By Tom Weston

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N and α1 , . . , αn of OK . Show that ∆(α1 , . . , αn ) = ∆(α1 , . . , αn ). 22. Let K = Q( d) be a quadratic number field with d a squarefree integer. Show that d d ≡ 1 (mod 4); ∆K = 4d d ≡ 2, 3 (mod 4). CHAPTER 3 Prime Splitting In this chapter we will investigate how to explicitly factor ideals in rings of integers of number fields. A common theme will be considering ideals of one ring in another. Specifically, we will often have the following situation: K and L will be number fields with K ⊆ L (so OK ⊆ OL ), a will be an ideal of OK , and we will consider the ideal aOL of OL generated by a.

Since a21 − b2 d ∈ Z 4 we must have b = b1 /2 where b1 in Z is also odd. Substituting this in, we find that a21 − b21 d ≡ 0 (mod 4), this being an ordinary congruence over the integers. Now, since a1 and b1 are both odd, a21 ≡ b21 ≡ 1 (mod 4). Substituting these in, we find that 1 − d ≡ a21 − b21 d ≡ 0 (mod 4), 2. ALGEBRAIC INTEGERS 41 so d≡1 (mod 4). Thus in the case that d ≡ 2, 3 (mod 4) there are no algebraic integers with a half an odd integer; if d ≡ 1 (mod 4), then there are additional integers of the form √ a1 + b1 d 2 where a1 and b1 are odd.

We have 5 = (2 + i)(2 − i) and 13 = (3 + 2i)(3 − 2i). 1. UNIQUE FACTORIZATION 31 To finish the factorization we simply have to figure out which of the prime factors of 13 divides α and whether one or both of the prime factors of 5 divide α. One finds that (2 + i)2 and 3 + 2i divide α. Up to a unit, then, the factorization of α is (1 + i) · (2 + i)2 · 7 · (3 + 2i); multiplying it out we find that the unit is i, so that −133 − 119i = i · (1 + i) · (2 + i)2 · 7 · (3 + 2i). Our analysis of factorization in Z[i] seems to have suggested some connection with primes of the form x2 + y 2 .

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Algebraic Number Theory [Lecture notes] by Tom Weston


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