Recall from the previous post (now with errata) that for every number field K, there exists a constant *m* depending on K such that every ideal of K is equivalent in the class group H to an ideal of norm <= *m*. In addition, every ideal of norm <= *m* is generated by prime ideals of norm <= *m*; hence, it’s enough to look at the behavior of prime integers less than *m* under decomposition into ideals.

If K = **Q**(SQRT(-19)), then O(K) = **Z**[0.5+0.5SQRT(-19)], with integral basis {1, 0.5+0.5SQRT(-19)}. With the identity conjugate, *m*1 = |1| + |0.5+0.5SQRT(-19)| = 1 + SQRT(5). Similarly, *m*2 = 1 + SQRT(5), so *m* = 6 + 2SQRT(5) ~= 10.42. So it’s enough to look at the behavior of 2, 3, 5, and 7.

The element 2 is prime in O(K). Let *s* = 0.5+0.5SQRT(-19), and note that *s*^2 = -4.5 + 0.5SQRT(-19) = *s* – 5. If 2 is not prime, it divides (*a* + *bs*)(*c* + *ds*) = *ac* – 5*bd* + (*ad* + *bc* + *bd*)*s*, but not *a* + *bs* or *c* + *ds*. If *b* is even, then *a* is odd, so as *ac* – 5*bd* is even, *c* is even; and as *ad* + *bc* + *bd* is even, *d* is even, which is a contradiction. Hence, *b* is odd. Similarly, *a*, *c*, and *d* must be odd. But in that case, *ad* + *bc* + *bd* is odd, which is a contradiction. So the ideal (2) is prime.

The element 3 is similarly prime. As with 2, the element 3 can’t divide *a*, *b*, *c*, or *d*. Modulo 3, we have equations *ac* – 5*bd* = 0, *ad* + *bc* + *bd* = 0. -5 = 1 mod 3, so the first equation becomes *ac* + *bd* = 0. If *a* = 1, then *c* = –*bd*, and we get *d* – *bbd* + *bd* = 0 –> *b*^2 – *b* – 1 = 0 which is impossible for any of the two possible values of *b*. If *a* = -1, then *c* = *bd*, and we get –*d* + *bbd* + *bd* = 0 –> *b*^2 + *b* – 1 = 0 which is again impossible.

Note that N(*s*) = 5, so 5 = *s***w*2(*s*) = *s**(1-*s*). The ideals (*s*) and (1-*s*) are prime since they have prime norm. Also, 7 = (1+*s*)(2-*s*), and N(1+*s*) = N(2-*s*) = 7, so the ideals they generate are prime. Therefore, all ideals of O(K) are equivalent to products of principal ideals, so that H is the trivial group, and K is a PID. In fact, it’s one of four PIDs known not to be Euclidean by any function, the other three being generated by SQRT(-43), SQRT(-67), and SQRT(-163).

If K = **Q**(SQRT(-5)), then *m* = 10.42 again, so we need to check 2, 3, 5, and 7 again. We have 5 = -SQRT(-5)^2. But the other elements don’t split so nicely. If *r* = SQRT(-5), then we have (2) = (2, 1+*r*)(2, 1+*r*), (3) = (3, 1+*r*)(3, 1-*r*), (7) = (7, 3+*r*)(7, 3-*r*). The ideal (2, 1+*r*) has a principal square, obviously. From way earlier, (3, 1+*r*) is equivalent to (2, 1+*r*). Thus, by division, so is (3, 1-*r*). Also, (7, 3+*r*)(2, 1+*r*) = (14, 7+7*r*, 6+2*r*, -2+4*r*) = (3+*r*), so by division so are (7, 3+*r*) and (7, 3-*r*). Hence H has just two elements, the class of principal ideals and the class of principal ideals times (2, 1+*r*). So H = **Z**/2**Z**. We say that *h* = 2, where *h* is the size of H.

In the next post, I’ll show how to use the fact that *h* = 2 in **Q**(SQRT(-5)) to partially recover unique factorization in the Diophantine equation *x*^2 + 5 = *y*^3.