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

By Robin Chapman

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Im are ideals and I1 · · · Im ⊆ P then Ik ⊆ J for some k. Proof Suppose, for a contradiction, that IJ ⊆ P but I ⊆ P and J ⊆ P . Then there exist β ∈ I, γ ∈ J with β ∈ / P and γ ∈ / P . But then βγ ∈ IJ, but βγ ∈ / P , since P is prime, contradicting the hypothesis IJ ⊆ P . The case of an m-term product I1 · · · Im now follows by induction. Primality is also equivalent to maximality. An ideal I of OK is maximal if I is nontrivial but the only ideals J of OK with I ⊆ J are J = I and J = OK .

2 Let K be a number field and let P be a prime ideal of OK . Then there is a fractional ideal J of K with P J = 1 . Proof We let P ∗ = {β ∈ K : βP ⊆ OK }. Then P ∗ is a fractional ideal of K, OK ⊆ P ∗ and P ⊆ P P ∗ ⊆ OK . By the maximality of the prime ideal P , either P P ∗ = P or P P ∗ = OK . We show that the latter is true, so to obtain a contradiction, suppose that P P ∗ = P . 8. Hence P ∗ ⊆ OK and we conclude that P ∗ = OK . To obtain the desired contradiction, it suffices to find an element in P ∗ but not in OK .

If f = a0 + a1 X + a2 X 2 + · · · + an X n with an = 0 then n is the degree of f , an is the leading coefficient of f and an X n is the leading term of f . The degree of f is denoted deg(f ). These concepts are not defined for the zero polynomial. A monic polynomial is one whose leading coefficient is 1. If R is an integral domain, then deg(f g) = deg(f ) + deg(g) whenever f and g are nonzero elements of R[X]. In particular f g = 0 and so R[X] is also an integral domain. 1 (The division algorithm) Let K be a field, and let f , g ∈ K[X] with g = 0.

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