Mobius Functions, Legendre Symbols, and Discriminants

Mobius Functions, Legendre Symbols, and Discriminants 1 Introduction Zev Chonoles, Erick Knight, Tim Kunisky Over the integers, there are two key number-theoretic functions that take on values of 1, 1, and 0: the Mobius function and the Legendre symbol. While over Z there is no aarent connection between them, we will develo formulae connecting the two functions over finite fields. Particularly, we will first relate the two over Z for an odd rime, which is easily generalized to finite fields of characteristic for odd rimes. The case = 2 will be a secial case that requires modification of the formula as well as generalization of the method of roof. The formula in both cases looks like: µ(f) = ( 1) deg(f) ( disc(f) where µ is the Mobius function, disc(f) is the discriminant, and the fraction notation is the Legendre symbol for odd rimes, generalized to the Kronecker symbol for the case of = 2. The discriminant will be the main tool we use to show the connection between the Legendre symbol and the Mobius function, so many intermediate roositions will involve its roerties. 2 Definitions Definition Let f(t ) 0 F[T ] for some finite field F. Then, the Mobius function µ(f(t )) is defined in a similar manner as in Z: { 0 if g(t ) µ(f(t )) = 2 f(t ) for some non-constant g(t ), ( 1) r if f(t ) has r distinct monic irreducible factors. Definition For monic f(t ) F[T ] for F a field, with roots α k in an extension K of F, disc(f) is the discriminant of f. If deg f < 2, the discriminant is one. Otherwise, it is defined as disc(f) = i<j (α i α j ) 2. ) 1

This formula matches the familiar formulas for discriminants of degree two and three monic olynomials, which will be roven in Proositions 3.1 and 3.2. We will also find it useful to define another quantity, denoted δ(f), the square root of the discriminant: δ(f) = i<j (α i α j ). Note that based on indexing of the roots, the value of this may vary u to sign, and therefore δ(f) is only well-defined u to sign. Finally, we define δ (f) as a slightly different version of δ(f): δ (f) = i<j (α i + α j ). Unlike δ(f), this is invariant under reindexing of the roots, so it is welldefined. Definition For any x Z/8, the Kronecker symbol (a generalization of the Legendre symbol) of x is defined as: ( x ) 0 if x 0, 2, 4, 6 (mod 8), = 1 if x 1, 7 (mod 8), 2 1 if x 3, 5 (mod 8). 3 Discriminant Proerties and Examles Proosition 3.1. (Monic Quadratic Discriminants) If f(t ) = T 2 + bt + c, then disc(f) = b 2 4c. Proof. Let the roots of f(t ) = T 2 + bt + c be α 1 and α 2. Then, in an extension of Z/, f(t ) = (T α 1 )(T α 2 ), so α 1 + α 2 = b and α 1 α 2 = c. Thus disc(f) = (α 1 α 2 ) 2 = α 2 1 2α 1 α 2 + α 2 2 = (α 1 + α 2 ) 2 4α 1 α 2 = b 2 4c. Proosition 3.2. (Monic Deressed Cubic Discriminants) If f(t ) = T 3 + at + b, then disc(f) = 4a 3 27b 2. 2

Proof. Let α 0, α 1, α 2 be the roots of f(t ) = T 3 + at + b. Then (T α 0 )(T α 1 )(T α 2 ) = T 3 + at + b. Differentiating both sides with resect to T gives: (T α 0 )(T α 1 ) + (T α 0 )(T α 2 ) + (T α 1 )(T α 2 ) = 3T 2 + a, and evaluating the equality at α 0, α 1, and α 2 yields: (α 0 α 1 )(α 0 α 2 ) = 3α 2 0 + a, (α 1 α 0 )(α 1 α 2 ) = 3α 2 1 + a, (α 2 α 0 )(α 2 α 1 ) = 3α 2 2 + a. Their roduct will give the negative of the discriminant of T : disc(f(t )) = (3α 2 0 + a)(3α 2 1 + a)(3α 2 2 + a). By maniulating the original olynomial, we can derive a olynomial with the roots 3α 2 k + a: Multilying by 27: t 3 + at = b t 6 + 2at 4 + a 2 t 2 = b 2 ( 3t 2 ) 3 6a( 3t 2 ) 2 + 9a 2 ( 3t 2 ) + 27b 2 = 0. Letting 3x 2 = a (with a change of sign) gives an exression for the discriminant: a 3 6a 3 + 9a 3 + 27b 2 = 4a 3 + 27b 2 disc(f(t )) = 4a 3 27b 2. Proosition 3.3. If f is a degree m monic olynomial with roots α k in an extension of Z/, then disc(f) = ( 1) m(m 1)/2 i j (α i α j ). 3

Proof. Let f satisfy the hyotheses. Then disc(f) = (α i α j ) 2 = (α i α j ) (α i α j ) i<j i<j i<j = (α i α j ) (α i α j )( 1) m(m 1)/2 i<j = ( 1) m(m 1)/2 (α i α j ), i j as we have m(m 1) 2 terms being negated. Examle Some examles of discriminant evaluation: disc(t 3 T 1) = 23, disc(t 3 + 3T 4) = 540, disc(t 7 + 2T 3 + 9T 1) = 235718099287, 4 Basic Results disc(t 10 + 8T 5 + 4T ) = 110357402604273664. Proosition 4.1. Let be an odd rime, and f(t ) = T 2 + bt + c be a disc(f(t )) monic quadratic olynomial in Z/. Then µ(f(t )) = ( ), where µ is the Mobius function on (Z/)[T ] and discf(t ) = b 2 4c. Proof. If f(t + k) factors as g(t )h(t ) in (Z/)[T ], then f(t ) must factor as g(t k)h(t k), so that adding a constant to the indeterminate T reserves the factorization of f(t ) and therefore the value of µ(f(t )). Thus, µ(f(t )) = µ(f(t + k)) for all k in Z/. Since Z/ is a field, we have that b 2 Z/ and therefore ( ( µ(f(t )) = µ f T b )) = µ ( T 2 (b 2 4c) ) 2 So: (1) f(t ) is irreducible if and only if b 2 4c is a quadratic non-residue. (2) f(t ) has a reeated root if and only if b 2 4c is 0. (3) f(t ) has two distinct roots if and only if b 2 4c is a quadratic residue. By the casework, µ(f(t )) = ( b2 4c ). 4

Proosition 4.2. (Viete s Formulas) If f(t ) F[T ] is monic and slits in a field E, E F, where f(t ) = a 0 +a 1 T + +a m T m = (T α 1 ) (T α m ) in E[T ], then a k = ( 1) m k α i1 α im k. 1 i 1 < <i m k m Proof. Let f(t ) = (T α 1 )(T α 2 )... (T α m ). In the exansion, every T m 1 term has coefficient of some α k, and each α k is multilied by m 1 factors of T. Therefore, the coefficient of T m 1 is the sum of all α k : α 1 α 2 α m = (α 1 + α 2 + + α m ) = a m 1 The constant term of the olynomial is, u to sign, just the roduct of all α k, which is: ( α 1 )( α 2 ) ( α m ) = ( 1) m (α 1 α m ) = a 0 In general, the coefficient of T k is a k. When the olynomial is exanded, each instance of T k will have a coefficient of a different roduct of m k roots, and therefore the coefficient of T k, u to sign, is the sum of all ossible roducts of m k roots. Each term in the sum will also have a coefficient of ( 1) m k, so we have the exression: a k = ( 1) m k α i1 α im k. 1 i 1 < <i m k m 5 Deriving the Formula Lemma 5.1. For any monic f (Z/)[T ], (disc(f)) = disc(f), so disc(f) Z/. Proof. Note that the discriminant is a symmetric olynomial in the roots, i.e. it is unchanged by any ermutation of the roots. By Corollary 2.4, disc(f) = i<j (α i α j ) 2 = i<j (α i α j )2. But x x is an automorhism of any finite field with characteristic. Thus disc(f) = (α i α j ) 2 = disc(f). i<j Consequently, disc(f) is in Z/ by Theorem 2.1. 5

Lemma 5.2. For degree m monic irreducible π(t ) (Z/)[T ], δ(π) = ( 1) m 1 δ(π), where δ(π) is the square root of discπ as defined in Section 1. Furthermore, for 2 ( ) discπ = ( 1) m 1. Proof. Let π(t ) be a monic irreducible olynomial of degree m over (Z/)[T ] with one root, α, in an aroriate extension field. Then the roots of π(t ) are α, α, α 2,..., α m 1, so δ(π) = 0 i<j m 1 (αi α j ). Consequently, δ(π) = = 0 i<j m 1 m (α j α) j=1 = ( 1) m 1 (α i+1 α j+1 ) 1 i<j m 1 0 i<j m 1 (α i α j ) (α i α j ) = ( 1) m 1 δ(π). And, noting that ( ) discπ = 1 δ(π) Z/ δ(π) = δ(π) ( 1) m 1 = 1 yields the second result of ( discπ ) = ( 1) m 1. Lemma 5.3. (Multilicativity of Discriminant) For monic f(t ), g(t ) (Z/)[T ], disc(fg) = disc(f)disc(g)c 2 f,g, where c f,g deends on f and g. Proof. Let f, g be olynomials in (Z/)[T ] of degree m, n, resectively. Let the roots of f be α i and the roots of g be β j. Then let c f,g = i,j (α i β j ). Since the Frobenius automorhism ermutes the roots of both f and g, and c f,g is symmetric in the roots of f and g, c f,g = i,j (α i β j ) = i,j (α i β j ) = c f,g, so c f,g Z/. Therefore, disc(fg) = i<j (α i α j ) 2 i<j (β i β j ) 2 i,j (α i β j ) 2 = disc(f)disc(g)c 2 f,g. Note that, if f and g are relatively rime, then none of the α i is a β j because otherwise f and g share a factor of (x k) where k is the shared root, so c f,g 0. 6

Theorem 5.4. (Mobius/Legendre Formula) For monic f(t ) (Z/)[T ] where 2, ( ) discf µ(f(t )) = ( 1) deg f. Proof. For relatively rime monic olynomials f and g in (Z/)[T ], ( ) ( ) disc(fg) (discf)(discg)c 2 ( ) ( ) f,g discf discg = =. Since c f,g Z/ and c f,g 0. Therefore, the function f(t ) ( 1) degf ( discf ) is multilicative on relatively rime olynomials in (Z/)[T ]. And, by Problem 8, for any monic irreducible olynomial π of degree m, ( 1) m ( disc(f) ) = ( 1) m ( 1) m 1 = 1 = µ(π). For irreducible π and k 2, ( µ(π k ) = 0 and ( 1) deg(πk ) disc(π k ) ) = 0. Since f(t ) µ(f(t )) and f(t ) ( 1) degf ( discf ) are multilicative functions that agree on owers of irreducibles, they must be the same function. Remark The same roof alies in all finite fields of characteristic > 2, if we relace the Legendre symbol with an aroriate quadratic character symbol (1 at quadratic residues, 1 at quadratic non-residues, and 0 at zero). 6 Quadratic Recirocity Theorem 6.1. (Law of Quadratic Recirocity) For rimes, q 2, ( ) ( ) q = ( 1) ( 1)(q 1)/4. q Proof. Let, q be distinct odds rimes. Viewing T q 1 as a olynomial in (Z/)[T ], we will comute disc(t q 1) in two different ways, first using the Mobius function, and then from the definition of the discriminant. 7

Let f(t ) = T q 1 T 1 viewed in (Z/)[T ]. Let the order of mod q be j and let α be a root of f(t ) in some extension of Z/. Then, by Theorem 2.9, we have that j 1 i=0 (T αi ) divides f(t ). Since that olynomial is irreducible and all divisors of f are of that form, we have that f, which is of degree q 1, has all of its irreducible factors of degree j, and so the number of irreducible factors of f is q 1 j. Therefore, µ(f(t )) = ( 1) q 1 j. Going into Z q, let x be a generator of U q. Let k = gcd(ind x (), q 1). Then ( 1) k = ( q ) and k = q 1 j. Thus disc(t q 1) = ( q ). Now we evaluate the discriminant directly from its definition. First, we need to introduce some new notation. Let ζ be a rimitive q th root of unity in some extension of Z/. Then define We can then derive Q i (T ) = T q q 1 1 T ζ i = ζ i(q k) T k Q i (ζ i ) = 1 = q. k=0 q 1 i=0 q 1 disc(t q 1) = ( 1) q(q 1)/2 (ζ i ζ j ) i=0 j i q 1 = ( 1) q(q 1)/2 Q i (ζ i ) = ( 1) (q 1)/2 q q. To rove the law of quadratic recirocity, we have ( ) = disc(t q 1) q ( disc(t = ( 1) q q ) 1) ( ) ( 1) (q 1)/2 q q = ( ) q = ( 1) ( 1)(q 1)/4. But all values of quadratic characters are one or negative one. Thus ( ) ( ) q = ( 1) ( 1)(q 1)/4. q i=0 8

7 Analogous Formula in (Z/2)[T ] Remark Several times in this section, we will use the fact that certain exressions in the roots of an integral olynomial are always integers. More secifically, rational olynomial exressions that are symmetric in the roots. Polynomial exressions in the roots of an integral olynomial must be algebraic integers, since algebraic integers are closed under sum and roduct. And, exressions symmetric in the roots must be rational, because ermutation of the roots fixes Q. But, rational algebraic integers are rational integers, and therefore this class of exressions will always be integral. Lemma 7.1. (Hensel s Lemma) Let f(t ) Z/ n, where f is irreducible over Z/. Let R n = Z/ n [T ]/(f(t )). Then, if g R n [x], r R 1, g(r) = 0, g (r) 0, then there exists a unique s R n such that s r mod and g(r) = 0. Remark Hensel s Lemma allows us to lift the roots of a olynomial over Z/2 to any Z/2 k, so initially it may seem that Z/4 would satisfy our needs. However, we assert that discf 0, 1 (mod 4) for any f Z[T ], and therefore Z/4 rovides us with no information different from that in Z/2. Proof. For some f Z[T ] of degree n with roots α k, consider δ(f) 2 δ(f) 2. First, we consider the relevant roducts as indexed from 1 to some k (irrelevant of the actual ordering, simly for convenience). We denote by (α 2 i ± 2α i α i + α 2 i ) the relevant term in each roduct, when the square is exanded. This gives: δ(f) 2 δ(f) 2 = k i=1 Define the following roducts: (α 2 i + 2α i α i + α 2 i ) P i = 1 j<i Q i = i<g k k i=1 (α 2 j 2α j α j + α 2 j ) (α 2 g + 2α g α g + α 2 g ) (α 2 i 2α i α i + α 2 i ) 9

Then, δ(f) 2 δ(f) 2 = = k i=1 P i ( (α 2 i + 2α i α i + α 2 i ) (α 2 i 2α i α i + α 2 i ) ) Q i k P i (4α i α i)q i = 4 i=1 k P i (α i α i)q i So, 4 (δ(f) 2 δ(f) 2 ) 4 (δ(f) 2 discf) discf δ(f) 2 (mod 4). We already know δ(f) Z, so the discriminant is a square modulo 4. But, the only squares modulo 4 are 0 and 1, so discf 0, 1 (mod 4). Thus, our olynomial roots will be lifted to Z/8 via Hensel s Lemma, where this roblem does not occur. Theorem 7.2. For any monic f(t ) (Z/8)[T ] that is irreducible in Z/2 of degree n, there exists an automorhism φ of (Z/8)[T ]/(f(t )) that acts as an n-cycle on the elements corresonding to roots of f and fixes recisely Z/8. Proof. Let f(t ) (Z/8)[T ] satisify the hyotheses. Let α i be the roots in Z/8[T ]/(f(t )) lifted from Z/2[T ]/(f(t )) with α i T 2i mod 2, where 0 i n 1. Define φ maing Z/8[T ]/(f(t )) to itself by φ(1) = 1 and φ(t ) = α 1. Then φ is a well-defined homomorhism, as is its restriction to Z/2[T ]/(f(t )). Since it sends T to T 2, it is the Frobenius automorhism. And, since φ is a homomorhism, it sends roots of f to roots of f, and, as noted before, since it is the Frobenius automorhism, φ sends α i α i+1 and α n 1 α 0. So, φ acts as an n-cycle on the roots. We know φ(1) = 1, so φ must fix Z/8. Suose g 0 is fixed by φ. Then, the reduction of g 0 to (Z/2)[T ]/(f(t )) must be an integer in Z/2, so there exists some g 0 Z/8 where 2 (g 0 g 0 ). So, for some g 1, 2g 1 = (g g ), and g 1 is fixed by φ as well. Alying the same argument to g 2 shows that g 1 g 1 for some g 1 Z/8 is equal to 2g 2 for some g 2. Iterating again allows us to find a g 2 where 2g 3 = g 2 g 2 for some g 3. So, we have that g 0 differs from an element of Z/8 by a multile of 8, and therefore g 0 Z/8. Lemma 7.3. (Multilicativity of discriminants) If f, g (Z/8)[T ] are monic, then disc(fg) = disc(f)disc(g)c 2 f,g where c f,g Z/8 deends on f and g. Moreover, c f,g = 0, 2, 4, 6 iff f and g have a non-constant common divisor. 10 i=1

Proof. The roof goes along the same lines as in the revious case. Let f, g be olynomials in (Z/8)[T ] of degree m, n, resectively. Let the roots of f be α i and the roots of g be β j. Then let c f,g = i,j (α i β j ), just as in the revious roof. But, we now use the reviously defined automorhism φ and aly it to c f,g. The result is exactly the same since φ ermutes the roots of both olynomials, therefore it must fix c f,g so we have c f,g Z/8. Note that: disc(fg) = i<j (α i α j ) 2 i<j (β i β j ) 2 i,j (α i β j ) 2 = disc(f)disc(g)c 2 f,g when f and g are relatively rime. If they are not, c f,g = c 2 f,g = 0, but the discriminant is zero if and only if f and g share a root, so the formula goes through in that case as well. Theorem 7.4. We have an analogous but not identical formula in Z/2: ( ) discf µ(f(t )) = ( 1) deg f 2 Where the Legendre symbol is just relaced with the Kronecker Symbol. Proof. The roof for irreducibles and owers of irreducibles goes by the same exact argument that it did in the odd rime case. And, the Kronecker symbol is multilicative in the same way that the Legendre symbol is, so the multilicative argument holds here as well, so the new formula is correct. Remark The roof for general finite fields of characteristic 2 is not so simle as it was for characteristic > 2. The entirety of the argument holds, excet for a small detail in the roof of multilicativity of discriminants. We must show that among the ossible discriminants, the roduct of two non-squares is a square. This fails in general; for examle, in Z/8, 5 7 = 3, but none of 3, 5, or 7 are squares. So, we must exclude some number of terms and only look at those that can be discriminants. Proosition 7.5. Suose we are looking at F = F 2 n, where F is a finite field of degree n of characteristic 2. Let R 2 be the ring containing Z/4 that is F lifted by Hensel s Lemma. And, let R 3 be the lifting of R 2 by Hensel s Lemma, containing Z/8. Then, the subset P of R 3 comosed of all elements that reduce to a square in R 3 (i.e. ossible discriminants) contains an equal number of squares and non-squares. 11

Proof. Since F has characteristic 2, all non-zero elements of F are squares. That is, there are 2 n 1 squares in F. Now, since (x + 2y) 2 x 2 mod 4, the number of squares in R 2 is the same as the number in F, namely 2 n 1. Thus, the number of ossible discriminants in R 3 is equal to 2 2n 2 n. Now, if x 2 = y 2 in R 3, then 8 (x 2 y 2 ) or 8 (x y)(x+y) and then either 4 (x y) or 4 (x + y). But each equivalence class under x y x 2 = y 2 has size 2 n+1, so there are 2 2n 1 2 n 1 erfect squares in R 3. Thus, our collection contains all ossible discriminants, and recisely half of the elements are erfect squares. So, within this collection the multilicativity argument alies, so we can comlete the roof as we did for Z/2, and our formula holds for all fields of characteristic 2. Combining with our revious result, we have that our formula holds in all finite fields. 12