Korff $F$-signatures of Hirzebruch surfaces

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by  Daisuke Hirose

KORFF F-SIGNATURES OF HIRZEBRUCH SURFACES
DAISUKE HIROSE AND TADAKAZU SAWADA
7
1
0
2

n Abstract. M.VonKorffintroducedthe F-signatureof a normal
a
J projective variety, and computed the F-signature of the product
of two projective lines P1 P1. In this paper, we compute F-
8
×
signatures of Hirzebruch surfaces.

]
G
A
Introduction
.
h
t
a Let R be a d-dimensional Noetherian local ring of prime characteris-
m tic p with perfect residue field. Let F : R R be the Frobenius map,
→
[ that is, F(x) = xp for all elements x of R. We denote by FeR the
∗

1 R-module whose abelian group structure is inherited by R, and whose
v R-module structure is given by the e-th iterated Frobenius map Fe. If
5
F R is finitely generated as an R-module, we say that R is F-finite.
0 ∗
9 Let the R-module FeR be decomposed as FeR = R⊕ae M , where M
∗ ∗ ∼ e e
1 ⊕
does not have free summands. In [HL02], C. Huneke and G. Leuschke
0
. introduced the notion of the F-signature s(R) of R as
1
0
a
7 s(R) = lim e .
1 e→∞ ped
:
v
Xi The existence of the limit lime→∞(ae/ped) is not trivial. In [Tuc12],
K. Tucker proved that the limit exists (under the assumption that
r
a R is F-finite). The notion of the F-signature is generalized by M.
Hashimoto and Y. Nakajima in [HN15], and A. Sannai in [San15]. As
aglobalanalogue, M. VonKorffintroducedtheF-signatureofanormal
projective variety, which we call the Korff F-signature, in [Kor12]. He
computed the Korff F-signature of the product of two projective lines
P1 P1. In this paper, we compute Korff F-signatures of Hirzebruch
×
surfaces.
In Section1, we review generalities onKorff F-signatures. InSection
2, we compute Korff F-signatures of projective spaces and Hirzebruch
surfaces. In Section 3 and 4, we gather together the figures and the
source code of wxMaxima in the computation of F-signatures of Hirze-
bruch surfaces.
1
2 Daisuke Hirose and Tadakazu Sawada
1. Preliminaries
In this section, we review generalities on Korff F-signatures. See
[Kor12] for details.
Let k be a field of positive characteristic p. Let X be a normal
projective variety over k, and let D be a Q-divisor on X. We denote
by Sec(X,D) the section ring Γ(X, (nD)) of D. Suppose that
n≥0 OX
Sec(X,D) is a finitely generated k-algebra of dimension at least two.
L
If c N is sufficiently divisible, then there exists an ample line bundle
∈
L on X such that Sec(X,cD) is isomorphic to a normal section ring
Sec(X,L) = Γ(X,L⊗n) of L. We assume that c is sufficiently
n≥0
divisible, and define the Korff F-signature s(X,D) of X along D to be
L
c s(Sec(X,cD)). (The F-signature of a graded ring is defined in the
·
same way as the case of local rings.)
Let N = Zn be a lattice, and let M be the dual lattice of N. We
∼
denote N R (resp. M R) by N (resp. M ). Let Σ be a fan in
Z Z R R
⊗ ⊗
N . We denote by X the toric variety corresponding to Σ.
R Σ
LetΣbeacompletefaninN withtheprimitivegeneratorsv ,...,v .
R 1 n
Let D ,...,D be the torus-invariant prime Weil divisors correspond-
1 n
n
ing to v ,...,v , respectively. Let D = a D be a torus-invariant
1 n i=1 i i
Weil divisor on X. The polytope associated to the divisor D is defined
P
to be the polytope u M u v a for all i , and denoted by
R i i
{ ∈ | · ≥ − }
P . We may assume that P contains the origin. Then we denote
D D
by M the lattice M (P R), where P R is the R-vector sub-
D D D
∩ · ·
space spanned by P in M . We denote by N the dual lattice of M .
D R D D
We define I to be the subset of 1,...,n such that the hyperplane
D
{ }
u u v = a determines a facet of P for i I , and P to be
i i D D Σ,D
{ | · − } ∈
the polytope (v,t) (M Z) 0 (v,t) (v ,a ) < 1 for all i I .
D R i i D
{ ∈ × | ≤ · ∈ }
Suppose that P is not full-dimensional. Let π : N N be the
D D
→
naturalprojection map. Let c betherational number such that π(c v )
i i i
is the primitive generator for its ray in (N ) . We define P′ to be
D R Σ,D
thepolytope (v,t) (M Z) 0 (v,t) c (v ,a ) < 1 for alli I .
D R i i i D
{ ∈ × | ≤ · ∈ }
Theorem 1.1 ([Kor12], Proposition 4.4.4, Corollary 4.4.7). Let X be
a d-dimensional projective toric variety over k, and let Σ be the fan
correspondingto X. Let v ,...,v be the primitive generators of Σ, and
1 n
let D ,...,D be the torus-invariant prime Weil divisors corresponding
1 n
n
to v ,...,v , respectively. Let D = a D be a (non-zero) effective
1 n i=1 i i
torus-invariant Weil divisor on X. Then:
P
(1) If P is full-dimensional, then s(X,D) = Vol(P ).
D Σ,D
(2) If P is not full-dimensional, then s(X,D) = Vol(P′ ).
D Σ,D
Korff F-signature of Hirzebruch surfaces 3
2. Examples
M. Von Korff computed Korff F-signatures of some toric varieties
including the product of two projective lines P1 P1 in [Kor12]. In
×
this section, we compute Korff F-signatures of projective spaces and
Hirzebruch surfaces.
Example 2.1 (Projective spaces Pn). Let N = Zn with standard basis
n
e ,...,e . Let e = e , and let Σ be the fan in N consisting
1 n 0 − i=1 i R
of the cones generated by all proper subsets of e ,e ,...,e . We see
0 1 n
that X is isomorphic Pto the n-dimensional pr{ojective space}Pn. Let
Σ
D be the torus-invariant prime Weil divisor corresponding to the ray
i
R e in N for 0 i n.
+ i R
n ≤ ≤
Let D = a D be a (non-zero) effective torus-invariant Weil
i=0 i i
divisor. We see that
P
n
x a ,
i=1 i ≤ 0
x a ,
PD = (x1,...,xn) MR(cid:12) P1 ≥ − 1 .
∈ (cid:12)
 
 (cid:12) ··· 
 (cid:12) xn an 
(cid:12) ≥ −
(cid:12)
 
 (cid:12) 
Since a = 0 for some i, P has an interior point, andthe equations
i D (cid:12)
n 6
x = a ,x = a ,...,x = a define the facets of P , respec-
i=1 i 0 1 − 1 n − n D
tively. Hence we have I = 0,1,...,n . In what follows, we compute
D
P { }
the volume of the n-dimensional parallelepipeds
n
0 x +a t < 1,
≤ − i=1 i 0
0 x +a t < 1,
PΣ,D = (x1,...,xn,t) (MD Z)R(cid:12) ≤ 1P 1 .
∈ × (cid:12)
 
 (cid:12) ··· 
 (cid:12) 0 xn +ant < 1 
(cid:12) ≤
(cid:12)
 
Let a = ni=0ai. The point of the i(cid:12)(cid:12)ntersection of the n+ 1 hyper-
n
planes defined by x +a t = 1,x +a t = 0,...,x +a t = 0
P − i=1 i 0 1 1 n n
is ( a /a,..., a /a,1/a). This is given by the following elementary
1 n
− − P
transformations of matrices:
1 1 1 a 1 0 0 0 a 1
0
− − ··· − ···
1 0 0 a 0 1 0 0 a 0
1 1
 ···   ··· 
→
··· ···
 0 0 1 an 0   0 0 1 an 0 
 ···   ··· 
 0 0 0 a 1   1 0 0 0 a /a
1
··· ··· −
1 0 0 0 a /a
 ··· − 1   ··· .
→ → 0 0 1 0 an/a
··· ··· −
 0 0 1 0 an/a   0 0 0 1 1/a 
 ··· −   ··· 
   
4 Daisuke Hirose and Tadakazu Sawada
Bythesameargument, weseethatthepointoftheintersectionofthe
n
n+1 hyperplanes defined by x +a t = 0, x +a t = 0 (l = k),
− i=1 i 0 l l 6
x +a t = 1is( a /a,...,1 a /a,..., a /a,1/a)foreach1 k n.
k k 1 k n
Let v = t( −a /a,..., a−/a,P1/a), a−nd ≤ ≤
0 1 n
− −
v = t( a /a,...,1 a /a,..., a /a,1/a)
k 1 k n
− − −
for 1 k n. Since P is spanned by v ,v ,...,v , we have
Σ,D 0 1 n
≤ ≤
s(Pn,D) = Vol(P )
Σ,D
= det(v ,v ,...,v )
0 1 n
| |
a /a 1 a /a a /a a /a
1 1 1 1
− − − ··· −
a /a a /a 1 a /a a /a
(cid:12)  − 2 − 2 − 2 ··· − 2 (cid:12)
= (cid:12)det (cid:12)
(cid:12) ··· (cid:12)
(cid:12)  a /a a /a a /a 1 a /a (cid:12)
n n n n
(cid:12)  − − − ··· − (cid:12)
(cid:12)  1/a 1/a 1/a 1/a (cid:12)
(cid:12)  ··· (cid:12)
(cid:12)  (cid:12)
(cid:12) a1 a a1 a1 a(cid:12)1
(cid:12) − − − ··· − (cid:12)
a a a a a
(cid:12)  − 2 − 2 − 2 ··· − 2 (cid:12)
= 1/an+1(cid:12)det (cid:12)
(cid:12) ··· (cid:12)
(cid:12)  a a a a a (cid:12)
n n n n
(cid:12)  − − − ··· − (cid:12)
(cid:12)  1 1 1 1 1 (cid:12)
(cid:12)  ··· (cid:12)
(cid:12)  (cid:12)
(cid:12) 0 a 0 0 (cid:12)
(cid:12) ··· (cid:12)
0 0 a 0
(cid:12)  ··· (cid:12)
= 1/an+1(cid:12)det (cid:12)
(cid:12) ··· (cid:12)
(cid:12)  0 0 0 a (cid:12)
(cid:12)  ··· (cid:12)
(cid:12)  1 1 1 1 (cid:12)
(cid:12)  ··· (cid:12)
= 1/a = 1(cid:12)/(a+a + +a ). (cid:12)
(cid:12) 0 1 n (cid:12)
···
(cid:12) (cid:12)
The following computation is the main result.
Example 2.2 (Hirzebruch surfaces). All figures of this example are in
Section 3. Let r be a positive integer, and let N = Z2 with standard
basis e and e . Let v = e ,v = e ,v = e , and v = e + re .
1 2 1 1 2 2 3 2 4 1 2
− −
We consider a fan Σ in N = R2. The fan Σ consists the four two-
r R r
dimensional cones v ,v , v ,v , v ,v , and v ,v , and all their
1 2 1 3 2 4 3 4
h i h i h i h i
faces. The Hirzebruch surface X = X is defined by the fan Σ (see,
Σr r
e.g., [CLS11], Example 3.1.16.). Let D be the torus-invariant prime
i
divisors of X corresponding to v . We have linear equivalents D D
i 1 4
∼
and D D + rD in the divisor class group of X. For an effective
3 2 4
∼
Q-divisor D = 4 a D , we have
i=1 i i
D a D +(a +a )D +(ra +a )D
P 1 1 2 3 2 3 4 4
∼
(a +ra +a )D +(a +a )D .
1 3 4 1 2 3 2
∼
Korff F-signature of Hirzebruch surfaces 5
Hence (D) = ((a + ra + a )D + (a + a )D ). Without loss of
∼ 1 3 4 1 2 3 2
O O
generality, we only compute the F-signature s(X,D) of X with respect
to a (non-zero) effective Q-divisor D = a D +a D .
1 1 2 2
The polytope P is to be
D
x a , y a
PD = (x,y) ∈ MR y ≥0−, 1 x+≥ry− 20 .
(cid:26) (cid:12) ≤ − ≥ (cid:27)
(cid:12)
By the figure 1, we see that(cid:12)P is a square if and only if ra < a .
(cid:12) D 2 1
In that case, I = 1,2,3,4 . On the other hand, if ra a , then P
D 2 1 D
{ } ≥
is a triangle. That implies I = 1,3,4 .
D
{ }
(1) Suppose that P is a square, i.e., ra < a . Then I = 1,2,3,4
D 2 1 D
{ }
Therefore
0 x+a t < 1,
1
≤
0 y +a t < 1,
PΣ,D = (x,y,t) ∈ (MD ×Z)R(cid:12)(cid:12) 0 ≤ y <21, .
 
 (cid:12) ≤ − 
 (cid:12) 0 x+ry < 1 
(cid:12) ≤ −
(cid:12)
 
We define polygons Q, R(x), and S(x)(cid:12)to be 
(cid:12)
0 y +a t < 1,
Q = (y,t) R2 ≤ 2
∈ 0 y < 1
(cid:26) (cid:12) ≤ − (cid:27)
(cid:12)
a t y < 1 a t,
= (y,t) R2(cid:12)(cid:12) − 2 ≤ − 2 ,
∈ 1 < y 0
(cid:26) (cid:12) − ≤ (cid:27)
(cid:12)
0 x+a t < 1,
R(x) = (y,t) R2(cid:12)(cid:12) ≤ 1
∈ 0 x+ry < 1
(cid:26) (cid:12) ≤ − (cid:27)
(cid:12)
x/a t < x/a +1/a ,
= (y,t) R2(cid:12)(cid:12) − 1 ≤ − 1 1 ,
∈ x/r y < x/r +1/r
(cid:26) (cid:12) ≤ (cid:27)
(cid:12)
(cid:12)
(cid:12)
and S(x) = Q R(x).(See the figure 2.)
∩
Then Q is a parallelogram, R(x) is a rectangle, and
P = (x,y,t) (M Z) x x x ,(y,t) S(x) ,
Σ,D D R min max
{ ∈ × | ≤ ≤ ∈ }
wherex = min x R S(x) = andx = max x R S(x) = .
min max
{ ∈ | 6 ∅} { ∈ | 6 ∅}
Therefore
xmax
s(X,D) = Vol(P ) = Area(S(x))dx.
Σ,D
Zxmin
We define some points and lines in the ty-plane appearing in our
argument. Let A,B,C, and D be the vertices of the rectangle R(x).
6 Daisuke Hirose and Tadakazu Sawada
In particular,
1 1 1 1 1
A = x, x ,B = x+ , x ,
−a r −a a r
(cid:18) 1 (cid:19) (cid:18) 1 1 (cid:19)
1 1 1 1 1 1 1
C = x+ , x+ ,and D = x, x+ .
−a a r r −a r r
(cid:18) 1 1 (cid:19) (cid:18) 1 (cid:19)
The point A moves along the line y = a t/r, denoted by l . The
1 1
−
points B and D move along the line y = a t/r +1/r, denoted by l .
1 2
−
The point C moves along the line y = a t/r+2/r, denoted by l .
1 3
−
We denote the vetices of the parallelogram Q except the origin O by
E,F, and G. That is,
1 2 1
E = , 1 ,F = , 1 ,and G = ,0 .
a − a − a
(cid:18) 2 (cid:19) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)
Let H and I be the intersection points of the t-axis with l and l ,
2 3
respectively. We denote the intersection points of the line y = 1 with
−
l , l , and l by J, K, and L, respectively. That is,
1 2 3
1 2
H = ,0 , I = ,0 ,
a a
(cid:18) 1 (cid:19) (cid:18) 1 (cid:19)
r r +1 r +2
J = , 1 , K = , 1 , and L = , 1 .
a − a − a −
(cid:18) 1 (cid:19) (cid:18) 1 (cid:19) (cid:18) 1 (cid:19)
For the point A, we denote the t-coordinate of A and y-coordinate of
A by A and A , respectively. We use the same notation for all points
t y
in the ty-plane. For example, B = x/a +1/a and J = 1.
t 1 1 y
− −
The shape of the polygon P depends on how the lines l ,l and
Σ,D 1 2
l across the parallelogram Q. We divide into five subcases from (1-1)
3
through (1-5):
(1-1) We assume that L E . That is, (r + 2)/a 1/a . This is
t t 1 2
≤ ≤
equivalent to (r + 2)a a . Then 2a a . This is equivalent to
2 1 2 1
≤ ≤
2/a 1/a . Hence I G . (See the figure 4.)
1 2 t t
≤ ≤
We denote the intersection points of the line OE with l and l by
2 3
T and U, respectively. Then
1 a 2 2a
2 2
T = , , and U = , .
a ra −a ra a ra −a ra
(cid:18) 1 − 2 1 − 2(cid:19) (cid:18) 1 − 2 1 − 2(cid:19)
If C = U, then ( x/a +1/a ,x/r+1/r) = (2/(a ra ), 2a /(a
1 1 1 2 2 1
− − − −
ra )). Hence x = (a +ra )/(a ra ). If D = T, then ( x/a ,x/r+
2 1 2 1 2 1
− − −
1/r) = (1/(a ra ), a /(a ra )). Hence x = a /(a ra ). If
1 2 2 1 2 1 1 2
− − − − −
B = T, then ( x/a + 1/a ,x/r) = (1/(a ra ), a /(a ra )).
1 1 1 2 2 1 2
− − − −
Hence x = ra /(a ra ). If C = I and D = H, then ( x/a +
2 1 2 1
− − −
Korff F-signature of Hirzebruch surfaces 7
1/a ,x/r + 1/r) = (2/a ,0). Hence x = 1. If A = O and B = H,
1 1
−
then ( x/a ,x/r) = (0,0). Hence x = 0.
1
−
Since 0 < ra < a , we have ra < a < a +ra and a ra > 0.
2 1 2 1 1 2 1 2
−
Hence (a + ra )/(a ra ) < a /(a ra ) < ra /(a ra ).
1 2 1 2 1 1 2 2 1 2
− − − − − −
Since a ra < a , we have a /(a ra ) < 1. If 2ra a , then
1 2 1 1 1 2 2 1
− − − − ≤
1 ra /(a ra ). Therefore
2 1 2
− ≤ − −
a +ra a ra
2 2 1 2
< < 1 < < 0.
−a ra −a ra − −a ra
1 2 1 2 1 2
− − −
On the other hand, if 2ra > a , then ra /(a ra ) < 1. Therefore
2 1 2 1 2
− − −
a +ra a ra
2 2 1 2
< < < 1 < 0.
−a ra −a ra −a ra −
1 2 1 2 1 2
− − −
Forvariousvaluesofx, weconsider shapesofS(x)andareasofthem.
If (a + ra )/(a ra ) < x < a /(a ra ), then S(x) is a
2 2 1 2 1 1 2
− − − −
triangle as in the figure 5. Hence
1 1 1 a a 1 1 1 1
2 2
Area(S(x)) = x+ x x+ x
2 r r − a − a · −a a − −ra − ra
(cid:26)(cid:18) (cid:19) (cid:18) 1 1(cid:19)(cid:27) (cid:26)(cid:18) 1 1(cid:19) (cid:18) 2 2(cid:19)(cid:27)
1
= (a ra )x+(a +ra ) 2.
2a2a r2 { 1 − 2 1 2 }
1 2
(1-1-1) Suppose that 2ra a .
2 1
≤
If a /(a ra ) < x < 1, then S(x) is a trapezoid as in the figure
1 1 2
− − −
6. Hence
1 1 1 a 1 1 a a 1
2 2 2
Area(S(x)) = x+ x + x+ x+
2 r r − a r r − a a · a
(cid:26)(cid:18) 1 (cid:19) (cid:18) 1 1(cid:19)(cid:27) 1
1
= 2(a ra )x+2a +ra .
2a2r { 1 − 2 1 2}
1
We denote this area by V .
1
If 1 < x < ra /(a ra ), then S(x) is a tropezoid as in the
2 1 2
− − −
figure 7. Hence
1 a a a 1
2 2 2
Area(S(x)) = x x+
2 −a − a a · a
(cid:18) 1 1 1(cid:19) 1
a
2
= (1 2x).
2a2 −
1
We denote this area by V .
2
8 Daisuke Hirose and Tadakazu Sawada
If ra /(a ra ) < x < 0, then S(x) is a pentagon as in the figure
2 1 2
− −
8. Hence
1 1 1 a 1 1 1
2
Area(S(x)) = x x x x x
−r · a − 2 a − r · −ra − −a
1 (cid:18) 1 (cid:19) (cid:26) 2 (cid:18) 1 (cid:19)(cid:27)
1 (a ra )2
= x 1 − 2 x2.
−ra − 2r2a2a
1 1 2
We denote this area by V .
3
Therefore
s(X,D) = Vol(P )
Σ,D
= −a1−a1ra2 1 (a ra )x+(a +ra ) 2dx
Z−aa12−+rraa22 2a21a2r2 { 1 − 2 1 2 }
−1 − ra2 0
a1−ra2
+ V dx+ V dx+ V dx
1 2 3
Z−a1−a1ra2 Z−1 Z−a1r−ar2a2
a
2
= .
a (a a r)
1 1 2
−
In the above calculation, we use wxMaxima (see Section 4 (%o1)).
(1-1-2) Suppose that 2ra > a . If a /(a ra ) < x < ra /(a
2 1 1 1 2 2 1
− − − −
ra ), then S(x) is a trapezoid as in the figure 9, which is the same
2
shape as that in the case where a /(a ra ) < x < 1 on (1-1-1).
1 1 2
− − −
Then Area(S(x)) = V .
1
If ra /(a ra ) < x < 1, then S(x) is a pentagon as in the
2 1 2
− − −
figure 10. Hence
1 1 1 1 1 a 1 1 1
2
Area(S(x)) = x+ x x x x
r r − r · a − 2 a − r · −ra − −a
(cid:18) (cid:19) 1 (cid:18) 1 (cid:19) (cid:26) 2 (cid:18) 1 (cid:19)(cid:27)
1 (a ra )2
= 1 − 2 x2.
ra − 2a2a r2
1 1 2
We denote this area by V .
4
If 1 < x < 0, then S(x) is a pentagon as in the figure 11, which is
−
the same shape as that in the case where ra /(a ra ) < x < 0 on
2 1 2
− −
(1-1-1). Then Area(S(x)) = V .
3
Korff F-signature of Hirzebruch surfaces 9
Therefore
s(X,D) = Vol(P )
Σ,D
= −a1−a1ra2 1 (a ra )x+(a +ra ) 2dx
Z−aa12−+rraa22 2a21a2r2 { 1 − 2 1 2 }
− ra2 −1 0
a1−ra2
+ V dx+ V dx+ V dx
1 4 3
Z−a1−a1ra2 Z−a1r−ar2a2 Z−1
a
2
= ,
a (a a r)
1 1 2
−
(see Section 4 (%o2)).
(1-2) We assume that K < E < L . That is, (r + 1)/a < 1/a <
t t t 1 2
(r + 2)/a . This is equivalent to (r + 1)a < a < (r + 2)a . Then
1 2 1 2
2a < a . We have I G . Since (r +1)a < a and a ra < a ,
2 1 t t 2 1 2 2 1
≤ ≤
we have (r +2)a < 2a . Therefore L < F . The point T denotes the
2 1 t t
same in the case (1-1). (See the figure 12.)
If C = L and D = K, then ( x/a + 1/a ,x/r + 1/r) = ((r +
1 1
−
2)/a , 1). Hence x = (r +1). If B = E , then (1/a )x+1/a =
1 t t 1 1
− − −
1/a . Hence x = (a a )/a . By the same argument in the case
2 1 2 2
− −
(1-1), we have the following four values of x. If D = T, then x =
a /(a ra ). If B = T, then x = ra /(a ra ). If C = I and
1 1 2 2 1 2
− − − −
D = H, then x = 1. If A = O and B = H, then x = 0.
−
Since (r + 1)a < a , we have r(r + 1)a + a < (r + 1)a . Hence
2 1 2 1 1
a < (r + 1)(a ra ). Therefore (r + 1) < a /(a ra ). If
1 1 2 1 1 2
− − − −
a /(a ra ) > (a a )/a , then a2 (r+2)a a +ra2 > 0. We have
1 1− 2 1− 2 2 1− 2 1 2
(r +2) √r2 +4 (r+2)+√r2 +4
− a < a < a .
2 1 2
2 2
By the assumption that (r +1)a < a < (r +2)a , we have a /(a
2 1 2 1 1
−
ra ) > (a a )/a in case (1-2). Therefore a /(a ra ) < (a
2 1 2 2 1 1 2 1
− − − − −
a )/a . Since 2a < a , we have a < a a . Hence (a a )/a <
2 2 2 1 2 1 2 1 2 2
− − −
1. The assumption that (r + 1)a < a gives the inequalities ra <
2 1 2
−
a a and a < a ra . Those imply ra /(a ra ) < (a
1 2 2 1 2 2 1 2 1
− − − −
a )/a . Therefore (a a )/a < ra /(a ra ). Finally, we com-
2 2 1 2 2 2 1 2
− − − −
pare the inequality between 1 and ra /(a ra ). Suppose that
2 1 2
− − −
1 < ra /(a ra ). Then a > 2ra . With the assumption that
2 1 2 1 2
− − −
a < (r+2)a , we have 2ra < (r+2)a . Hence r < 2. That is, r = 1.
1 2 2 2
Therefore if r = 1, then
a a a ra
1 1 2 2
(r +1) < < − < 1 < < 0.
− −a ra − a − −a ra
1 2 2 1 2
− −
10 Daisuke Hirose and Tadakazu Sawada
By the same argument, if ra /(a ra ) < 1, then 1 < r. Therefore
2 1 2
− − −
if r 2, then
≥
a a a ra
1 1 2 2
(r +1) < < − < < 1 < 0.
− −a ra − a −a ra −
1 2 2 1 2
− −
We remark that ra /(a ra ) = 1, because r is integer.
2 1 2
− − 6 −
Forvariousvaluesofx, weconsider shapesofS(x)andareasofthem.
If (r+1) < x < a /(a ra ), then S(x) is a trapezoid as in the
1 1 2
− − −
figure 13. Hence
1 1 1 1 1 1 1 1
Area(S(x)) = x+ x + x+
2 −a a − −ra − ra −a a − a
(cid:26)(cid:18) 1 1 (cid:18) 2 2(cid:19)(cid:19) (cid:18) 1 1 2(cid:19)(cid:27)
1 1
x+ ( 1)
× r r − −
(cid:18) (cid:19)
1
= (a 2a r)x+2a r+a a r (x+r +1).
2a a r2 { 1 − 2 2 1 − 1 }
1 2
If a /(a ra ) < x < (a a )/a , then S(x) is a pentagon as
1 1 2 1 2 2
− − − −
in the figure 14. Hence
1 1 1 1 1 1 a
2
Area(S(x)) = x+ ( 1) x x ( 1)
r r − − · a − 2 a − −a · a − −
(cid:26) (cid:27) 1 (cid:26) 2 (cid:18) 1 (cid:19)(cid:27) (cid:26) 1 (cid:27)
1 1
= (x+r +1) (a x+a )2.
a r − 2a2a 2 1
1 1 2
We denote this area by V .
5
(1-2-1) Suppose that r = 1. If (a a )/a < x < 1, then S(x) is a
1 2 2
− − −
trapezoidasinthefigure15, which isthesameshapeasthatinthecase
where a /(a ra ) < x < 1 on (1-1-1). Hence Area(S(x)) = V .
1 1 2 1
− − −
If 1 < x < ra /(a ra ), then S(x) is a trapezoid as in the
2 1 2
− − −
figure 16, which is the same shape as that in the case where 1 < x <
−
ra /(a ra ) on (1-1-1). Hence Area(S(x)) = V .
2 1 2 2
− −
If ra /(a ra ) < x < 0, then S(x) is a pentagon as in the figure
2 1 2
− −
17, which isthesameshape asthat inthecasewhere ra /(a ra ) <
2 1 2
− −
x < 0 on (1-1-1). Hence Area(S(x)) = V .
3