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#1 |
"Sam"
Nov 2016
5068 Posts |
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Let K be a number field K = Q(w) where w is the root of some irreducible monic polynomial (refer to this whenever polynomial is mentioned throughout this post), and Ok its ring of integers. If e is an algebraic integer (or an element) in K, let N(e) be the norm function of any element in K.
For instance, in the ring Gaussian integers, we have elements e = a*i+b where i^2 = 1. The norm function is given by N(a*w+b) = a^2 + b^2, the corresponding number field is Q(i) and i the root of the polynomial x^2+1. For Eisenstein integers, w is a root of x^2+x+1, and elements e = a*w+b. The norm function N(e) = a^2 - a*b + b^2. This process is uneasy to go backwards for quadratic fields, for example say we have elements e of the form e = a*f+b, e has norm function N(e) = a^2 + 2*b^2, and f is a root of an unknown polynomial. Find the polynomial f is a root of. In this case, all we would do is construct a polynomial using the norm function a^2 + 2*b^2, by setting b = 1 (as an example). We have the polynomial a-->x, x^2+2, showing that f must be a root of this polynomial. Now suppose we are given a cubic number field. For instance let r be a root of x^3+2, and the number field K = Q(r). Then each element e in Ok = a*r^2+b*r+c, and e has norm function N(a*r^2+b*r+c) = 4*a^3 - 2*b^3 + c^3 + 6*a*b*c. If we were only given the norm function N(a*r^2+b*r+c) = 4*a^3 - 2*b^3 + c^3 + 6*a*b*c, given r is the root of some polynomial, and asked to find the field K that is defined by this polynomial, would this be a hard task? Since we already know that r is a root of x^3+2, try this example: r is a root of an unknown cubic polynomial, hence the field Q(r) is unknown. Elements e are of the form a*r^2+b*r+c. The norm function of an element is N(a*r^2+b*r+c) = a^3 + b^3 + c^3 + 3*a^2*c^2 + 4*a*b*c Find the associated field K such that each element e is an algebraic integer in K. |
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#2 |
Dec 2012
The Netherlands
27×13 Posts |
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Finding rings of integers is in general a tricky problem. For example, \(\mathbb{Z}[\sqrt{-3}]\) is not the ring of integers of \(\mathbb{Q}(\sqrt{-3})\).
Perhaps the best way forwards with your question is this. Let \(K=\mathbb{Q}(w)\) be a number field and \(c\in K\). Then K is a finite-dimensional vector space over \(\mathbb{Q}\) and the function \(T:K\rightarrow K\) given by \(T(x)=cx\) is linear. So, by choosing a basis for K over \(\mathbb{Q}\), you can represent T as a matrix, and the norm of c (with respect to K over \(\mathbb{Q}\)) is the determinant of that matrix. Last fiddled with by Nick on 2018-01-05 at 10:34 Reason: Fixed typo |
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