By Sergey Shpectorov

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**Additional resources for Algebraic Number Theory [Lecture notes]**

**Example text**

1 with C = √12 we get that every ideal class √ √ contains an integral ideal of norm at most √12 20 = 10 < 4. Hence we need to find all ideals of norm 1, 2, and 3. Clearly, there is only one ideal of norm 1—ok itself. 5) every ideal whose norm is a prime number is prime. So we are looking at prime ideals only. We will need the following fact about prime ideals. 3 If I is a nonzero prime ideal of ok then I ∩ Z is a nonzero prime ideal of Z. Proof: Let J = I ∩ Z. If a, b ∈ Z \ J then a, b ∈ ok \ I and hence, since I is prime, we have that ab ∈ I.

If a ∈ I and a = 0 then normk/Q (a) ∈ J and normk/Q (a) = 0. √ We now continue with our example k = Q( −5). The above lemma means that, say, an ideal I of ok of norm 2 would have to contain pZ for a prime p ∈ Z. Consequently, p ∈ I and so (p) ⊆ I. 4, we compute Normk/Q ((p)) = |normk/Q (p)| = p2 . By multiplicativity, we must have that 2 divides p2 . Hence p = 2. So I contains (2). Similarly, if I is an ideal of I of norm 3 then it contains (3). Thus we simply need to determine all ideals of ok above (that is, properly containing) (2) and all ideals above (3).

Hence (3) is prime. For p = 5 we have a similar situation: ok /(5) ∼ = Z5 [x]/(x2 + 3), which is a field. Thus ok contains no ideals of norm 5 and (5) is prime. We now turn to the last possibility: ideals of norm 4. We claim that there is only one such ideal, namely, (2), and it is clearly principal. Indeed, suppose J is an ideal of ok of norm 4. If J is prime then J ∩ Z = (p) and, clearly, we must have p = 2. So J contains (2) and then we must have J = (2), as they both have the same norm 4. However, (2) is not prime, as 30 √ it is properly contained in I = (2, 1 + −13).