Please use this identifier to cite or link to this item: http://hdl.handle.net/123456789/3273
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dc.contributor.authorKhanduja, S.K.-
dc.date.accessioned2020-12-21T07:05:30Z-
dc.date.available2020-12-21T07:05:30Z-
dc.date.issued2020-
dc.identifier.citationJournal of Pure and Applied Algebra, 224(7)en_US
dc.identifier.otherhttps://doi.org/10.1016/j.jpaa.2019.106281-
dc.identifier.urihttps://www.sciencedirect.com/science/article/pii/S0022404919302944?via%3Dihub-
dc.identifier.urihttp://hdl.handle.net/123456789/3273-
dc.descriptionOnly IISERM authors are available in the record.-
dc.description.abstractLet K=Q(θ) be an algebraic number field with θ in the ring AK of algebraic integers of K having minimal polynomial f(x) over Q. For a prime number p, let ip(f) denote the highest power of p dividing the index [AK:Z[θ]]. Let f¯(x)=ϕ¯1(x)e1⋯ϕ¯r(x)er be the factorization of f(x) modulo p into a product of powers of distinct irreducible polynomials over Z/pZ with ϕi(x)∈Z[x] monic. Let the integer l≥1 and the polynomial N(x)∈Z[x] be defined by f(x)=∏i=1rϕi(x)ei+plN(x),N‾(x)≠0¯. In this paper, we prove that ip(f)≥∑i=1ruideg⁡ϕi(x), where ui is a constant defined only in terms of l,ei and the highest power of the polynomial ϕ¯i(x) dividing N‾(x). Further a class of irreducible polynomials is described for which the above inequality becomes equality. The results of the paper quickly yield the well known Dedekind criterion which gives a necessary and sufficient condition for ip(f) to be zero. In fact, these results are proved in a more general set up replacing Z by any Dedekind domain.en_US
dc.language.isoenen_US
dc.publisherElsevier B.V.en_US
dc.subjectRings of algebraic integersen_US
dc.subjectDedekind domainsen_US
dc.subjectValued fieldsen_US
dc.titleOn the index of an algebraic integer and beyonden_US
dc.typeArticleen_US
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