# Algebraic Number Fields by Gerald J. Janusz

By Gerald J. Janusz

The e-book is directed towards scholars with a minimum heritage who are looking to study category box idea for quantity fields. the one prerequisite for analyzing it really is a few easy Galois thought. the 1st 3 chapters lay out the required historical past in quantity fields, such the mathematics of fields, Dedekind domain names, and valuations. the subsequent chapters speak about category box conception for quantity fields. The concluding bankruptcy serves for example of the suggestions brought in prior chapters. specifically, a few fascinating calculations with quadratic fields exhibit using the norm residue image. For the second one version the writer additional a few new fabric, multiplied many proofs, and corrected error present in the 1st variation. the most aim, in spite of the fact that, is still just like it was once for the 1st variation: to provide an exposition of the introductory fabric and the most theorems approximately category fields of algebraic quantity fields that may require as little historical past practise as attainable. Janusz's booklet could be a superb textbook for a year-long direction in algebraic quantity conception; the 1st 3 chapters will be appropriate for a one-semester path. it's also very compatible for autonomous learn.

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Extra info for Algebraic Number Fields

Example text

Y k ( 0 ) NL/K(O>= n T i Y k ( 8 ) = nTi(NL/E(e)>= NE/K(NL/E(e)). 2. 5 Theorem. ,u, be a basis of L over K . , u, with the property T L / K ( u i u j ) = 6, where 6, = 1 if i = j and 6 , = 0 if i + j . PROOF. This is a standard result about vector spaces. Every K-linear function from L to K has the form x -+ ( x , y )for some unique y in L. We let ui be the element with the property x -+ ( x ,ui) is the function equal to zero at uj,,j # i and I at u i . It is easy to verify the ui give a basis of L over K .

In both of these cases it follows p must contain the discriminant ideal. Next we consider some means by which a factorization of pR' can be computed. This procedure will not cover all possible cases but is still rather general. 6 Theorem. Let R' denote the integral closure of the Dedekind ring R in a finite-dimensional extension L of the quotient field K of R . Let p be a nonzero prime ideal of R . Suppose there is an element 6 such that the integral closure of R, in L is R,[B]. LetJ'(X) be the minimal polynomial of 0 over K .

Since not all aiare zero, there is a highest 28 1 SUBRINGS OF FIELDS power of II which divides all the a,. After removing this highest power we find at least one of the as, say a , , which is not in nR,. Thus a , has an inverse in R, and XI = - l / q aixi. c i#l This contradicts the choice of the xi as a minimal set of generators. Next we prove the xi in fact give a basis for L over K. If this were not so, there would exist in L an element y such that K y n C K x i = 0. However there is some elements # 0 in R such that sy is integral over R .