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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 polynomial and its possible factors are supposed to belong.
To see this, choose a monic irreducible polynomial f(X 1, ..., X n, Y) whose root generates N over E. If f(a 1, ..., a n, Y) is irreducible for some a i, then a root of it will generate the asserted N 0.) Construction of elliptic curves with large rank. [2] Hilbert's irreducibility theorem is used as a step in the Andrew Wiles proof of Fermat's ...
The fact that the polynomial after substitution is irreducible then allows concluding that the original polynomial is as well. This procedure is known as applying a shift. For example consider H = x 2 + x + 2, in which the coefficient 1 of x is not divisible by any prime, Eisenstein's criterion does not apply to H.
In abstract algebra, irreducible can be an abbreviation for irreducible element of an integral domain; for example an irreducible polynomial. In representation theory, an irreducible representation is a nontrivial representation with no nontrivial proper subrepresentations. Similarly, an irreducible module is another name for a simple module.
Cohn's irreducibility criterion is a sufficient condition for a polynomial to be irreducible in [] —that is, for it to be unfactorable into the product of lower-degree polynomials with integer coefficients.
If f(x) is irreducible, there is no lower-degree polynomial (other than the zero polynomial) that shares any root with it. For example, x 2 − 2 is irreducible over the rational numbers and has 2 {\displaystyle {\sqrt {2}}} as a root; hence there is no linear or constant polynomial over the rationals having 2 {\displaystyle {\sqrt {2}}} as a root.
For example, Swan noted in 1962 (for reasons unrelated to Hypothesis H) that the polynomial x 8 + u 3 {\displaystyle x^{8}+u^{3}\,} over the ring F 2 [ u ] is irreducible and has no fixed prime polynomial divisor (after all, its values at x = 0 and x = 1 are relatively prime polynomials) but all of its values as x runs over F 2 [ u ] are composite.
Imperfect fields cause technical difficulties because irreducible polynomials can become reducible in the algebraic closure of the base field. For example, [4] consider (,) = + [,] for an imperfect field of characteristic and a not a p-th power in k.
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