By Ash R.B.

This can be a textual content for a uncomplicated direction in algebraic quantity conception.

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Extra info for A Course In Algebraic Number Theory

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2 Proposition Let b be any nonzero element of the ideal I of B, and let m = NL/Q (b) ∈ Z. Then m ∈ I and |B/mB| = mn , where n = [L : Q]. Proof. 6), m = bc where c is a product of conjugates of b. But a conjugate of an algebraic integer is an algebraic integer. ) Thus c ∈ B, and since b ∈ I, we have m ∈ I. 9), B is the direct sum of n copies of Z, hence by the ﬁrst isomorphism theorem, B/mB is the direct sum of n copies of Z/mZ. Consequently, |B/mB| = mn . 3 Corollary If I is any nonzero ideal of B, then N (I) is ﬁnite.

Let z be an element of H whose order n is the exponent of H, that is, the least common multiple of the orders of all the elements of H. Then y n = 1 for every y ∈ H, so H consists of roots of unity. Since the polynomial X n − 1 has at most n distinct roots, we have |H| ≤ n. But 1, z, . . , z n−1 are distinct elements of H, because z has order n. Thus H is cyclic. ♣ For our group G, even more is true. 6 Proposition The group G consists exactly of all the roots of unity in the ﬁeld L. Proof. 5), every element of G is a root of unity.

We claim that (2)√= (2, (1√+ m)/2) (2, (1 − m)/2). To verify this note that the right side is (2, 1 − m, 1 + √ m, (1 − m)/4). This coincides with (2) because (1 − m)/4 is an even √ integer and 1 − m + 1 + m = 2. If m ≡ 3 or 7 mod 8, then m ≡ 3 mod 4, so there is only one remaining case. (c3) Let m ≡ 5 mod 8, hence m ≡ 1 mod 4, so D = m and 2 does not ramify. Consider f (x) = x2 − x + √(1 − m)/4 over B/P , where P is any prime ideal lying over (2). The roots of f are (1 ± m)/2, so f has a root in B, hence in B/P .