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HOMEWORK SET 4
DUE 12.12.2012, WEDNESDAY
Remember: a, b, etc. stand for integers! p stand ALWAYS for a prime number. O√ stands for the ring of integers of the quadratic number field Q( ∆). α, β etc. stand for elements of O√ .
(1) Show that the intersection of two distinct quadratic number fields, Q( ∆) and Q( ∆ ) is Q.
Solution 1. Recall that ∆ and ∆ have to be square-free integers. Suppose that α ∈ (Q( ∆) ∪ Q( ∆ )) \ Q.
Then α ∈ Q( ∆) implies that there are rationals a, b ∈ Q so that α = a + b ∆ and similarly there are rationals c, d ∈ Q so that α = c + d ∆ . Then we have a + b ∆ = c + d ∆ ⇔ a − c + b ∆ − d ∆ = 0.
As a − c is a rational number as well as . But b/d is a rational number, hence we must have 0, the difference b ∆ − d ∆ must be a rational number, hence must forces b to be 0 in which case we are done. So suppose d = 0. Then we that e2|∆ and f2|∆ , contradiction.
(2) For α ∈ Q( ∆), we define the norm of α to be the rational number iii. If α ∈ O√ the N(α) ∈ Z. However, show the converse to this statement is false by giving three examples of non-integers whose norm is rational integer, i.e. find three elements in the set Q( ∆) \ O√ whose Solution 2.
i. Say N(α) = 0. Write α = a + b ∆. Then N(α) = αα = a2 − ∆b2. Assume, to the contrary, that ao and bo be a non-zero solution to the equation a2 − ∆b2 = 0 ⇔ a2 = ∆b2 square-free; contradiction. Conversely, of α = 0 then N(α) = 00 = 0 ii. N(αβ) = αβαβ = αβαβ = (αα)(ββ) = N(α)N(β).
iii. There are two cases. If ∆ is not congruent to 1 (mod 4) then elements of O√ are of the form α = a + b ∆ b ∈ Z. Hence N(α) = a2 − ∆b2 ∈ Z. If ∆ ≡ 1 (4) then integers of Q( ∆) are of the form ∆ with a, b ∈ Z and a ≡ b (2). N(α) = 1 (a2 − ∆b2). Hence it is enough to show that a2 − ∆b2 ≡ a2 − b2 ≡ 0 (4). If a and b are even, we are done. If both are odd, then a, b are congruent to1 (mod 4).
(3) Show that each of the following numbers are primes in Q( −5): Solution 3.
i. N(3 + 2 −5) = (3 + 2 −5)(3 − 2 −5) = 9 + 5 · 4 = 29 is a prime, hence 3 + 2 −5 must be a ii. Assume that there are integers α, β ∈ O√ with 37 = αβ. Then N(37) = 37 · 37 = N(αβ) = N(α)N(β). As 37 is a prime, we must have N(α) = 37 = N(β). Now, write α = a + b −5. Then N(α) = a2 + 5b2 = 37.
But this equation has no solution in rational integers. So there are no elements of O√ having norm 37.
iii. N(1 + 2 −5) = (1 + 2 −5)(1 − 2 −5) = 21 = 3 · 7. Thus, if 1 + 2 −5 is not a prime, then there should exist two integers α and β of norm 3 and 7. However, there are no integers in Q( −5) of norm 3 because there are no solutions to the equation a2 + 5b2 = 3. Hence 1 + (4) Show that 2 and 3 are not primes in Q( 6). (In fact, they both can be written as a product of two associate Solution 4. 6 ≡ 2 (4), hence integers in Q( 6) are of the form α = a + b 6 for a, b ∈ Z. And N(α) = a2 − 6b2.
6) with both being primes as they have norm 3; hence 3 is not a prime. As for factoring 2, although there are no integers of norm 2 there are integers of norm −2, the reason being that theDiophantine equation x2 − 6y2 = 2 has no solutions(can you prove this?) whereas x2 − 6y2 = −2 does have infinitely many solutions! We can write 2 = (−1)(2 + (5) Show that there is no integer in Q( 7) of norm 3 but that 3 is not a prime in Q( 7).(We have seen in class that 2 is a prime in Q( −47) because there are no elements in O√ with norm 2.) Solution 5. Once again 7 ≡ 3 (4) integers are of the form a + b 7. Let α = a + b 7 be one. Then N(α) =
a2 − 7b2. The equation x2 − 7y2 = 3 has no solution in Z, because if it had one solution then we would have:
Here is a list of squares modulo 7: 0, 1, 4, 2, 2, 4, 1. In other words, the equation x2 ≡ 3 (7) have no solution,hence the given equation cannot have (6) Suppose that ε is a unit in Q( ∆) and any integers solutions. Therefore, O√ does not contain any element of 7 does have norm −3 hence they are primes, and the product ε is an integer(i.e. an element of O√ ). Show that Solution 6. ε is a unit implies that N(ε) = ±1. As
(N( ε))2 = 1, hence N( ε) = ±1, i.e.
ε is a unit, by Theorem 16 of our notes.
(7) Prove that O√ is not a UFD for ∆ =: i. −17 (Hint: Try to factor 18 to see that the number of factors in two decompositions may be different.) ii. −26 (Hint: Try to factor 27 to see that the number of distinct primes appearing in two decompositions Solution 7.
i. 18 = 2 · 3 · 3 = (1 + −17)(1 − −17). We have to show that each factor is a prime. 2 and 3 are primes because there are no integers of norm ±2 and ±3 in Q( −17). Indeed, N(α) = N(a + b −17) =a2 + 17b2, and the two Diophantine equations x2 + 17y2 = ±2 and x2 + 17y2 = ±3 have no solution. Since −17) = 18 = 2 · 3 · 3 and there are no integers of norm 2 or 3 in O√ is a prime. Similar argument works for 1 − the number of prime factors in decompositions may be different! −26). 3 is a prime simply because there are no integers of norm ±3 in −26 is prime by more or less the same arguments made in part i. Hence O√ We also observe that there are cases where the number of distinct factors in a prime factorization (8) Show that 2 and 3 are primes in Q( 10). (Hint: Try to compute their norms. Then reduce modulo 10.) Solution 8. Suppose 2 is not a prime and write 2 = αβ. We must have N(α) = N(a + b 10) = a2 − 10b2 = 2.
Then modulo 10 we have a2 = 2 (10). Similarly, suppose 3 is not a prime and write 3 = α β . We must have
10) = a 2 − 10b 2 = 3. Then modulo 10 we have a2 = 3 (10). Here is a list of squares modulo 10: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1; contradiction! Notes: You may write your solutions in the language you find appropriate.

Source: http://math.gsu.edu.tr/azeytin/pdfs/2012-1-371/HW4-Key.pdf