THEORY PART II (SOLUTIONS)

1. Eisenstein Exercise 1. Let √ −1 + −3 ω = . 2 Verify that ω2 + ω + 1 = 0. Solution. We have √ 2 √ √ √ −1 + −3 −1 + −3 1 + −3 −1 + −3 + + 1 = − + + 1 2 2 2 2  1 1   1 1 √ = − − + 1 + − + −3 2 2 2 2 = 0.

Exercise 2. The set Z[ω] = {a + bω : a, b ∈ Z} is called the of Eisenstein integers. Prove that it is a ring. That is, show that for any Eisenstein integers a + bω, c + dω the numbers (a + bω) + (c + dω), (a + bω) − (c + dω), (a + bω)(c + dω) are Eisenstein integers as well. Solution. For the sum and difference the calculations are easy: (a + bω) + (c + dω) = (a + c) + (b + d)ω, (a + bω) − (c + dω) = (a − c) + (b − d)ω. 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 ω2 = −1−ω: (a + bω)(c + dω) = ac + adω + bcω + bdω2 = ac + adω + bcω + bd(−1 − ω) = (ac − bd) + (ad + bc − bd)ω. 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 .

Exercise 3. For any Eisenstein integer a + bω define the norm N(a + bω) = a2 − ab + b2. Prove that the norm is multiplicative. That is, for any Eisenstein integers a + bω, c + dω the equality N ((a + bω)(c + dω)) = N(a + bω)N(c + dω) 1 2 PART II (SOLUTIONS) holds. Verify that the norm is non-negative: N(a + bω) ≥ 0 for any a, b and N(a + bω) = 0 if and only if a = b = 0. Solution. We have N(a + bω) = a2 − ab + b2,N(c + dω) = c2 − cd + d2. Also, from the previous exercise we know that N ((a + bω)(c + dω)) = N ((ac − bd) + (ad + bc − bd)ω) = (ac − bd)2 − (ac − bd)(ad + bc − bd) + (ad + bc − bd)2 = (a2 − ab + b2)(c2 − cd + d2) = N(a + bω)N(c + dω). To see that the norm is non-negative, we need to show that N(a + bω) = 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 + bω) = 0, we just need to solve the equation  b 2 3 a − + b2 = 0 2 4 in integers a and b. If |b| ≥ 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 if γ | 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 | 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 + bω. In order to show that γ is a unit we need to prove that γ | 1, i.e. there exists some Eisenstein integer c + dω such that (a + bω)(c + dω) = 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+bω) = 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 |b| ≥ 2, then the left hand side is at least 3, so it must be the case that |b| ≤ 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 − ω, −1, −ω, ω, 1, 1 + ω. 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 = ω3 = ω · ω2 = ω(−1 − ω). Since −1 − ω is an Eisenstein integer, we see that ω | 1, so it is a unit. From the above equation we can also see that −1 − ω | 1, because ω is an Eisenstein integer. Finally, we can also write the above equality as 1 = (−ω)(1 + ω), which means that −ω | 1 and 1 + ω | 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 γ = αβ with α, β ∈ Z[ω] 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 + bω)(c + dω) for some Eisenstein integers a + bω, c + dω. In order to prove that 2 is prime we need show that either a + bω or c + dω is a unit. Write 4 = N(2) = N ((a + bω)(c + dω)) = N(a + bω)N(c + dω). Since N(a + bω) 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 + dω is equal to 1, so it is a unit. Thus it remains to check the case N(a + bω) = 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 |b| ≥ 2 then the left hand side is at least 3, so it must be the case that |b| ≤ 1. We can easily verify that there are no integer a’s that correspond to b = −1, 0 or 1. Therefore N(a + bω) 6= 2, which means that N(a + bω) = 1 or 4, so either a + bω 4 ALGEBRAIC NUMBER THEORY PART II (SOLUTIONS) or c + dω 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−ω. We can now find c+dω such that (1−ω)(c+dω) = 3. We have c = 2, d = 1, so (1 − ω)(2 + ω) = 3. We see that neither 1 − ω nor 2 + ω 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 − 2ω)(3 + 2ω), 13 = (1 − 3ω)(4 + 3ω). Remark 1.1. As an exercise, try proving that every 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 + ω)(1 − ω)2, where 1 + ω is a unit. Therefore 3 is divisible by a square of an Eisenstein prime 1 − ω. 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 √  √ Z[ −5] = a + b −5: a, b ∈ Z √ 2 2 along with the norm map N(a + b −5) = a +√ 5b , which is known to be multi- plicative. Prove that ±1 are the only√ units in Z[ −5]. Solution. Suppose that a + b −5 is a unit, so it divides 1. Therefore √ √ (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 √ √ N(a + b −5)N(c + d −5) = 1. √ √ 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. Clearly it cannot be negative, so we need to solve the equation a2 + 5b2 = 1 in integers a and b. If |b| ≥ 1, then the left hand side is at least 5, so it must be the case that b = 0. Therefore a2 = 1, which means that a = ±1. ALGEBRAIC NUMBER THEORY PART II (SOLUTIONS) 5 √ √ √ Exercise 2. Prove that the√ numbers 2, 3√, 1+ −5, 1− −5 are prime in Z[ −5]. Solution. Write 2 = (a + b −5)(c + d −5). Then √ √ √ √ 4 = N(2) = N (a + b −5)(c + d −5) = N(a + b −5)N(c + d −5). √ √ Therefore N(a + b −5) is either√ equal to 1, 2 or 4. If it is 1 then a + b −5 is a unit, while√ if it is 4 then c + d −5 is a unit. It remains to consider the case N(a + b −5) = 2. However, it is quite easy to see that the equation a2 + 5b2 = 2 √ has no solutions in integers a and b. Therefore 2 is prime in Z[ −5]. Analogously√ we can prove this statement for 3 by proving that the equation N(a + b −5) = 3 has no solutions√ in integers. √ √ √ For 1 + −5 the logic is similar: we write 1 + −5 = (a + b −5)(c + d −5) and observe that √ √ √ √ √ 6 = N(1 + 5) = N (a + b −5)(c + d −5) = N(a + b −5)N(c + d −5). √ √ Therefore N(a + b −5) is equal√ to either 1, 2, 3 or 6. If it is 1 then a + b −5 is a unit√ and if it 6 then c + d −√5 is a unit. Thus we need consider the cases N(a + b −5) = 2 and N(a + b −5) = 3. However, they have already√ been considered, and we proved that they have no solutions.√ We conclude that 1 + −5 is prime and analogously we can prove that 1 − −5 is prime. √ Exercise 3. Using Exercise 2 prove that the unique factorization fails in Z[ −5]. Solution. Note that √ √ 6 = 2 · 3 = (1 + −5)(1 − −5), so we obtain two of√ a number 6, so we obtained two different prime factorizations of a number 6 in Z[ −5].

3. Rings with Infinitely Many Units √ √Exercise 1. Find at least one unit√ different from ±1 in the rings Z[ 2] and 2 2 Z[ 5]. Hint: the norm map√ is N(a+b d) = a −db and it is multiplicative. Con- vince yourself that N(a+b d) = ±1 and then find one solution to the Diophantine equation you obtained. √ √ Solution. Suppose that a + b 2 is a unit, so a + b 2 | 1. Then there exist integers c and d such that √ √ (a + b 2)(c + d 2) = 1.

Therefore √ √ N(a + b 2)N(c + d 2) = 1. √ √ Since N(a + b 2) is an integer, we must have N(a + b 2) = ±1. Thus we need to solve the equation a2 − 2b2 = ±1 in integers a and b. By√ observation, a = b√= 1 is a solution. √ Analogously, if a + b 5 is a unit in Z[ 5], then N(a + b 5) = ±1, so we need to solve the equation a2 − 5b2 = ±1 in integers a and b. By observation, a = 2, b = 1 is a solution. 6 ALGEBRAIC NUMBER THEORY PART II (SOLUTIONS) √ Exercise 2. Suppose that you have√ found a unit a + b d. Prove that for any positive integer n the number (a + b d)n is also a unit. Prove that there are infinitely many units. √ Proof. For d = 2 we have found a unit 1 + 2. Note that √ √  √ √ n (1 + 2)n(−1 + 2)n = (1 + 2)(−1 + 2) = 1. √ √ √ Since ( 2 − 1)n ∈ Z[ 2] we see that (1 + 2)n | 1, so by definition it is a unit. To see that there are infinitely many units, just note that the sequence √ √ √ 1 + 2 < (1 + 2)2 < (1 + 2)3 < . . . is strictly increasing, so the numbers in it do not repeat. Analogously, we observe that √ √  √ √ n (2 + 5)n(−2 + 5)n = (2 + 5)(−2 + 5) = 1. √ √ √ Since (−2 + 5)n ∈ Z[ 5], we conclude that (2 + 5)n | 1, so it is a unit. Further, √ √ √ 2 + 5 < (2 + 5)2 < (2 + 5)3 < . . . is a strictly increasing sequence, so there are infinitely many units.