =1 C + 2F − E − E , j ∈{1, 2,..., r } r 0 P Q 2 2 i =1 i j 4 3 2 C + 3F − E − E , j , j , j ∈{1, 2,..., r } 0 P Q 1 2 3 2 i =1 i =1 j 2C + F − E − E , j ∈{1, 2,..., r } r 0 P Q 2 2 i j i =1 4,2 4,2 (−2)-curves of X Cardinality of the set of (−2)-curves on X 0 0 C − E − E 1 0 Q Q 1 2 4,r 4,r 2 2 (−2)-curves of X with r = 1, 2 Cardinality of the set of (−2)-curves on X with r = 1, 2 2 2 0 0 4,r There are no (−2)-curves on X with r = 1,2Zero 0 Arab. J. Math. 5,r Table 2 List of (−1) and (−2)-curves of X 5,0 5,0 (−1)-curves of X Cardinality of the set of (−1)-curves on X 1 1 C 1 F − E , i ∈{1, 2,..., 5} 5 E , i ∈{1, 2,..., 5} 5 C + F − E , i , i ∈{1, 2,... , 5} 10 1 P 1 2 =1 i C + 2F − E , i , i , i , i ∈{1, 2,... , 5} 5 1 P 1 2 3 4 =1 i 2C + 2F − E 1 1 P i =1 i 5,0 5,0 (−2)-curves of X Cardinality of the set of (−2)-curves on X 1 1 5,0 There are no (−2)-curves on X Zero 5,r 5,r 2 2 (−1)-curves of X with r > 0 Cardinality of the set of (−1)-curves on X with r > 0 2 2 1 1 F − E , i ∈{1, 2,..., 5} 5 F − E , j ∈{1, 2,..., r } r Q 2 2 E , i ∈{1, 2,..., 5} 5 E , j ∈{1, 2,..., r } r Q 2 2 C + F − E , i , i ∈{1, 2,... , 5} 10 1 P 1 2 =1 i C + 2F − E − E , i , i , i ∈{1, 2,..., 5}, j ∈{1, 2,..., r } 10r 1 P Q 1 2 3 2 2 =1 i j 4 2 C + 3F − E − E , i , i , i , i ∈{1, 2,... , 5}, j , j ∈{1, 2,..., r } 5 1 P Q 1 2 3 4 1 2 2 =1 i =1 j 5 3 C + 4F − E − E , j , j , j ∈{1, 2,..., r } 1 P Q 1 2 3 2 i =1 i =1 j C + 2F − E , i , i , i , i ∈{1, 2,... , 5} 5 1 P 1 2 3 4 =1 i C + 3F − E − E , j ∈{1, 2,..., r } r 1 P Q 2 2 i =1 i j 2C + 2F − E 1 1 P i =1 5,1 5,1 (−2)-curves of X Cardinality of the set of (−2)-curves on X 1 1 C − E 1 1 Q 5,r 5,r 2 2 (−2)-curves of X with r > 1 Cardinality of the set of (−2)-curves on X with r > 1 2 2 1 1 Tere are no (−2)-curves Zero Arab. J. Math. n+4,r Table 3 List of (−1) and (−2)-curves of X with n ≥ 2and r ≥ 0 n 2 n+4,r n+4,r 2 2 (−1)-curves of X with n ≥ 2and r ≥ 0 Cardinality of the set of (−1)-curves on X with n ≥ 2and r ≥ 0 n 2 n 2 F − E , i ∈{1, 2,..., n + 4} n + 4 F − E , j ∈{1, 2,..., r } r Q 2 2 E , i ∈{1, 2,..., n + 4} n + 4 E , j ∈{1, 2,..., r } r Q 2 2 n + 4 n+1 C + nF − E , i ,..., i ∈{1, 2,..., n + 4} n P 1 n+1 =1 i n + 1 i ,..., i ∈{1, 2,..., n + 4} n + 4 n+2 1 n+2 C + (n + 1)F − E − E , r n P Q 2 =1 i j j ∈{1, 2,..., r } n + 1 i ,..., i ∈{1, 2,..., n + 4} r n+3 2 1 n+3 2 C + (n + 2)F − E − E , (n + 4) n P Q =1 i =1 j j , j ∈{1, 2,..., r } 2 1 2 2 n+4 3 2 C + (n + 3)F − E − E , j , j , j ∈{1, 2,..., r } n P Q 1 2 3 2 i =1 j i =1 n+3 C + (n + 1)F − E , i ,..., i ∈{1, 2,..., n + 4} n + 4 n P 1 n+3 =1 i n+4 C + (n + 2)F − E − E , j ∈{1, 2,..., r } r n P Q 2 2 i =1 i j 6,0 6,0 (−2)-curves of X Cardinality of the set of (−2)-curves on X 2 2 C 1 6,r 6,r 2 2 (−2)-curves of X with r > 0 Cardinality of the set of (−2)-curves on X with r > 0 2 2 2 2 6,r There are no (−2)-curves on X with r >0Zero n+4,0 n+4,0 (−2)-curves of X with n > 2 Cardinality of (−2)-curves on X with n > 2 n n n + 4 n+2 C + nF − E , i ,..., i ∈{1, 2,..., n + 4} n P 1 n+2 =1 i n + 2 n+4,r n+4,r 2 2 (−2)-curves of X with n > 2and r > 0 Cardinality of the set of (−2)-curves on X with n > 2and r > 0 n 2 n 2 i ,..., i ∈{1, 2,..., n + 4} n+3 1 n+3 C + (n + 1)F − E − E , r (n + 4) n P Q 2 =1 i j j ∈{1, 2,..., r } 2 Arab. J. Math. r ,r r r 1 2 1 2 Proof Let D be a class of a nef divisor in NS(X ), so D = aC + bF − γ E − μ E , n n j P Q j =1 =1 for some integers a, b,γ ,γ ,...,γ ,μ ,μ ,...,μ . Therefore, D ≥ 0, D · (C − E ) = 0, and 1 2 r 1 2 r n Q 1 2 =1 D · G = 0. From these, we have the following: r r 1 2 2 2 2 2ab − a n − γ − μ ≥ 0, (3) j =1 =1 2a + b − γ = 0, and (4) j =1 b − an − μ = 0. =1 2a + b b − an Now, let γ¯ = γ − for all j ∈{1, 2,..., r }, and let μ ¯ = μ − for each ∈{1, 2,..., r }. j j 1 2 r r 1 2 Therefore r r 1 2 2 2 (2a + b) (b − an) 2 2 2 0 ≤ γ¯ + μ ¯ ≤ 2ab − a n − − . r r 1 2 j =1 =1 This implies that 2 2 (2a + b) (b − an) 2ab − a n − − ≥ 0. r r 1 2 Consequently, after completing the square in a, we obtain that 2 2 2 2 (r r − 2r + nr )b (r r − 2r + nr ) b (r + r )b 1 2 2 1 1 2 2 1 1 2 a − ≤ − . 2 2 2 2 nr r + 4r + n r (nr r + 4r + n r ) nr r + 4r + n r 1 2 2 1 1 2 2 1 1 2 2 1 Therefore 2 2 −((n + 2) + nr + 4r − nr − r r )b ≥ 0. 2 2 1 1 2 Therefore, using our numerical condition, we infer that b is equal to zero, and from Eqs. (3)and (4), we get that a is equal to zero. Thus, D = 0. Therefore, we are done. In the following theorem, the numerical condition (n + 2) + nr + 4r − nr − r r > 0 gives us a family 2 2 1 1 2 of smooth projective rational surfaces whose Cox rings are ﬁnitely generated. Theorem 4.2 With the above notation, if (n + 2) + nr + 4r − nr − r r is a positive integer, then the Cox 2 2 1 1 2 r ,r 1 2 ring of the surface X is ﬁnitely generated. Proof It follows from Theorem 3.1, Lemma 4.1, and Theorem 1 of [7]. Acknowledgements The authors are extremely grateful to the referees for their suggestions to improve the readability of our paper. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. 123 Arab. J. Math. Funding Gloria Andablo-Reyes and Mustapha Lahyane acknowledge a partial support from Coordinación de la Investigación Cientíﬁca de la Universidad Michoacana de San Nicolás de Hidalgo (UMSNH) during 2022. Brenda Leticia De La Rosa-Navarro was supported by Programa para el Desarrollo Profesional Docente, para el Tipo Superior under Grant No. UABC-PTC-558. 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Arabian Journal of Mathematics – Springer Journals
Published: Mar 13, 2023
Keywords: 14J26; 14C20; 14C22; 14C17; 14Q20
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