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Additional info for Algebraic Number Theory [Lecture notes]
3 implies that N(α) = N(β) = p. Writing α = a + bi, this implies that p = a2 + b2 . But this is impossible, since modulo 4 all sums of two squares are congruent to 0, 1 or 2. Thus p is still prime as an element of Z[i]. Now let p be a positive prime such that p ≡ 1 (mod 4) and suppose that p does not factor in Z[i]. 2 we have that there exists a ∈ Z such that a2 ≡ −1 (mod p). Thus p divides a2 + 1 in Z. Factoring a2 + 1 as (a + i)(a − i) over Z[i], we have that p divides the product (a + i)(a − i).
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.
10. Show that a Dedekind domain is a UFD if and only if it is a PID. ) 5. 11. Compute the ring of integers in K = Q( 2, 3). (Hint: Note that 6 ∈ K. First show that α ∈ K is an algebraic integer if and only if TrK/Q(√2) (α) and NK/Q(√2) (α) are algebraic integers. Next show that the same √ √ √ is true if 2 is replaced by 3 or 6. Now consider an arbitrary element √ √ √ α=a+b 2+c 3+d 6∈K with a, b, c, d ∈ Q, and use the trace conditions to restrict a, b, c, d when α is an integer. 5. Problems on cyclotomic fields.