A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS. 1. Abstract

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1 A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS CASEY BRUCK 1. Abstract The goal of this aer is to rovide a concise way for undergraduate mathematics students to learn about how rime numbers behave in quadratic fields. This aer will rovide students with some basic number theory background required to understand the material being resented. We start with the toic of quadratic fields, number fields of degree two. This section includes some basic roerties of these fields and definitions which we will be using later on in the aer. The next section introduces the reader to rime numbers and how they are different from what is taught in earlier math courses, secifically the difference between an irreducible number and a rime number. We then move onto the majority of the discussion on rime numbers in quadratic fields and how they behave, secifically when a rime will ramify, slit, or be inert. The final section of this aer will detail an exlicit examle of a quadratic field and what haens to rime numbers within it. The secific field we choose is Q( 5) and we will be looking at what forms rimes will have to be of for each of the three ossible outcomes within the field. 2. Quadratic Fields One of the most imortant concets of algebraic number theory comes from the factorization of rimes in number fields. We want to construct Date: March 17,

2 2 CASEY BRUCK a way to observe the behavior of elements in a field extension, and while number fields in general may be a very comlicated subject beyond the scoe of this aer, we can fully analyze quadratic number fields. Definition 1. A number is said to be square free if when decomosed into a roduct of rime numbers there are no reeated factors. Definition 2. A quadratic field is a field of the form ( ) { Q d = a + b } d a, b Q where d is a square free integer. Our definition of a quadratic field does not include a way to analyze the different values of d. We introduce this through a set of olynomials that are deendent on the value of k in the congruence d k mod 4. This aears in [1, Proosition 1.2, Chater ] and we aroriate it as a definition in our aer. Definition. The minimal olynomial for a quadratic number field Q( d) is defined as x 2 d d 2, mod 4; f d (x) = x 2 x + 1 d 4 d 1 mod 4. This is imortant because the roots of the olynomial f d (x), ± d, are used to define the quadratic field Q( d). Next, we need to define the ring of integers in a quadratic field, denoted ( d ) O K, for K = Q. We use discussion from [5, Introduction] in order to come u with the definitions that follow.

3 A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS Definition 4. If α O K is an element of a quadratic field we define the ring Z[α] to be the set Z[α] = {a + b α a, b Z}. We can aly the above definition to determine how we construct the integers in a quadratic field. We define the ring of integers in a quadratic field in terms of d via [1, Proosition 2.24, Chater 2], ( d ) Definition 5. Let K = Q be a quadratic field. If d 2, mod 4, then O K = Z[ d]. If d 1 mod 4, then O K = Z[ 1+ d 2 ]. To distinguish between integers of O K and the more familiar integers in Z, we refer to the latter as rational integers. Next, we want to introduce the norm of a quadratic field. The norm is used to ma elements of a larger field into a subfield. Since quadratic fields are extensions of the rational numbers, we want to send elements of the form a + b d to Q. Definition 6. Let K be a quadratic number field. We define the norm of an element α = a + b d K as N K/Q (α) = a 2 b 2 d. Norms hel us identify elements of quadratic fields called units. These will hel us when we talk about the factorization of numbers in a field and what we describe as associates in Definition 11. Definition 7. An element u Z[x] is a unit if and only if N K/Q (u) = 1. The final iece of information we need for quadratic fields is the discriminant of the quadratic field extension. The discriminant of an algebraic number field is a value that determines the relative size of that number field.

4 4 CASEY BRUCK Discriminants are very imortant and are used in one of the comutations that can be used to find out how rime numbers act in quadratic fields, secifically we will be using the discriminant in Lemma 2 in Section.1 below. In a quadratic field extension the discriminant is defined as: Definition 8. Let K=Q( d) be a quadratic number field. We define the discriminant, K, as d, if d 1 mod 4; K = 4d, if d 2, mod 4.. Prime Numbers The first definition of rime numbers that we encounter in mathematics is: A number Z is rime if and only its only divisors are 1 and itself. This is true in the integers, but in more general rings of integers O K, this definition is actually for that of an irreducible number, not a rime number. Definition 9. We define a number n O K to be irreducible if there exist a, b O K such that n = ab imlies that a or b is a unit. If n is not irreducible, then it is said to be reducible. It just so haens that these grous of numbers are identical when dealing with the rational integers. In general, the set of all rime numbers is contained within the set of all irreducibles so all rime numbers are irreducible, but not all irreducible numbers are rime. This means that, in articular, any number that is reducible cannot be rime. One of the first results a student of number theory learns is Euclid s Lemma: Lemma 1. If a rime divides a roduct of integers ab, then must divide a or b.

5 A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS 5 We take this lemma as the definition of rime numbers in quadratic fields. Definition 10. Let a, b, O K. We say that is rime if ab imlies a or b. Let s introduce some easy examles of how secific rimes will behave in secific fields. Examle. Take = 5 in Z[ 5]. Clearly in this field we still have 1 and 5 so we can write 5 = 5 1. But, in this field, is there another way to comute the number 5? Based on revious definitions Z[ 5] = { a + b } 5 a, b Z. In order to show 5 is reducible in Z[ 5] we use Definition 9. Since N( 5) = 5, which is greater than 1, we know that 5 is not a unit and 5 is reducible therefore not rime. We need to show that 5 is rime in order to comlete showing that 5 is not rime in the field. To show that 5 is a rime in Z[ 5] we need to show it is irreducible. We do this through the norm, N( 5) = 5. We can right this as the roduct of two norms such that N 5) = N(α)N(β). This can be written as 5 = N(α)N(β) and by Euclid s Lemma, 5 N(α or 5 N(β). Since the norm mas elements to the rational numbers and 5 is rime in the rational numbers we have that 5 is a rime, as desired. Definition 11. We define two rimes of O K to be associates if they differ by a unit. That is if we have a unit, u and two rimes, 1 and 2, we consider these rimes to be associates if 1 = u 2. If we can factor a number in two ways we do not consider these to be distinct factorizations if they involve swaing out associates to achieve this

6 6 CASEY BRUCK factorization. This hels us in removing some inconsistencies in the logic behind our definitions in Definition 12. Now that we have enough rior knowledge we can move onto the main section of this aer, how rimes behave in quadratic fields..1. Behavior of Primes in Quadratic Fields. First, let us examine what to look for regarding rimes in quadratic fields. Primes behave in one of three ways when looked at in a quadratic field extension: either they slit, they ramify, or they become inert. Definition 12. Let Z be a rational rime, and let K = Q( d) be a quadratic field. (1) If there exist distinct rimes (i.e. not associates) 1, 2 O K such that = 1 2, then is slit. (2) If there exists a rime O K such that = u 2, then is ramified, where u is a unit a discussed in definition 7. () If there exists a rime O K such that =, then is inert. Examle. Now take = 2 in Z[ı], this will allow us to give an examle of when a unit must be allowed to obtain the true behavior of a rime. Then 2 = 2 1, since 2, 1 Z[ı]. Z[ı] = {a + bı a, b Z} But we also have 2 = (1 + ı)(1 ı). Then 2 factors non-trivially in Z[ı], but we will introduce some definitions later that show = 2 must be ramified. This is solved by looking at the equation 2 = ı(1 ı) 2.

7 A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS 7 If we exand the right hand side of this equation we see that it is indeed equal to 2. In this case we have ı acting as a unit for the associates ı and ı. Additionally in the last section, we introduced what f d (x) looks like for the different values of d k mod 4. Using this, we can give an alternative characterization of inert, ramified, and slit rimes according to how f d (x) behaves mod based on the discussion on [1, g. 56]. Theorem 1. A rational rime is inert in O K if and only if f d (x) is irreducible modulo ; it slits if and only if f d (x) modulo factors into two different linear olynomials; and it ramifies if and only if f d (x) modulo is the square of a linear olynomial. These relationshis allow us to introduce a lemma that allows us to easily determine the ramified rimes in a quadratic field. Lemma 2. Let K = Q( d) be a quadratic field. Then a rational rime is ramified in O K if and only if K. Proof. By Theorem 1, we want to show that f d (x) factors as the square of a linear olynomial if and only if K. First consider the case when d 2, mod 4. ( ) Let f d (x) modulo be the square of a linear olynomial modulo. Then x 2 d (x + α) 2 x 2 + 2αx + α 2 mod, by matching coefficients we have 2α 0 mod and α 2 d mod. By Lemma 1 we know that 2α imlies that 2 or α. Since we are working in rational rimes if 2 then secifically = 2. Since K = 4d if = 2 then K as desired. If α then α 2 as well so d by the congruences above, so it follows that K. So in either case of 2 or α we have that K.

8 8 CASEY BRUCK ( ) Let K Then we have d or 2. If 2 then x 2 d x mod 2 or x 2 d x 2 mod 2. If we have the second case then we have f d (x) as the square of a linear olynomial modulo 2 as desired. If we have x 2 d x mod 2 then we have x 2 d (x + 1) 2 mod 2 since if we exand this olynomial on the right we get x 2 + 2x + 1. When reduced modulo 2 we have x which again is equivalent to (x + 1) 2 mod 2 and f d (x) is the square of a linear olynomial modulo 2 as desired. If d then x 2 d x 2 mod and f d (x) is the square of a linear olynomial modulo as desired. Now consider the case when d 1 mod 4. ( ) Let f d (x) modulo be the square of a linear olynomial modulo. Then x 2 x + 1 d 4 (x + α) 2 x 2 + 2αx + α 2 mod. By matching coefficients we have 2α 1 mod and α 2 1 d 4 mod. If 2α 1 mod then we can square both sides and get 4α 2 1 mod. One of our congruences above is that α 2 1 d 4 mod. We can multile by 4 on both sides to get 4α 2 1 d mod. Since we know that 4α 2 1 mod then 1 d 1 mod so we have that d 0 mod and K as desired. ( ) Let K. Then we have d and f d (x) x 2 x (x 1 2 )2 mod. So f d (x) is the square of a linear olynomial modulo as desired. One corollary to Lemma 2 is that it imlies only finitely many rimes ramify in any quadratic field. This is because the discriminant is finite so it has finitely many rime divisors. Examle. Let K = Q ( 6 ), so O K = Z [ 6 ]. Take = 2. Then by Lemma 2 we know that 2 ramifies since K = 24 and 2 divides 24.

9 A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS 9 For = 5, x x 2 + 5x + 6 mod 5. (x + 2)(x + ) mod 5. so 5 slits in O K. In this examle we showed how two rimes act in the quadratic field Q ( 6 ), and it was fairly simle to do. The difficulty comes in when we take an arbitrary rime and need to determine how it behaves within a quadratic field. Since there are infinitely many rime numbers and thus infinitely many quadratic field extensions it is infeasible to brute force a solution for all cases. So we need a way to determine the behavior of all rimes in any secific quadratic field. This is done through the Legendre symbol. Definition 1. Let be an odd rime. We define the Legendre symbol by ( ) a = 1 if a is a non-zero square modulo ; 0 if a 0 mod ; 1 if a is not a square modulo. ( We comute the Legendre Symbol as a ) a 1 2 mod. One of the ( ) roerties this has is that if n = ab and we are looking for n then we ( ) ) ( can decomose n as its divisors and n = b ) in other words the Legendre Symbol is a multilicative function. Also if a b mod then ( ( ) a ) = b. Now that we have a way our way of determining the behavior of rimes we need to formalize a way to concetually analyze all rimes in any quadratic field. ( a

10 10 CASEY BRUCK The next determining theorem we have to use is described in [1, Theorem.6, Chater ] on age 72. Theorem 2. If,q are distinct odd rimes then ( q ) ( ) q = ( 1) ( 1) (q 1) 2 2. (1) We can rearrange this equation to be in the following form: ( ) q = ( 1) ( 1) (q 1) 2 2 ( ). q This theorem gives us a much simler way to determine the value of ( d ). Based on discussion in [2, Quadratic Recirocity] we can introduce a Proosition to deal with the secial cases of this theorem namely when = 2 and when = 1. Proosition 1. (2) ( ) 1 1 if 1 mod 4, = ( 1) 1 2 = 1 if 1 mod 4. () ( ) 2 = ( 1) if ±1 mod 8, 8 = 1 if ± mod 8. We can determine the roerties of a rime by using the Legendre symbol through a roosition that aears in [1, Proosition 1.6, Chater ]. Proosition 2. Let K = Q( d) be a quadratic number field and let be ( an odd rational rime. Then slits in K if and only if d ) = 1; it ramifies ( ( in K if and only if d ) = 0; and it is inert in K if and only if d ) = 1.

11 A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS 11 Proof. From 5, in Q( d) we have O K = Z[ d+ d 2 ] as the ring of integers. By the minimal olynomial for this ring is f d (x) = x 2 dx + d2 d 4 = (x 2 d + d 2 )(x 2 d d ). 2 Let be an odd rime. If d 0 mod then f d (x) x 2 mod, so is ramified by Theorem 1. If d n 2 mod then f d (x) (x n2 + n 2 )(x n2 n ) mod, 2 so is slit by Theorem 1 as it is the roduct of two distinct linear olynomials. If d n 2 mod then f d (x) modulo cannot be factored and f d (x) (x α)(x β), so by Theorem 1 is inert. We can also use this same quadratic field and ring of integers to deal with when we have a rime that is not odd, i.e. = 2. Proosition. Let K = Q( d) be a quadratic number field and let = 2. ( Then slits in K if and only if d ) = 1; it ramifies in K if and only if ( ( d ) = 0; and it is inert in K if and only if d ) = 1.

12 12 CASEY BRUCK Proof. We focus on d mod 8 instead of d mod 2 due to 1. If d 0 mod 8 then d 2 d 0 mod 8 so d2 d 4 0 mod 2 and f d (x) x 2 mod 2, so 2 is ramified. If d 1 mod 8 then d 2 d 0 mod 8 and f d (x) x 2 + x x(x + 1) mod 2, so 2 is slit. If d 4 mod 8 then d d 4 mod 8 and f d (x) x (x + 1) 2 mod 2, so 2 is ramified. Ifd 5 mod 8 then d 2 d 4 mod 8, and d2 d 4 1 mod 2 and f d (x) x 2 + x + 1 mod 2, so 2 is inert. ( d ) Now we can look at any Q with discriminant D = ( 1) a 2 b 1 n, in which all the i are distinct rimes. From the above discussion that the Legendre symbol is a multilicative function we have that ( ) D = ( 1 ) a ( 2 ) b ( ) 1... ( ) n. Based on our revious work we can see that this equation deends only on ( ) a ( ) mod 4 and mod 8 for 1 and 2 b. For the rest of the terms in the equation we can see that they are deendent on mod i. The combination of these relationshis allows us to come to the following conclusion.

13 A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS 1 Corollary 1. The factorization of an odd rime in Q( d) deends only on the residue mod 4d. With this corollary, we can easily describe a rime in any quadratic field by factoring 4d into the roduct of distinct rimes. Now that we have everything in terms of rime numbers, we can use quadratic recirocity to solve for each mod 4d and determine whether it slits or stays inert. Primes that ramify are the rimes which divide the discriminant and are already being accounted for. 4. An Exlicit Examle We are interested in the secific quadratic field: Q ( 5 ) and determining what rimes slit, which ramify, and which stay inert. We start with determining the discriminant of our secific fields, since 5 mod 4 then the discriminant is 20. Now we need to factor our discriminant into the roduct of distinct rimes. This is very easily done, 20 = So based on Lemma 2 we have resented all rimes will ramify and we will be looking at all other odd rimes modulo 20 which is equivalent to rimes modulo 20. All of our comutations are of the form: (4) ( ) 20 = ( 1 ) ( 2 ) 2 ( ) 5.

14 14 CASEY BRUCK Examle. Take (7) as above and =. ( ) 20 = ( 1 ) ( 2 ) 2 ( ) 5 Since 1 mod 4 then ( ) ( ) 2 1 = 1, for every, 2 is equal to 1, and we aly quadratic recirocity to our equation to the resultant equation: ( ) 20 We can simlify this equation to get = ( 1)( 1) ( ) ( ) 20 = ( 1), 5 comuting ( 5) to arrive at our final answer: ( ) 20 = ( 1)( 1) = 1. So slits in Q ( 5 ). ( ) 5 Now take = 11. ( ) ( ) ( ) ( ) 5 = ( ) ( ) = ( 1)( 1) ( ) 20 = ( 1)(1)(1) = 1 11 Since ( ) ( ) = 1 then 11 is inert in Q 5. Doing all of those comutations for each rime modulo 20 we receive the following results:

15 A CONCRETE EXAMPLE OF PRIME BEHAVIOR IN QUADRATIC FIELDS 15 mod 20 ( ) 20 = rime behavior 1 1 slits 1 slits 7 1 slits 9 1 slits 11-1 inert 1-1 inert 17-1 inert 19-1 inert This table allows u to look at all odd rimes modulo 20 and see how they will behave in Q ( 5 ). We see that 15 is not on the table and this because there are no rimes that are 15 modulo 20 since they will always be divisible by 5. Now we have shown comutations for comuting the rime behavior in a secific quadratic field we can aroriate this for any quadratic field to discover the behavior of rimes in that field. References [1] T. Weston, Algebraic Number Theory htts:// mmwood/748fall2016/weston.df [2] S. Mack-Crane, Prime Slitting in Quadratic Fields, t. 2 htts://math.berkeley.edu/ sander/blog/rime-slitting-in-quadratic-fields-art-ii [] S. Mack-Crane, Prime Slitting in Quadratic Fields, t. 1 htts://math.berkeley.edu/ sander/blog/rime-slitting-in-quadratic-fields-art-i [4] D. Tall and I. Stewart, Algebraic Number Theory and Fermat s Last Theorem, Third Edition [5] K. Conrad, Factoring in Quadratic Fields htt:// kconrad/blurbs/gradnumthy/quadraticgrad.df