algebraic integer वाक्य
उदाहरण वाक्य
मोबाइल
- From the point of view of algebraic number theory it is of interest to study " normal integral bases ", where we try to replace " L " and " K " by the rings of algebraic integers they contain.
- Every square-free integer ( different from 0 and 1 ) defines a "'quadratic integer ring "', which is the integral domain of the algebraic integers contained in \ mathbf { Q } ( \ sqrt { D } ).
- Since the square root of an algebraic integer is again an algebraic integer, it is not possible to factor any nonzero nonunit algebraic integer into a finite product of irreducible elements, which implies that \ overline { \ textbf { Z } } is not Noetherian!
- Since the square root of an algebraic integer is again an algebraic integer, it is not possible to factor any nonzero nonunit algebraic integer into a finite product of irreducible elements, which implies that \ overline { \ textbf { Z } } is not Noetherian!
- Since the square root of an algebraic integer is again an algebraic integer, it is not possible to factor any nonzero nonunit algebraic integer into a finite product of irreducible elements, which implies that \ overline { \ textbf { Z } } is not Noetherian!
- In Algebraic number theory, an "'algebraic integer "'is a complex number that is a root of some monic polynomial ( a polynomial whose leading coefficient is 1 ) with coefficients in \ mathbb { Z } ( the set of integers ).
- The special case of an integral element of greatest interest in number theory is that of complex numbers integral over "'Z "'; in this context, they are usually called algebraic integers ( e . g ., \ sqrt { 2 } ).
- This property stems from the fact that for each " n ", the sum of " n " th powers of an algebraic integer " x " and its conjugates is exactly an integer; this follows from an application of Newton's identities.
- However, if " R " is the ring of algebraic integers in an algebraic number field, or more generally a Dedekind domain, the multiplication defined above turns the set of fractional ideal classes into an abelian group, the "'ideal class group "'of " R ".
- It was shown that while rings of algebraic integers do not always have unique factorization into primes ( because they need not be principal ideal domains ), they do have the property that every proper ideal admits a unique factorization as a product of prime ideals ( that is, every ring of algebraic integers is a Dedekind domain ).
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