Search results
Results from the WOW.Com Content Network
Irreducible polynomial. In mathematics, an irreducible polynomial is, roughly speaking, a polynomial that cannot be factored into the product of two non-constant polynomials. The property of irreducibility depends on the nature of the coefficients that are accepted for the possible factors, that is, the ring to which the coefficients of the ...
For a concrete example one can take R = Z[i√5], p = 1 + i√5, a = 1 − i√5, q = 2, b = 3. In this example the polynomial 3 + 2X + 2X 2 (obtained by dividing the right hand side by q = 2) provides an example of the failure of the irreducibility statement (it is irreducible over R, but reducible over its field of fractions Q[i√5]).
Irreducibility (mathematics) In mathematics, the concept of irreducibility is used in several ways. A polynomial over a field may be an irreducible polynomial if it cannot be factored over that field. In abstract algebra, irreducible can be an abbreviation for irreducible element of an integral domain; for example an irreducible polynomial.
For example, the minimal polynomial (over the reals as well as over the rationals) of the complex number i is +. The cyclotomic polynomials are the minimal polynomials of the roots of unity . In linear algebra , the n × n square matrices over K form an associative K -algebra of finite dimension (as a vector space).
For example, is algebraic over the rational numbers, because it is a root of If an element x of L is algebraic over K, the monic polynomial of lowest degree that has x as a root is called the minimal polynomial of x. This minimal polynomial is irreducible over K.
In particular, is irreducible if and only if p is a primitive root modulo n, that is, p does not divide n, and its multiplicative order modulo n is (), the degree of . [ 13 ] These results are also true over the p -adic integers , since Hensel's lemma allows lifting a factorization over the field with p elements to a factorization over the p ...
is irreducible, because if it could be factored there would be a linear factor giving a rational solution, while none of the possible roots given by the rational root test are actually roots. Since its discriminant is positive, it has three real roots, so it is an example of casus irreducibilis. These roots can be expressed as
This polynomial is irreducible over the rationals and so the three cosines are conjugate algebraic numbers. Likewise, tan 3 π / 16 , tan 7 π / 16 , tan 11 π / 16 , and tan 15 π / 16 satisfy the irreducible polynomial x 4 − 4x 3 − 6x 2 + 4x + 1 = 0, and so are conjugate algebraic integers. This is the ...