MATH 361: NUMBER THEORY ELEVENTH LECTURE

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1 MATH 361: NUMBER THEORY ELEVENTH LECTURE The subjects of this lecture are characters, Gauss sums, Jacobi sums, and counting formulas for olynomial equations over finite fields. 1. Definitions, Basic Proerties Let be an odd rime. (However, essentially everything to follow here works verbatim uon relacing by q = e.) Definition 1.1. The character grou (or dual grou) modulo is F = homomorhisms : F C } = χ : F C χ(ab) = χ(a)χ(b) for all a, b F }. The grou law on the character grou is that for all χ, λ F, the roduct χλ is given by Examles of characters are The trivial character (χλ)(a) = χ(a)λ(a) for all a F. ε : F C, ε(a) = 1 for all a F. The quadratic character ( ) : F C, a ( ) a. (Here if we change to q then the Legendre symbol becomes the Jacobi symbol.) Recall that F is cyclic of order 1. Choose a generator g of F, and let ζ 1 = e 2πi/( 1). Define χ o : F C, χ o (g n ) = ζ n 1, n = 0, 1,, 2. Note that χ o is not canonical, but deends on the choice of g. Proosition 1.2 (Basic Character Proerties). For any character χ modulo, the following roerties hold. F (1) χ(1 F ) = 1 C. (2) χ(a) 1 = 1 C for all a F. (3) χ(a 1 ) = χ(a) 1 = χ(a) for all a F, and χ is again a character. The roerties follow immediately from the facts that χ is a homomorhism and is finite. Proosition 1.3. The character grou F 1 is cyclic.

2 2 MATH 361: NUMBER THEORY ELEVENTH LECTURE Proof. Let g generate F. Then any χ F is determined by its value on g, and this value must be χ(g) = ζ 1 k for some k 0,, 2}. Thus χ = χ k o, showing that χ o generates F. Since F and F are both cyclic of order 1, they are isomorhic. But they are noncanonically isomorhic, meaning that there is no one referred way to choose the isomorhism between them. 2. An Image and a Kernel Consider a finite cyclic grou, written additively, G = Z/qZ. Let e be a ositive integer, and consider the endomorhsim x ex of G. To study its image and its kernel, let ê = gcd(e, q). Thinking in terms of ideals quickly shows that the endomorhism has as its image e + qz = me + nq + qz} = ê + qz, the unique order-q/ê subgrou of G. Consequently its kernel is the unique order-ê subgrou, q/ê + qz, and the endomorhism is ê-to-1 to its image, Note that the image, the kernel, and the multilicity deend only on ê = gcd(e, q), rather than on the original datum e. For examle, let q = 6 and let e = 10, so that ê = gcd(10, 6) = 2. The endomorhism x 10x of Z/6Z has image 0, 2, 4} and kernel 0, 3}, and these are the image and the kernel of the variant endomorhism x 2x of Z/6Z. This tells us multilicatively that the endomorhism x x 10 of (Z/7Z) has image 1, 3 2, 3 4 } = 1, 2, 4} (noting that 3 is generator modulo 7) and kernel 1, 3 3 } = 1, 6}, and these are the image and the kernel of the variant endomorhism x x 2 of (Z/7Z). However, this does not say that the endomorhisms x x 10 and x x 2 of (Z/7Z) are equal. 3. A Basic Counting Formula Let e be a ositive integer, and let u F. This section will use characters to count the solutions x modulo of the equation x e = u. Let the symbol N denote solution-count, N(x e = u) = x F : x e = u}. We want to exress N(x e = u) in terms of characters. The revious section has shown that the kernel and the image of the endomorhism x x e of F deend only on gcd(e, 1). So we freely assume that e 1. Since the endomorhism x x e has kernel of order e, it follows that N(x e = u) 0, e} for all u F. Consider the order-e subgrou of F, consisting of the characters χ such that χ e = ε, ε, χ ( 1)/e o, χ 2( 1)/e o,, χ o (e 1)( 1)/e }. (For examle, if e = 2 then the subgrou is ε, ( /)}.) If x e = u for some x then 1 = e = N(x e = u). χ e =ε χ(u) = χ e =ε χ(x e ) = χ e =ε χ(x) e = χ e =ε

3 MATH 361: NUMBER THEORY ELEVENTH LECTURE 3 On the other hand, if x e u for all x F 0 < R < e. Therefore, χ ( 1)/e o then u takes the form u = g Qe+R where (u) = χ ( 1)/e o (g Qe+R ) = ζ (Qe+R)( 1)/e 1 = ζ R( 1)/e 1 1, and thus the general identity (1) χ(a) = χ ( 1)/e o (a) χ(a) for any a F χ e =ε χ e =ε (because multilying by χ ( 1)/e o of F ) shows that in articular χ e =ε ermutes the characters in the order-e subgrou χ(u) = 0 = N(x e = u). Finally, extend characters modulo to all of F by defining Then And so we have in all cases, ε(0) = 1, χ(0) = 0 if χ ε. χ(0) = 1 = N(x e = 0). χ e =ε If = 1 mod e then χ e =ε χ(u) = N(x e = u) for any u F. As exlained above, the formula in the box contains the information to comute N(x e = u) for all all ositive values of e, not only for divisors e of The Orthogonality Relations Proosition 4.1. The following two relations hold. 1 if χ = ε, χ(a) = 0 if χ ε and a F χ F χ(a) = 1 if a = 1 F, 0 if a 1 F. Proof. Both identities are roved essentially as we roved identity (1) in the revious section. The first identity is clear if χ = ε. Otherwise χ(a o ) 1 for some a o F, and so (since multilying by a o ermutes F ) χ(a) = χ(a o a) = χ(a o ) χ(a), a F a F showing that the sum vanishes. The second identity is clear if a = 1 F. Otherwise χ o (a) 1 because χ o sends only 1 F to 1 C, and so (since multilying by χ o ermutes F ) a F χ(a) = (χ o χ)(a) = χ o (a) χ(a), χ F χ F χ F

4 4 MATH 361: NUMBER THEORY ELEVENTH LECTURE and again the sum vanishes. The same methods aly to additive characters ψ : F C, meaning that ψ(a+b) = ψ(a)ψ(b) for all a, b F. For examle, the character ψ(a) = e 2πia/ = ζ a for a F = Z/Z is additive. Thus we have if ψ = ε, ψ(a) = 0 if ψ ε a F and ψ F ψ(a) = if a = 0 F, 0 if a 0 F. 5. Gauss Sums Again Every character χ modulo has an associated Gauss sum, τ(χ) = t F χ(t)ζ t. Note that χ is a character of the multilicative grou F while t ζ t is a nontrivial character of the additive grou F. By orthogonality of additive characters, τ(ε) = 0. For χ ε we may sum over t F because χ(0) = 0. In this case comute, exchanging the order of summation and relacing t by tu in the inner sum at the second ste, exchanging the order of summation again at the third, and using orthogonality for the fourth, τ(χ)τ(χ) = t,u F = χ(tu 1 )ζ t+u = χ(t) u F ζ (1+t)u u,t F χ(t). χ(u 1 tu)ζ u+tu t F t F By orthogonality of characters of F, the first term is χ( 1), and by orthogonality of characters of F, the second term is 0. Thus overall, τ(χ)τ(χ) = χ( 1). Also, still for χ ε, noting for the last ste that χ( 1) = χ( 1), τ(χ) = χ(t)ζ t = χ( 1) χ( t)ζ t = χ( 1)τ(χ). t F t F This gives the third of our results, τ(ε) = 0, τ(χ)τ(χ) = χ( 1), τ(χ) =.

5 MATH 361: NUMBER THEORY ELEVENTH LECTURE 5 6. More Counting Formulas; Jacobi Sums Still working over F, we now want the solution-count N(a 1 x e1 1 + a 2x e a rx er r = b) where each a i is nonzero and each e i divides 1. We exect the solution-count to be roughly r 1 since the condition imoses one constraint on r variables from F. The following two quantities will arise in the course of calculating the solutioncount. Definition 6.1. Let χ 1,, χ r be characters modulo. The corresonding Jacobi sums are J 0 (χ 1,, χ r ) = χ 1 (u 1 ) χ r (u r ) and J(χ 1,, χ r ) = u : u i=0 u : u i=1 Recall the basic counting formula for e 1, N(x e = u) = χ e =ε χ(u). χ 1 (u 1 ) χ r (u r ). Using the basic formula, comute a srawling exression for the solution-count that we seek, N(a 1 x e1 1 + a 2x e a rx er r = b) = u 1,,u r u 1+ +u r=b = u : u i=b i=1 = N(x e1 1 = a 1 1 u 1) N(x er r = a 1 r u r ) r r N(x ei i u : u i=b i=1 χ i:χ e i i = = u : u i=b (χ 1,,χ r) each χ e i i =ε = a 1 i u i ) =ε χ i (a 1 i u i ) χ 1 (a 1 1 u 1) χ r (a 1 r u r ) χ 1 (a 1 1 ) χ r(a 1 χ: each χ e i i =ε r ) u : u i=b χ 1 (u 1 ) χ r (u r ) And now insecting the definition of the two tyes of Jacobi sum shows that the desired counting formula is N(a 1 x e1 1 + a 2x e a rx er r = b) = χ 1 (a 1 1 ) χ r(a 1 r ) χ: each χ e i i =ε J 0 (χ 1,, χ r ) if b = 0, (χ 1 χ r )(b)j(χ 1,, χ r ) if b 0.

6 6 MATH 361: NUMBER THEORY ELEVENTH LECTURE 7. A Quadratic Examle Let be an odd rime. We count the oints of the unit circle modulo, The general counting formula gives N(x 2 + y 2 = 1) = x 2 + y 2 = 1. χ 2 1 =χ2 2 =ε J(χ 1, χ 2 ). The only relevant characters are ε and ( /). Thus in fact, ( ) ( N(x 2 + y 2 = 1) = J(ε, ε) + 2J(ε, ) + J( ), ( ) ). But J(ε, ε) = (and we exect this to be the dominant term in the answer), while J(ε, ( /)) = 0 by the second orthogonality relation, and finally, ( ) ( ) J(, ) = ( ) ( ) u1 u2 = ( ) u1 (1 u 1 ). u 1+u 2=1 u 1 0,1 Since we are working with the quadratic character, we may relace the first u 1 in the numerator by u 1 1 to get ( ) ( ) J(, ) = ( u 1 ) ( ) =. In sum, N(x 2 + y 2 = 1) = u 1 0,1 ( ) 1 1 if = 1 mod 4, = + 1 if = 3 mod 4. What s secretly haening here is that the unit circle modulo world really should lie in rojective sace, where it has +1 oints for all. Deending on the quadratic character of 1 modulo (i.e., deending on mod 4), two of the oints are rojective or all of them are affine. We exlain this next. 8. A Generalization of the Quadratic Examle by Other Means Let d Z be squarefree. So in articular, d 0. The quadratic curve homogenizes to The mas and Q hom P 1, Q : x 2 dy 2 = 1 Q hom : x 2 dy 2 = z 2. P 1 Q hom, [s, t] [s 2 + dt 2, 2st, s 2 dt 2 ] [x, y, z] [x + z, y] [1, 0, 1] [0, 1] if [x, y, z] [1, 0, 1], are readily checked to be inverses rovided that 2 0 and d 0. Let 2d be rime and work over the field F. Then Q hom (F ) = P 1 (F ) = + 1.

7 MATH 361: NUMBER THEORY ELEVENTH LECTURE 7 Furthermore, all oints of P 1 (F ) ma to affine oints [,, 1] of Q hom excet for the oints [s, 1] : s 2 = d}. There are no excetional oints if (d/) = 1 and there are two excetional oints if (d/) = 1. Thus the number of affine oints is Q(F ) = 1 if (d/) = 1, + 1 if (d/) = 1 = (d/). This is the formula that we obtained by Jacobi sums for d = 1. We will establish the following table. 9. Analysis of the Jacobi Sums χ J( χ) J( χ) J 0 ( χ) J 0 ( χ) ε r 1 r 1 r 1 r 1 ( ε s, χ r s ) i χ τ(χ 1 ) τ(χ r ) i ε (r 1)/2 0 0 τ(χ 1 χ r ) i χ i = ε τ(χ 1) τ(χ r ) r/2 1 ( 1) τ(χ 1) τ(χ r ) ( 1) r/2 1 The table shows that N(a 1 x e1 1 + a 2x e a rx er r = b) r 1 M 0 r/2 1 + M 1 (r 1)/2 if b 0, M 0 ( 1) r/2 1 if b = 0, where there are e i 1 ossibilities for each χ i, and M 0 = χ : i χ i = ε} and M 1 = χ : i χ i ε}. To derive the various results in the table, begin by noting that its to row is clear because both J( ε) and J 0 ( ε) sum the value 1 over r-tules u such that i u i = 1 or i u i = 0. In both cases, the first r 1 constants u i are free and then u r is determined. The second row of the table follows similarly from the second orthogonality relation. For examle, J( ε s, χ r s ) = u 2 ε(u 2 ) u r χ r (u r )ε(1 u 2 u r ) = 0.

8 8 MATH 361: NUMBER THEORY ELEVENTH LECTURE Next comute that when none of the characters is trivial, J 0 ( χ) = = u r F u r F χ 1 (u 1 ) χ r 1 (u r 1 ) χ r (u r ) u 1+ +u r 1= u r u 1+ +u r 1=1 = (χ 1 χ r 1 )( 1)J(χ 1,, χ r 1 ) = χ 1 (u 1 ) χ r 1 (u r 1 ) (χ 1 χ r 1 )( 1)(χ 1 χ r )(u r ) u r F (χ 1 χ r )(u r ) 0 if i χ i ε, ( 1)χ r ( 1)J(χ 1,, χ r 1 ) if i χ i = ε. This gives the right half of the third row, and since χ 1 χ r 1 ε it reduces the right half of the fourth row to the left half of the third row. We will return to the right half of the fourth row below. For the left half of the third row, comute that when i χ i ε (quoting the J 0 calculation just carried out), τ(χ 1 ) τ(χ r ) = χ 1 (t 1 ) χ r (t r )ζ t1+ +tr t 1,,t r F = χ 1 (t 1 ) χ r (t r )ζ u u F t: t i=u = J 0 ( χ) + J( χ) (χ 1 χ r )(u)ζ u u F = J( χ)τ(χ 1 χ r ) since the J 0 term vanishes. This establishes the left half of the third row. Also, returning to the right half of the fourth row, we have χ 1 χ r 1 ε, and so we may now quote the left half of the third row as we continue the calculation of J 0 in the nonzero case, J 0 ( χ) = ( 1)χ r ( 1)J(χ 1,, χ r 1 ) = ( 1)χ r ( 1) τ(χ 1) τ(χ r 1 ) τ(χ 1 χ r 1 ) = ( 1)χ r ( 1) τ(χ 1) τ(χ r ) τ(χ r )τ(χ r ) = ( 1) τ(χ 1) τ(χ r ). τ(χ r) τ(χ r ) Finally, for the left half of the fourth row, modify the calculation of the roduct τ(χ 1 ) τ(χ r ) to take into account the relation i χ i = ε, using the fact that now (χ 1 χ r )(u)ζ u = τ(ε) 1 = 1, u F

9 MATH 361: NUMBER THEORY ELEVENTH LECTURE 9 and using the relevant J 0 -value now that we know it, τ(χ 1 ) τ(χ r ) = J 0 ( χ) J( χ) = ( 1) τ(χ 1) τ(χ r ) From here basic algebra gives the desired value of J( χ). 10. A Cubic Examle J( χ). Let be rime. We want to count the oints of the cubic Fermat curve modulo, x 3 + y 3 = 1. If = 2 mod 3 then cubing is an automorhism modulo, and the counting roblem reduces to x + y = 1, which trivially has solutions. So from now on we assume that we are in the interesting case, = 1 mod 3. Again referring to the general counting formula, we have N(x 3 + y 3 = 1) = J(χ 1, χ 2 ). χ 3 1 =χ3 2 =ε This time the relevant characters are ε, χ, and χ, where χ(g) = ζ 3. Exand the formula, N(x 3 + y 3 = 1) = J(ε, ε) + 2(J(ε, χ) + J(ε, χ)) + 2J(χ, χ) + J(χ, χ) + J(χ, χ). According to the table, N(x 3 + y 3 = 1) = 2τ(χ)τ(χ)/ + 2Re(J(χ, χ)). We know that τ(χ)τ(χ) = χ( 1), and in fact χ( 1) = 1 because 1 is a cube and χ 3 = ε (i.e., χ( 1) = χ(( 1) 3 ) = χ 3 ( 1) = ε( 1) = 1). Therefore, N(x 3 + y 3 = 1) = 2 + 2Re(J(χ, χ)). Gauss reasoned as follows. We know that J(χ, χ) = a + bω, a, b Z, ω = ζ 3. We are looking for 2Re(J(χ, χ)) = 2a b. By the table, J(χ, χ) 2 =, i.e., a 2 ab + b 2 =. Knowing and knowing that a + bω 2 = does not uniquely determine a and b. Indeed, the six values α + βω in values ±(a + bω), ±ω(a + bω), ±ω 2 (a + bω) satisfy α + βω 2 =. But as we will see soon in our studies of the Eisenstein integer ring Z[ω], exactly one of these six values α + βω is such that α = 2 mod 3 and β = 0 mod 3. These congruences give 4 = (2α β) 2 + 3β 2 = A B 2, A = 2α β = 1 mod 3, B = b/3. Note that finding the integers A and B is an easy search because both terms of A B 2 are ositive. (By contrast, searching for α and β such that = α 2 αβ + β 2 is not so simle because of the minus sign, and searching for α and β such that = (α β/2) 2 + 3β 2 /4 involves quarter-integers.) Below we will show that the conditions 4 = A B 2, A = 1 mod 3 determine A uniquely. Returning to the Jacobi sum J(χ, χ), we want to show that among the six Eisenstein integers

10 10 MATH 361: NUMBER THEORY ELEVENTH LECTURE a + bω with = a 2 ab + b 2, the Jacobi sum is in fact the distinguished one such that a = 2 mod 3 and b = 0 mod 3. Comute, a + bω = J(χ, χ) = τ(χ)2 τ(χ) = τ(χ)3 τ(χ)τ(χ) = τ(χ)3 χ( 1) = τ(χ)3. Consider the resulting equality a + bω = τ(χ) 3, working modulo 3. On the one hand, a + bω 3 a + bω, and on the other hand, τ(χ) 3 3 t F χ(t) 3 ζ 3t = t F ζ 3t = 1, Thus a = 2 mod 3 and b = 0 mod 3, as claimed. This shows that the desired value 2Re(J(χ, χ)) = 2a b is A. We have roved Theorem 10.1 (Gauss). Let = 1 mod 3. The number of oints on the cubic Fermat curve mod is N(x 3 + y 3 = 1) = 2 + A where 4 = A B 2, A = 1 mod 3. For examle, let = 103, so that 4 = 412. The only values 27B 2 less than 4 are 27, 108, 243. Next, and are not squares but = 169 = (±13) 2. Thus A = 13. The equation x 3 + y 3 = 1 mod 103 has = 114 solutions. The homogeneous Fermat equation of degree 3 is x 3 + y 3 = z 3. For a rime 1 mod 3 the equation has + 1 rojective solutions: the affine solutions [x : y : 1] arising from the circumstance that the cubing ma is an automorhism modulo, and one non-affine solution [ 1 : 1 : 0] arising from the circumstance that 1 is the unique cube root of 1 modulo. Certainly [1 : 0 : 0] is not a solution. On the other hand, for a rime = 1 mod 3, the equation has 2 + A affine solutions according to Gauss, and also it has three non-affine solutions [x : 1 : 0] because now 1 has three cube roots modulo. Again [1 : 0 : 0] is not a solution. So a revision of Gauss s result is that counting rojectively modulo any rime (and relacing A from above by its additive inverse here), N(x 3 + y 3 = z 3 4 = A B 2, A = 2 mod 3 if = 1 mod 3, ) = + 1 A, A = 0 if 1 mod 3. The cubic Fermat curve is an ellitic curve whose conductor is 27. According to the Modularity Theorem, the value A = A in the revious dislay must also be the th Fourier coefficient of a certain modular form, a cus form whose level matches the conductor. There exists only one suitable cus form of level 27, whose Fourier coefficients can be found in tables online, and one can check that indeed each th Fourier coefficent matches the value A in the revious dislay. Finally, we discuss the fact that for = 1 mod 3, the condition 4 = A B 2, A = 1 mod 3 holds for a unique A. Soon we will see that there exists an Eisenstein integer π = a + bω with a = 2 mod 3 and b = 0 mod 3 such that = a 2 ab + b 2,

11 MATH 361: NUMBER THEORY ELEVENTH LECTURE 11 and furthermore the only other such Eisenstein integer is π = (a b) bω. Rearranging the revious dislay, = (a b/2) 2 + 3(b/2) 2, a = 2 mod 3, b = 0 mod 3. Note that (a b) ( b)/2 = a b/2, showing that the quantity a b/2 in the revious dislay deends only on, not on a choice between π and π. Multilying the dislay by 4 gives a condition of the desired form 4 = A B 2 with A = 1 mod 3. On the other hand, consider a reresentation 4 = A 2 +27B 2 with A = 1 mod 3. Set b = 3B to get 4 = A 2 + 3b 2. Note that b has the same arity as A. Now set a = (A + b)/2, so that 2a = A + b = 1 mod 3, giving a = 2 mod 3. This gives a reresentation of as in the revious dislay. Thus A = 2a b for a suitable a+bω, and we have seen that this quantity is unique to.