Download e-book for iPad: Algebraic number fields by Gerald J. Janusz

By Gerald J. Janusz

ISBN-10: 0821804294

ISBN-13: 9780821804292

The publication is directed towards scholars with a minimum historical past who are looking to study classification box thought for quantity fields. the single prerequisite for interpreting it really is a few basic Galois thought. the 1st 3 chapters lay out the mandatory history in quantity fields, such the mathematics of fields, Dedekind domain names, and valuations. the following chapters talk about category box idea for quantity fields. The concluding bankruptcy serves for instance of the thoughts brought in earlier chapters. specifically, a few fascinating calculations with quadratic fields express using the norm residue image. For the second one version the writer further a few new fabric, multiplied many proofs, and corrected blunders present in the 1st variation. the most goal, even though, is still kind of like it was once for the 1st version: to provide an exposition of the introductory fabric and the most theorems approximately classification fields of algebraic quantity fields that might require as little history practise as attainable. Janusz's booklet might be a very good textbook for a year-long path in algebraic quantity thought; the 1st 3 chapters will be appropriate for a one-semester path. it's also very compatible for self reliant research

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1 2 CHAPTER 7. CYCLOTOMIC EXTENSIONS Proof. 2), (1 − ζ ) = p = Φpr (1) = ζ ( ζ r 1−ζ )(1 − ζ) = v(1 − ζ)ϕ(p ) 1−ζ where v is a unit in Z[ζ]. The result follows. ♣ We can now give a short proof of a basic result, but remember that we are operating under the restriction that n = pr . 4 Proposition The degree of the extension Q(ζ)/Q equals the degree of the cyclotomic polynomial, namely ϕ(pr ). Therefore the cyclotomic polynomial is irreducible over Q. Proof. 3), (p) has at least e = ϕ(pr ) prime factors (not necessarily distinct) in the ring of algebraic integers of Q(ζ).

If N (I) is prime, then I is a prime ideal. Proof. Suppose I is the product of two ideals I1 and I2 . 7), N (I) = N (I1 )N (I2 ), so by hypothesis, N (I1 ) = 1 or N (I2 ) = 1. Thus either I1 or I2 is the identity element of the ideal group, namely B. Therefore, the prime factorization of I is I itself, in other words, I is a prime ideal. 9 Proposition N (I) ∈ I, in other words, I divides N (I). ] 6 CHAPTER 4. FACTORING OF PRIME IDEALS IN EXTENSIONS Proof. Let N (I) = |B/I| = r. If x ∈ B, then r(x + I) is 0 in B/I, because the order of any element of a group divides the order of the group.

Then (p) = P12 for some prime ideal P1 , and we say that p ramifies in L. We will examine all possibilities systematically. (a) Assume p is an odd prime not dividing m. Then p does not divide the discriminant, so p does not ramify. 2 2 (a1) If m is a quadratic residue mod p, then p splits. √ Say m ≡√n mod p. Then x − m factors mod p as (x + n)(x − n), so (p) = (p, n + m) (p, n − m). (a2) If m is not a quadratic residue mod p, then x2 − m cannot be the product of two linear factors, hence x2 − m is irreducible mod p and p remains prime.

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Algebraic number fields by Gerald J. Janusz

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