ALGEBRAIC NUMBER THEORY PART II SOLUTIONS) 1. Eisenstein Integers Exercise 1. Let ω = 1 + 3. Verify that ω + ω + 1 = 0. Solution. We have ) 1 + 3 + 1 + 3 + 1 = 1 + 3 + 1 + 3 + 1 = 1 1 ) + 1 + 1 + 1 3 ) = 0. Exercise. The set Z[ω] = {a + bω : a, b Z} is called the ring 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 ω = 1 ω: a + bω)c + dω) = ac + adω + bcω + bdω = 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 integer. Exercise 3. For any Eisenstein integer a + bω define the norm Na + bω) = a ab + b. Prove that the norm is multiplicative. That is, for any Eisenstein integers a + bω, c + dω the equality N a + bω)c + dω)) = Na + bω)nc + dω) 1
ALGEBRAIC NUMBER THEORY PART II SOLUTIONS) holds. Verify that the norm is non-negative: Na + bω) 0 for any a, b and Na + bω) = 0 if and only if a = b = 0. Solution. We have Na + bω) = a ab + b, Nc + dω) = c cd + d. Also, from the previous exercise we know that N a + bω)c + dω)) = N ac bd) + ad + bc bd)ω) = ac bd) ac bd)ad + bc bd) + ad + bc bd) = a ab + b )c cd + d ) = Na + bω)nc + dω). To see that the norm is non-negative, we need to show that Na + bω) = a ab + b 0 for all a, b. This becomes evident once we try to use the trick called completing the square : a ab + b = a a b + b = a a b + b + ) b = a a b + b ) + b = a b ) 4 b. b b Now it suddenly becomes very clear that this number is always non-negative, because it is a sum of two non-negative quantities a b/) and 3b /4. To see for which a, b it is possible to have Na + bω) = 0, we just need to solve the equation a b ) 4 b = 0 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/) = a = 0, which means that a = 0. Exercise 3. We say that γ is an Eisenstein unit 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 Nβγ) = N1). Then we use the multiplicative property of the norm to conclude that Nβ)Nγ) = 1. Since Nγ) 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 Nγ) = 1. Conversely, suppose that Nγ) = 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 Na+bω) = 1. This is equivalent to solving the equation a ab + b = 1 in integers. We recall that this equation is the same as a b ) 4 b = 1. Clearly, if b, 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 = ω 3 = ω ω = 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 factorization γ = αβ with α, β Z[ω] forces one of α or β to be a unit. Find Eisenstein primes among rational primes, 3, 5, 7, 11, 13. Solution. Suppose that = a + bω)c + dω) for some Eisenstein integers a + bω, c + dω. In order to prove that is prime we need show that either a + bω or c + dω is a unit. Write 4 = N) = N a + bω)c + dω)) = Na + bω)nc + dω). Since Na + bω) is an integer dividing 4, it must be equal to either 1, 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 Na + bω) =. For this purpose we need to solve the equation a ab + b = in integers a and b. This equation is equivalent to a b ) 4 b =. If b 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 Na + bω), which means that Na + bω) = 1 or 4, so either a + bω
4 ALGEBRAIC NUMBER THEORY PART II SOLUTIONS) or c + dω is a unit. We conclude that 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 a b ) 4 b = 3. 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 =, d = 1, so 1 ω) + ω) = 3. We see that neither 1 ω nor + ω have norm equal to 1, which means that 3 is not an Eisenstein prime. Analogously, we can find integer solutions to equations a ab + b = 7, a ab + b = 13 and prove that neither 7 nor 13 are primes by factoring them into Eisenstein integers that are not units: 7 = 1 ω)3 + ω), 13 = 1 3ω)4 ω). Remark 1.1. As an exercise, try proving that every prime number of the form 3k +, like, 5, 11 or 17, is not an Eisenstein prime. This can be done by showing that every integer of the form a ab + b has to be either of the form 3k or of the form 3k + 1, but it can never take the form 3k +. Remark 1.. In fact, a bit more can be said about the rational prime 3: we can factor it as 3 = 1 + ω)1 ω), 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.. Failure of Unique Factorization Exercise 1. Consider the ring Z[ 5] = { a + b 5: a, b Z } along with the norm map Na + b 5) = a + 5b, which is known to be multiplicative. 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 Na + b 5)Nc + d 5) = 1. Since Na + b 5) is an integer, it must be the case that Na + b 5) = a + 5b is equal to either 1 or 1. Clearly it cannot be negative, so we need to solve the equation a + 5b = 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 a = 1, which means that a = ±1.
ALGEBRAIC NUMBER THEORY PART II SOLUTIONS) 5 Exercise. Prove that the numbers, 3, 1+ 5, 1 5 are prime in Z[ 5]. Solution. Write = a + b 5)c + d 5). Then 4 = N) = N a + b 5)c + d 5) ) = Na + b 5)Nc + d 5). Therefore Na + b 5) is either equal to 1, 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 Na + b 5) =. However, it is quite easy to see that the equation a + 5b = has no solutions in integers a and b. Therefore is prime in Z[ 5]. Analogously we can prove this statement for 3 by proving that the equation Na + 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 = N1 + 5) = N a + b 5)c + d 5) ) = Na + b 5)Nc + d 5). Therefore Na + b 5) is equal to either 1,, 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 Na + b 5) = and Na + 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 prove that the unique factorization fails in Z[ 5]. Solution. Note that 6 = 3 = 1 + 5)1 5), so we obtain two factorizations 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[ ] and Z[ 5]. Hint: the norm map is Na+b d) = a db and it is multiplicative. Convince yourself that Na+b d) = ±1 and then find one solution to the Diophantine equation you obtained. Solution. Suppose that a + b is a unit, so a + b 1. Then there exist integers c and d such that Therefore a + b )c + d ) = 1. Na + b )Nc + d ) = 1. Since Na + b ) is an integer, we must have Na + b ) = ±1. Thus we need to solve the equation a b = ±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 Na + b 5) = ±1, so we need to solve the equation a 5b = ±1 in integers a and b. By observation, a =, b = 1 is a solution.
6 ALGEBRAIC NUMBER THEORY PART II SOLUTIONS) Exercise. 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 = we have found a unit 1 +. Note that 1 + ) n 1 + ) n = 1 + ) 1 + n )) = 1. Since 1) n Z[ ] we see that 1 + ) n 1, so by definition it is a unit. To see that there are infinitely many units, just note that the sequence 1 + < 1 + ) < 1 + ) 3 <... is strictly increasing, so the numbers in it do not repeat. Analogously, we observe that + 5) n + 5) n = + 5) + n 5)) = 1. Since + 5) n Z[ 5], we conclude that + 5) n 1, so it is a unit. Further, + 5 < + 5) < + 5) 3 <... is a strictly increasing sequence, so there are infinitely many units.