By Ash R.B.

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

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**Example text**

We assume the AKLB setup with A = Z, so that B is a number ring, that is, the ring of algebraic integers of a number ﬁeld L. If I is a nonzero ideal of B, we deﬁne the norm of I by N (I) = |B/I|. We will show that the norm is ﬁnite, so if P is a nonzero prime ideal of B, then B/P is a ﬁnite ﬁeld. Also, N has a multiplicative property analogous to the formula N (xy) = N (x)N (y) for elements. 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. 3), λ(B ∗ ) is a discrete subgroup of Rr1 +r2 . 1. PRELIMINARY RESULTS 3 λ(B ∗ ) is a lattice in Rs , hence a free Z-module of rank s, for some s ≤ r1 + r2 . The proof of this is outlined in the exercises. Now by the ﬁrst isomorphism theorem, λ(B ∗ ) ∼ = B ∗ /G, with λ(x) corresponding to the coset xG. If x1 G, . . , xs G form a basis for B ∗ /G and x ∈ B ∗ , then xG is a ﬁnite product of powers of the xi G, so x is an element of G times a ﬁnite product of powers of the xi . Since the λ(xi ) are linearly independent, so are the xi , provided we translate the notion of linear independence to a multiplicative setting.

But y ∈ S ⇒ 2y ∈ S, and z ∈ S ⇒ 2z ∈ S ⇒ −2z ∈ S by symmetry about the origin. Thus y − z ∈ S and since y and z are distinct, y − z ∈ H \ {0}. (b) We apply (a) to (1+1/m)S, m = 1, 2, . . Since S, hence (1+1/m)S, is a bounded set, it contains only ﬁnitely many points of the lattice H. Consequently, for every positive integer m, Sm = (1 + 1/m)S ∩ (H \ {0}) is a nonempty ﬁnite, hence compact, subset of Rn . Since Sm+1 ⊆ Sm for all m, the sets Sm form a nested sequence, and therefore ∞ ∩∞ m=1 Sm = ∅.