By Robin Chapman

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Certainly P1 ⊇ I = Q1 Q2 · · · Qs . 3, P1 ⊇ Qk for some k. We reorder the Qj so that P1 ⊇ Q1 . By maximality of Q1 then P1 = Q1 . Multiplying (∗) by P1−1 gives P2 · · · Pr = Q2 · · · Qs and the result now follows from the inductive hypothesis. We can also repair an earlier omission; we can show that the ideal norm is multiplicative. 5 Let K be a number field, and let I and J be nonzero ideals of OK . Then N (IJ) = N (I)N (J). Proof If I = OK there is nothing to prove, so we can write I as a product of prime ideals.

Each symmetric function can be expressed in terms of elementary symmetric polynomials. 2 (Newton) Let R be a ring, and let f ∈ R[T1 , . . , Tn ] be a symmetric polynomial. Then there is a polynomial g ∈ R[X1 , . . , Xn ] with the property that f (T1 , . . , Tn ) = g(E1 , . . , En ) where Er denotes er (T1 , . . , Tn ). Proof (outline) Order the terms of f “lexicographically”, that is write the term aT1i1 · · · Tnin ahead of bT1j1 · · · Tnjn when ik > jk for the least k where ik = jk . Let aT1i1 · · · Tnin be the first term of f in lexicographic ordering.

More precisely R2 is the disjoint union of the sets a + F = {a + t : t ∈ F} 48 for a ∈ Λ. The set X is bounded and so meets only finitely many of these a + F. Also X = (a + F) ∩ X a∈Λ is a disjoint union. Hence area((a + F) ∩ X ) area(X ) = a∈Λ where only a finite number of terms in the sum are nonzero. Consider the set (a + F) ∩ A. Its elements are a + t where t ∈ F and t ∈ {x − a : x ∈ X } = X − a using the obvious notation. Hence (a + F) ∩ A = a + (F ∩ (X − a)) and clearly area((a + F) ∩ A) = area(F ∩ (X − a)).

### Algebraic Number Theory: summary of notes [Lecture notes] by Robin Chapman

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