Algebraic number theory M 3/4 P 15

Answer Sheet 1

1. (a) 1 mark. By applying the description of A(d) one computes A(10) = Z[\sqrt{10}]; A(−7) = Z[\frac{1}{2} + \frac{1}{2} \sqrt{−7}]; but −12 is not square free, so A(−12) = A(−3) = Z[\frac{1}{2} + \frac{1}{2} \sqrt{−3}].

(b) 1 mark. By computing the norms one sees that 4+\sqrt{17}, \sqrt{−1}, 2+\sqrt{5} are units, whereas 2 + \sqrt{−5} is not.

(c) 3 marks. First of all, A(−5) = Z[\sqrt{−5}]. Next, (we have checked this in the lectures) 2 and 3 are not representable as a^{2} + 5b^{2} with integers a and b, hence there are no elements in A(−5) with such norms. The norm of n + \sqrt{−5} is prime for n = 0 and 6, hence then this number is irreducible. For n = 1, 2, 3, 4 the norm is 6, 9, 14, 21, and any non-trivial factorization of these integers contains either 2 or 3, hence here again n+\sqrt{−5} is irreducible. Finally, 5+\sqrt{−5} factorizes as \sqrt{−5}(1−\sqrt{−5}), and 7+\sqrt{−5} = (1+\sqrt{−5})(2− \sqrt{−5}), where neither factor is a unit. Hence these two are reducible.

2. 4 marks.We observe that A(−2) = Z[\sqrt{−2}]. The first axiom in the definition of size function is clear since norm is a non-negative integer. To check the second axiom we have to find r and s in Z[\sqrt{−2}] such that x = sy+r with N(r) < N(y) or r = 0. For this is enough to approximate x/y by an element of Z[\sqrt{−2}], call it s, such that N(x/y − s) < 1. Take s to be such that its coordinates are the nearest integers to the coordinates of x/y. Then the absolute values of the coordinates of x/y − s are at most \frac{1}{2}, hence N(x/y − s) < \frac{1}{4} + 2\frac{1}{4} < 1.

3. 5 marks. (a) If y is even, then the LHS is congruent to 2 mod 4, then the RHS is even, then the RHS is congruent to 0 mod 4. Contradiction.

(b) If a + b\sqrt{−2} is a common divisor of y + \sqrt{−2} and y − \sqrt{−2}, it divides their sum and difference, that is, 2y and 2\sqrt{−2}. On taking norms we get a^{2} + 2b^{2}|4y^{2} and a^{2} + 2b^{2}|8. By (a) it follows that a^{2} + 2b^{2}|4, hence (a, b) = (±1, 0) or (a, b) = (±2, 0) or (a, b) = (0, ±1). The first solution corresponds to units. Since y is odd, y + \sqrt{−2} is not divisible by \sqrt{−2}, neither is it divisible by 2. Thus the second and the third solutions do not lead to divisors of y + \sqrt{−2}. Hence y + \sqrt{−2} and y − \sqrt{−2} are coprime.

(c) Comparing the coefficients at \sqrt{−2} in y + \sqrt{−2} = (c + d\sqrt{−2})^{3} one gets 1 = d(3c^{2} − 2d^{2}). The rest is immediate.

4. (a) 3 marks. If I = (3 + \sqrt{−5}, 7) ⊂ A(−5) were principal, say I = (x), then N(x)|N(3+\sqrt{−5}) = 14 and N(x)|N(7) = 49. One possibility is N(x) = 7 which is impossible since 7 = a^{2} + 5b^{2} has no solutions in integers. The only other possibility is N(x) = 1, then I = A(−5) and contains 1. To rule out this possibility we show that 1 = (3 + \sqrt{−5})(a + b\sqrt{−5}) + 7(c + d\sqrt{−5}) has no solutions in integers. Comparing the coefficients of the LHS and the RHS we obtain that 3a − 5b is congruent to 1 mod 7, and a + 3b is congruent to 0 mod 7. Then −14b is congruent to 1 mod 7, which is nonsense.

(b) 3 marks. By definition, I^{2} = (4+6\sqrt{−5}, 49, 21+7\sqrt{−5}). If an ideal is principal, it is generated by its any element of minimal norm. From the hint we try x = 2 + 3\sqrt{−5}. We have N(x) = 49, so 49 is the multiple of x, and so is 4 + 6\sqrt{−5}. Since the norm of 21 + 7\sqrt{−5} is 2.7^{3}, if we need y ∈ A(−5) such that xy = 21 + 7\sqrt{−5}, we must look for elements with norm 14 (there are 8 such elements). By trial and error, y = 3 − \sqrt{−5}. (Alternatively, xy = 21 + 7\sqrt{−5} is a system of two linear equations in two variables y_{1} and y_{2}, where y = y_{1} + y_{2}\sqrt{−5}.)