ALGEBRAIC NUMBER THEORY PART II (SOLUTIONS) 1. Eisenstein Integers Exercise 1. Let p −1 + −3 ! = : 2 Verify that !2 + ! + 1 = 0. Solution. We have p 2 p p p −1 + −3 −1 + −3 1 + −3 −1 + −3 + + 1 = − + + 1 2 2 2 2 1 1 1 1 p = − − + 1 + − + −3 2 2 2 2 = 0: Exercise 2. The set Z[!] = fa + b! : a; b 2 Zg is called the ring of Eisenstein integers. Prove that it is a ring. That is, show that for any Eisenstein integers a + b!, c + d! the numbers (a + b!) + (c + d!); (a + b!) − (c + d!); (a + b!)(c + d!) are Eisenstein integers as well. Solution. For the sum and difference the calculations are easy: (a + b!) + (c + d!) = (a + c) + (b + d)!; (a + b!) − (c + d!) = (a − c) + (b − d)!: Since a; b; c; d are integers, we see that a + c; b + d; a − c; b − d are integers as well. Therefore the above numbers are Eisenstein integers. The product is a bit more tricky. Here we need to use the fact that !2 = −1−!: (a + b!)(c + d!) = ac + ad! + bc! + bd!2 = ac + ad! + bc! + bd(−1 − !) = (ac − bd) + (ad + bc − bd)!: Since a; b; c; d are integers, we see that ac − bd and ad + bc − bd are integers as well, so the above number is an Einstein integer. Exercise 3. For any Eisenstein integer a + b! define the norm N(a + b!) = a2 − ab + b2: Prove that the norm is multiplicative. That is, for any Eisenstein integers a + b!, c + d! the equality N ((a + b!)(c + d!)) = N(a + b!)N(c + d!) 1 2 ALGEBRAIC NUMBER THEORY PART II (SOLUTIONS) holds. Verify that the norm is non-negative: N(a + b!) ≥ 0 for any a; b and N(a + b!) = 0 if and only if a = b = 0. Solution. We have N(a + b!) = a2 − ab + b2;N(c + d!) = c2 − cd + d2: Also, from the previous exercise we know that N ((a + b!)(c + d!)) = N ((ac − bd) + (ad + bc − bd)!) = (ac − bd)2 − (ac − bd)(ad + bc − bd) + (ad + bc − bd)2 = (a2 − ab + b2)(c2 − cd + d2) = N(a + b!)N(c + d!): To see that the norm is non-negative, we need to show that N(a + b!) = a2 − ab + b2 ≥ 0 for all a; b. This becomes evident once we try to use the trick called \completing the square": b a2 − ab + b2 = a2 − 2a + b2 2 b b 2 b 2 = a2 − 2a + b2 + − 2 2 2 b b 2 b 2 = a2 − 2a + + b2 − 2 2 2 b 2 3 = a − + b2: 2 4 Now it suddenly becomes very clear that this number is always non-negative, be- cause it is a sum of two non-negative quantities (a − b=2)2 and 3b2=4. To see for which a; b it is possible to have N(a + b!) = 0, we just need to solve the equation b 2 3 a − + b2 = 0 2 4 in integers a and b. If jbj ≥ 1, then clearly the left hand side will be greater than zero, so it must be the case that b = 0. Therefore (a−b=2)2 = a2 = 0, which means that a = 0. Exercise 3. We say that γ is an Eisenstein unit if γ j 1. Prove that if γ is an Eisenstein unit then its norm is equal to one. Solution. Suppose that γ is a unit, so γ j 1. By definition, there exists an Eisenstein integer β such that βγ = 1. Applying the norm function on both sides of this equation, we get N(βγ) = N(1). Then we use the multiplicative property of the norm to conclude that N(β)N(γ) = 1: Since N(γ) is an integer, it has to be equal to either +1 or −1. However, in the previous exercise we established that the norm is non-negative, so N(γ) = 1. Conversely, suppose that N(γ) = 1. Write γ = a + b!. In order to show that γ is a unit we need to prove that γ j 1, i.e. there exists some Eisenstein integer c + d! such that (a + b!)(c + d!) = 1: ALGEBRAIC NUMBER THEORY PART II (SOLUTIONS) 3 Exercise 4. Find all Eisenstein integers of norm one (there are six of them) and show that all of them are Eisenstein units. Solution. Let us find all integers a; b such that N(a+b!) = 1. This is equivalent to solving the equation a2 − ab + b2 = 1 in integers. We recall that this equation is the same as b 2 3 a − + b2 = 1: 2 4 Clearly, if jbj ≥ 2, then the left hand side is at least 3, so it must be the case that jbj ≤ 1. Thus we need to check if integer solutions exist for b = −1, b = 0 or b = 1. In fact, to each of those b's correspond two values of a. The complete list of solutions is: (a; b) = (−1; −1); (−1; 0); (0; −1); (0; 1); (1; 0); (1; 1): They correspond to Eisenstein integers −1 − !, −1, −!, !, 1, 1 + !. So far, we produced a complete list of Eisenstein integers that have norm one. Now we will prove that they are all units. For 1 and −1 this is obviously the case. We have 1 = !3 = ! · !2 = !(−1 − !): Since −1 − ! is an Eisenstein integer, we see that ! j 1, so it is a unit. From the above equation we can also see that −1 − ! j 1, because ! is an Eisenstein integer. Finally, we can also write the above equality as 1 = (−!)(1 + !); which means that −! j 1 and 1 + ! j 1. Therefore all six Eisenstein integers of norm one that we have found are units. Exercise 5. An Eisenstein integer γ is prime if it is not a unit and every factorization γ = αβ with α; β 2 Z[!] forces one of α or β to be a unit. Find Eisenstein primes among rational primes 2; 3; 5; 7; 11; 13: Solution. Suppose that 2 = (a + b!)(c + d!) for some Eisenstein integers a + b!; c + d!. In order to prove that 2 is prime we need show that either a + b! or c + d! is a unit. Write 4 = N(2) = N ((a + b!)(c + d!)) = N(a + b!)N(c + d!): Since N(a + b!) is an integer dividing 4, it must be equal to either 1, 2 or 4. If it is equal to 1 then the previous exercise tells us that it is a unit. If it is equal to 4 then the norm of c + d! is equal to 1, so it is a unit. Thus it remains to check the case N(a + b!) = 2. For this purpose we need to solve the equation a2 − ab + b2 = 2 in integers a and b. This equation is equivalent to b 2 3 a − + b2 = 2: 2 4 If jbj ≥ 2 then the left hand side is at least 3, so it must be the case that jbj ≤ 1. We can easily verify that there are no integer a's that correspond to b = −1; 0 or 1. Therefore N(a + b!) 6= 2, which means that N(a + b!) = 1 or 4, so either a + b! 4 ALGEBRAIC NUMBER THEORY PART II (SOLUTIONS) or c + d! is a unit. We conclude that 2 is a Gaussian prime. Analogously, we can find out that 5 and 11 are Gaussian primes. To see that 3 is not a Gaussian prime, we need to solve the equation b 2 3 a − + b2 = 3: 2 4 This equation has six solutions, one of which is (a; b) = (1; −1), which corresponds to the Eisenstein integer 1−!. We can now find c+d! such that (1−!)(c+d!) = 3. We have c = 2, d = 1, so (1 − !)(2 + !) = 3: We see that neither 1 − ! nor 2 + ! have norm equal to 1, which means that 3 is not an Eisenstein prime. Analogously, we can find integer solutions to equations a2 − ab + b2 = 7; a2 − ab + b2 = 13 and prove that neither 7 nor 13 are primes by factoring them into Eisenstein integers that are not units: 7 = (1 − 2!)(3 + 2!); 13 = (1 − 3!)(4 + 3!): Remark 1.1. As an exercise, try proving that every prime number of the form 3k + 2, like 2; 5; 11 or 17, is not an Eisenstein prime. This can be done by showing that every integer of the form a2 − ab + b2 has to be either of the form 3k or of the form 3k + 1, but it can never take the form 3k + 2. Remark 1.2. In fact, a bit more can be said about the rational prime 3: we can factor it as 3 = (1 + !)(1 − !)2; where 1 + ! is a unit. Therefore 3 is divisible by a square of an Eisenstein prime 1 − !. This means that 3 is a ramified prime. In fact, it is the only rational prime that ramifies. 2. Failure of Unique Factorization Exercise 1. Consider the ring p p Z[ −5] = a + b −5: a; b 2 Z p 2 2 along with the norm map N(a + b −5) = a +p 5b , which is known to be multi- plicative. Prove that ±1 are the onlyp units in Z[ −5]. Solution. Suppose that a + b −5 is a unit, so it divides 1. Therefore p p (a + b −5)(c + d −5) = 1 for some integers c and d. Applying the norm function on both sides and then using its multiplicative property, we see that p p N(a + b −5)N(c + d −5) = 1: p p Since N(a + b −5) is an integer, it must be the case that N(a + b −5) = a2 + 5b2 is equal to either 1 or −1.