Citation: Aleksandar Vasilev. Can shocks to the discount factor explain business cycle fluctuations in Bulgaria (1999–2018)?[J]. National Accounting Review, 2020, 2(3): 285-296. doi: 10.3934/NAR.2020016
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Let Fq be the finite field of q elements with characteristic p, where q=pr, p is a prime number. Let F∗q=Fq∖{0} and Z+ denote the set of positive integers. Let s∈Z+ and b∈Fq. Let f(x1,…,xs) be a diagonal polynomial over Fq of the following form
f(x1,…,xs)=a1xm11+a2xm22+⋯+asxmss, |
where ai∈F∗q, mi∈Z+, i=1,…,s. Denote by Nq(f=b) the number of Fq-rational points on the affine hypersurface f=b, namely,
Nq(f=b)=#{(x1,…,xs)∈As(Fq)∣f(x1,…,xs)=b}. |
In 1949, Hua and Vandiver [1] and Weil [2] independently obtained the formula of Nq(f=b) in terms of character sum as follows
Nq(f=b)=qs−1+∑ψ1(a−11)⋯ψs(a−ss)J0q(ψ1,…,ψs), | (1.1) |
where the sum is taken over all s multiplicative characters of Fq that satisfy ψmii=ε, ψi≠ε, i=1,…,s and ψ1⋯ψs=ε. Here ε is the trivial multiplicative character of Fq, and J0q(ψ1,…,ψs) is the Jacobi sum over Fq defined by
J0q(ψ1,…,ψs)=∑c1+⋯+cs=0,ci∈Fqψ1(c1)⋯ψs(cs). |
Though the explicit formula for Nq(f=b) are difficult to obtain in general, it has been studied extensively because of their theoretical importance as well as their applications in cryptology and coding theory; see[3,4,5,6,7,8,9]. In this paper, we use the Jacobi sums, Gauss sums and the results of quadratic form to deduce the formula of the number of Fq2-rational points on a class of hypersurfaces over Fq2 under certain conditions. The main result of this paper can be stated as
Theorem 1.1. Let q=2r with r∈Z+ and Fq2 be the finite field of q2 elements. Let f(X)=a1xm11+a2xm22+⋯+asxmss, g(Y)=y1y2+y3y4+⋯+yn−1yn+y2n−2t−1+… +y2n−3+y2n−1+bty2n−2t+⋯+b1y2n−2+b0y2n, and l(X,Y)=f(X)+g(Y), where ai,bj∈F∗q2, mi≠1, (mi,mk)=1, i≠k, mi|(q+1), mi∈Z+, 2|n, n>2, 0≤t≤n2−2, TrFq2/F2(bj)=1 for i,k=1,…,s and j=0,1,…,t. For h∈Fq2, we have
(1) If h=0, then
Nq2(l(X,Y)=0)=q2(s+n−1)+∑γ∈F∗q2(s∏i=1((γai)mimi−1)(qs+2n−3+(−1)tqs+n−3)). |
(2) If h∈F∗q2, then
Nq2(l(X,Y)=h)=q2(s+n−1)+(qs+2n−3+(−1)t+1(q2−1)qs+n−3)s∏i=1((hai)mimi−1)+∑γ∈F∗q2∖{h}[s∏i=1((γai)mimi−1)(q2n+s−3+(−1)tqn+s−3)]. |
Here,
(γai)mi={1,ifγaiisaresidueofordermi,0,otherwise. |
To prove Theorem 1.1, we need the lemmas and theorems below which are related to the Jacobi sums and Gauss sums.
Definition 2.1. Let χ be an additive character and ψ a multiplicative character of Fq. The Gauss sum Gq(ψ,χ) in Fq is defined by
Gq(ψ,χ)=∑x∈F∗qψ(x)χ(x). |
In particular, if χ is the canonical additive character, i.e., χ(x)=e2πiTrFq/Fp(x)/p where TrFq/Fp(y)=y+yp+⋯+ypr−1 is the absolute trace of y from Fq to Fp, we simply write Gq(ψ):=Gq(ψ,χ).
Let ψ be a multiplicative character of Fq which is defined for all nonzero elements of Fq. We extend the definition of ψ by setting ψ(0)=0 if ψ≠ε and ε(0)=1.
Definition 2.2. Let ψ1,…,ψs be s multiplicative characters of Fq. Then, Jq(ψ1,…,ψs) is the Jacobi sum over Fq defined by
Jq(ψ1,…,ψs)=∑c1+⋯+cs=1,ci∈Fqψ1(c1)⋯ψs(cs). |
The Jacobi sums Jq(ψ1,…,ψs) as well as the sums J0q(ψ1,…,ψs) can be evaluated easily in case some of the multiplicative characters ψi are trivial.
Lemma 2.3. ([10,Theorem 5.19,p. 206]) If the multiplicative characters ψ1,…,ψs of Fq are trivial, then
Jq(ψ1,…,ψs)=J0q(ψ1,…,ψs)=qs−1. |
If some, but not all, of the ψi are trivial, then
Jq(ψ1,…,ψs)=J0q(ψ1,…,ψs)=0. |
Lemma 2.4. ([10,Theorem 5.20,p. 206]) If ψ1,…,ψs are multiplicative characters of Fq with ψs nontrivial, then
J0q(ψ1,…,ψs)=0 |
if ψ1⋯ψs is nontrivial and
J0q(ψ1,…,ψs)=ψs(−1)(q−1)Jq(ψ1,…,ψs−1) |
if ψ1⋯ψs is trivial.
If all ψi are nontrivial, there exists an important connection between Jacobi sums and Gauss sums.
Lemma 2.5. ([10,Theorem 5.21,p. 207]) If ψ1,…,ψs are nontrivial multiplicative characters of Fq and χ is a nontrivial additive character of Fq, then
Jq(ψ1,…,ψs)=Gq(ψ1,χ)⋯Gq(ψs,χ)Gq(ψ1⋯ψs,χ) |
if ψ1⋯ψs is nontrivial and
Jq(ψ1,…,ψs)=−ψs(−1)Jq(ψ1,…,ψs−1)=−1qGq(ψ1,χ)⋯Gq(ψs,χ) |
if ψ1⋯ψs is trivial.
We turn to another special formula for Gauss sums which applies to a wider range of multiplicative characters but needs a restriction on the underlying field.
Lemma 2.6. ([10,Theorem 5.16,p. 202]) Let q be a prime power, let ψ be a nontrivial multiplicative character of Fq2 of order m dividing q+1. Then
Gq2(ψ)={q,ifmoddorq+1meven,−q,ifmevenandq+1modd. |
For h∈Fq2, define v(h)=−1 if h∈F∗q2 and v(0)=q2−1. The property of the function v(h) will be used in the later proofs.
Lemma 2.7. ([10,Lemma 6.23,p. 281]) For any finite field Fq, we have
∑c∈Fqv(c)=0, |
for any b∈Fq,
∑c1+⋯+cm=bv(c1)⋯v(ck)={0,1⩽k<m,v(b)qm−1,k=m, |
where the sum is over all c1,…,cm∈Fq with c1+⋯+cm=b.
The quadratic forms have been studied intensively. A quadratic form f in n indeterminates is called nondegenerate if f is not equivalent to a quadratic form in fewer than n indeterminates. For any finite field Fq, two quadratic forms f and g over Fq are called equivalent if f can be transformed into g by means of a nonsingular linear substitution of indeterminates.
Lemma 2.8. ([10,Theorem 6.30,p. 287]) Let f∈Fq[x1,…,xn], q even, be a nondegenerate quadratic form. If n is even, then f is either equivalent to
x1x2+x3x4+⋯+xn−1xn |
or to a quadratic form of the type
x1x2+x3x4+⋯+xn−1xn+x2n−1+ax2n, |
where a∈Fq satisfies TrFq/Fp(a)=1.
Lemma 2.9. ([10,Corollary 3.79,p. 127]) Let a∈Fq and let p be the characteristic of Fq, the trinomial xp−x−a is irreducible in Fq if and only if TrFq/Fp(a)≠0.
Lemma 2.10. ([10,Lemma 6.31,p. 288]) For even q, let a∈Fq with TrFq/Fp(a)=1 and b∈Fq. Then
Nq(x21+x1x2+ax22=b)=q−v(b). |
Lemma 2.11. ([10,Theorem 6.32,p. 288]) Let Fq be a finite field with q even and let b∈Fq. Then for even n, the number of solutions of the equation
x1x2+x3x4+⋯+xn−1xn=b |
in Fnq is qn−1+v(b)q(n−2)/2. For even n and a∈Fq with TrFq/Fp(a)=1, the number of solutions of the equation
x1x2+x3x4+⋯+xn−1xn+x2n−1+ax2n=b |
in Fnq is qn−1−v(b)q(n−2)/2.
Lemma 2.12. Let q=2r and h∈Fq2. Let g(Y)∈Fq2[y1,y2,…,yn] be a polynomial of the form
g(Y)=y1y2+y3y4+⋯+yn−1yn+y2n−2t−1+⋯+y2n−3+y2n−1+bty2n−2t+⋯+b1y2n−2+b0y2n, |
where bj∈F∗q2, 2|n, n>2, 0≤t≤n2−2, TrFq2/F2(bj)=1, j=0,1,…,t. Then
Nq2(g(Y)=h)=q2(n−1)+(−1)t+1qn−2v(h). | (2.1) |
Proof. We provide two proofs here. The first proof is as follows. Let q1=q2. Then by Lemmas 2.7 and 2.10, the number of solutions of g(Y)=h in Fq2 can be deduced as
Nq2(g(Y)=h)=∑c1+c2+⋯+ct+2=hNq2(y1y2+y3y4+⋯+yn−2t−3yn−2t−2=c1)⋅Nq2(yn−2t−1yn−2t+y2n−2t−1+bty2n−2t=c2)⋯Nq2(yn−1yn+y2n−1+b0y2n=ct+2)=∑c1+c2+⋯+ct+2=h(qn−2t−31+v(c1)q(n−2t−4)/21)(q1−v(c2))⋯(q1−v(ct+2))=∑c1+c2+⋯+ct+2=h(qn−2t−21+v(c1)q(n−2t−2)/21−v(c2)qn−2t−31−v(c1)v(c2)q(n−2t−4)/21)⋅(q1−v(c3))⋯(q1−v(ct+2))=∑c1+c2+⋯+ct+2=h(qn−t−21+v(c1)q(n−2)/21−v(c2)qn−t−31+⋯+(−1)t+1v(c1)v(c2)⋯v(ct+2)q(n−2t−4)/21)=qn−11+q(n−2)/21∑c1∈Fq2v(c1)+⋯+(−1)t+1∑c1+c2+⋯+ct+2=hv(c1)v(c2)⋯v(ct+2)q(n−2t−4)/21. | (2.2) |
By Lamma 2.7 and (2.2), we have
Nq2(g(Y)=h)=qn−11+(−1)t+1v(h)q(n−2)/21=q2(n−1)+(−1)t+1v(h)qn−2. |
Next we give the second proof. Note that if f and g are equivalent, then for any b∈Fq2 the equation f(x1,…,xn)=b and g(x1,…,xn)=b have the same number of solutions in Fq2. So we can get the number of solutions of g(Y)=h for h∈Fq2 by means of a nonsingular linear substitution of indeterminates.
Let k(X)∈Fq2[x1,x2,x3,x4] and k(X)=x1x2+x21+Ax22+x3x4+x23+Bx24, where TrFq2/F2(A)=TrFq2/F2(B)=1. We first show that k(x) is equivalent to x1x2+x3x4.
Let x3=y1+y3 and xi=yi for i≠3, then k(X) is equivalent to y1y2+y1y4+y3y4+Ay22+y23+By24.
Let y2=z2+z4 and yi=zi for i≠2, then k(X) is equivalent to z1z2+z3z4+Az22+z23+Az24+Bz24.
Let z1=α1+Aα2 and zi=αi for i≠1, then k(X) is equivalent to α1α2+α23+α3α4+(A+B)α24.
Since TrFq2/F2(A+B)=0, we have α23+α3α4+(A+B)α24 is reducible by Lemma 2.9. Then k(X) is equivalent to x1x2+x3x4. It follows that if t is odd, then g(Y) is equivalent to x1x2+x3x4+⋯+xn−1xn, and if t is even, then g(Y) is equivalent to x1x2+x3x4+⋯+xn−1xn+x2n−1+ax2n with TrFq2/F2(a)=1. By Lemma 2.11, we get the desired result.
From (1.1), we know that the formula for the number of solutions of f(X)=0 over Fq2 is
Nq2(f(X)=0)=q2(s−1)+d1−1∑j1=1⋯ds−1∑js=1¯ψj11(a1)⋯¯ψjss(as)J0q2(ψj11,…,ψjss), |
where di=(mi,q2−1) and ψi is a multiplicative character of Fq2 of order di. Since mi|q+1, we have di=mi. Let H={(j1,…,js)∣1≤ji<mi, 1≤i≤s}. It follows that ψj11⋯ψjss is nontrivial for any (j1,…,js)∈H as (mi,mj)=1. By Lemma 2, we have J0q2(ψj11,…,ψjss)=0 and hence Nq2(f(X)=0)=q2(s−1).
Let Nq2(f(X)=c) denote the number of solutions of the equation f(X)=c over Fq2 with c∈F∗q2. Let V={(j1,…,js)|0≤ji<mi,1≤i≤s}. Then
Nq2(f(X)=c)=∑γ1+⋯+γs=cNq2(a1xm11=γ1)⋯Nq2(asxmss=γs)=∑γ1+⋯+γs=cm1−1∑j1=0ψj11(γ1a1)⋯ms−1∑js=0ψjss(γsas). |
Since ψi is a multiplicative character of Fq2 of order mi, we have
Nq2(f(X)=c)=∑γ1c+⋯+γsc=1∑(j1,…,js)∈Vψj11(γ1c)ψj11(ca1)⋯ψjss(γsc)ψjss(cas)=∑(j1,…,js)∈Vψj11(ca1)⋯ψjss(cas)∑γ1c+⋯+γsc=1ψj11(γ1c)⋯ψjss(γsc)=∑(j1,…,js)∈Vψj11(ca1)⋯ψjss(cas)Jq2(ψj11,…,ψjss). |
By Lemma 2.3,
Nq2(f(X)=c)=q2(s−1)+∑(j1,…,js)∈Hψj11(ca1)⋯ψjss(cas)Jq2(ψj11,…,ψjss). |
By Lemma 2.5,
Jq2(ψj11,…,ψjss)=Gq2(ψj11)⋯Gq2(ψjss)Gq2(ψj11⋯ψjss). |
Since mi|q+1 and 2∤mi, by Lemma 2.6, we have
Gq2(ψj11)=⋯=Gq2(ψjss)=Gq2(ψj11⋯ψjss)=q. |
Then
Nq2(f(X)=c)=q2(s−1)+qs−1m1−1∑j1=1ψj11(ca1)…ms−1∑js=1ψjss(cas)=q2(s−1)+qs−1(m1−1∑j1=0ψj11(ca1)−1)⋯(ms−1∑js=0ψjss(cas)−1). |
It follows that
Nq2(f(X)=c)=q2(s−1)+qs−1s∏i=1((cai)mimi−1), | (3.1) |
where
(cai)mi={1,ifcai is a residue of ordermi,0,otherwise. |
For a given h∈Fq2. We discuss the two cases according to whether h is zero or not.
Case 1: h=0. If f(X)=0, then g(Y)=0; if f(X)≠0, then g(Y)≠0. Then
Nq2(l(X,Y)=0)=∑c1+c2=0Nq2(f(X)=c1)Nq2(g(Y)=c2)=q2(s−1)(q2(n−1)+(−1)t+1(q2−1)qn−2)+∑c1+c2=0c1,c2∈F∗q2Nq2(f(X)=c1)Nq2(g(Y)=c2). | (3.2) |
By Lemma 2.12, (3.1) and (3.2), we have
Nq2(l(X,Y)=0)=q2(s+n−2)+(−1)t+1q2(s−1)+hn−(−1)t+1q2(s−2)+n+∑c1∈F∗q2[q2(s+n−2)−(−1)t+1q2(s−2)+n+s∏i=1((c1ai)mimi−1)(q2n+s−3−(−1)t+1qn+s−3)]=q2(s+n−2)+(−1)t+1q2(s−1)+n−(−1)t+1q2(s−2)+n+q2(s+n−1)−(−1)t+1q2(s−1)+n−q2(s+n−2)+(−1)t+1q2(s−2)+n+∑c1∈F∗q2[s∏i=1((c1ai)mimi−1)(q2n+s−3−(−1)t+1qn+s−3)]=q2(s+n−1)+∑c1∈F∗q2[s∏i=1((c1ai)mimi−1)(q2n+s−3−(−1)t+1qn+s−3)]. | (3.3) |
Case 2: h∈F∗q2. If f(X)=h, then g(Y)=0; if f(X)=0, then g(Y)=h; if f(X)∉{0,h}, then g(Y)∉{0,h}. So we have
Nq2(l(X,Y))=h)=∑c1+c2=hNq2(f(X)=c1)Nq2(g(Y)=c2)=Nq2(f(X)=0)Nq2(g(Y)=h)+Nq2(f(X)=h)Nq2(g(Y)=0)+∑c1+c2=hc1,c2∈F∗q2∖{h}Nq2(f(X)=c1)Nq2(g(Y)=c2). | (3.4) |
By Lemma 2.12, (3.1) and (3.4),
Nq2(l(X,Y)=h)=2q2(s+n−2)+(−1)t+1q2s+n−2−(−1)t+12q2s+n−4+(qs+2n−3+(−1)t+1(q2−1)qs+n−3)s∏i=1((hai)mimi−1)+∑c1∈F∗q2∖{h}[q2(s+n−2)−(−1)t+1q2s+n−4+s∏i=1((c1ai)mimi−1)(q2n+s−3−(−1)t+1qn+s−3)]. |
It follows that
Nq2(l(X,Y)=h)=2q2(s+n−2)+(−1)t+1q2s+n−2−(−1)t+12q2s+n−4+(qs+2n−3+(−1)t+1(q2−1)qs+n−3)s∏i=1((hai)mimi−1)+∑c1∈F∗q2∖{h}[q2(s+n−2)−(−1)t+1q2s+n−4+s∏i=1((c1ai)mimi−1)(q2n+s−3−(−1)t+1qn+s−3)]=q2(s+n−1)+(qs+2n−3+(−1)t+1(q2−1)qs+n−3)s∏i=1((hai)mimi−1)+∑c1∈F∗q2∖{h}[s∏i=1((c1ai)mimi−1)⋅(q2n+s−3+(−1)tqn+s−3)]. | (3.5) |
By (3.3) and (3.5), we get the desired result. The proof of Theorem 1.1 is complete.
There is a direct corollary of Theorem 1.1 and we omit its proof.
Corollary 4.1. Under the conditions of Theorem 1.1, if a1=⋯=as=h∈F∗q2, then we have
Nq2(l(X,Y)=h)=q2(s+n−1)+(qs+2n−3+(−1)t+1(q2−1)qs+n−3)s∏i=1(mi−1)+∑γ∈F∗q2∖{h}[s∏i=1((γh)mimi−1)(q2n+s−3+(−1)tqn+s−3)], |
where
(γh)mi={1,ifγhisaresidueofordermi,0,otherwise. |
Finally, we give two examples to conclude the paper.
Example 4.2. Let F210=⟨α⟩=F2[x]/(x10+x3+1) where α is a root of x10+x3+1. Suppose l(X,Y)=α33x31+x112+y23+α10y24+y1y2+y3y4. Clearly, TrF210/F2(α10)=1, m1=3, m2=11, s=2, n=4, t=0, a2=1. By Theorem 1.1, we have
N210(l(X,Y)=0)=10245+(327+323)×20=1126587102265344. |
Example 4.3. Let F212=⟨β⟩=F2[x]/(x12+x6+x4+x+1) where β is a root of x12+x6+x4+x+1. Suppose l(X,Y)=x51+x132+y23+β10y24+y1y2+y3y4. Clearly, TrF212/F2(β10)=1, m1=5, m2=13, s=2, n=4, t=0, a1=a2=1. By Corollary 1, we have
N212(l(X,Y)=1)=25×12+(647−643×4095)×48=1153132559312355328. |
This work was jointly supported by the Natural Science Foundation of Fujian Province, China under Grant No. 2022J02046, Fujian Key Laboratory of Granular Computing and Applications (Minnan Normal University), Institute of Meteorological Big Data-Digital Fujian and Fujian Key Laboratory of Data Science and Statistics.
The authors declare there is no conflicts of interest.
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