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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.
As 2 and 3 are coprime, the intersection of GF(4) and GF(8) in GF(64) is the prime field GF(2). The union of GF(4) and GF(8) has thus 10 elements. The remaining 54 elements of GF(64) generate GF(64) in the sense that no other subfield contains any of them. It follows that they are roots of irreducible polynomials of degree 6 over GF(2).
every element x of GF(2) satisfies x + x = 0 and therefore −x = x; this means that the characteristic of GF(2) is 2; every element x of GF(2) satisfies x 2 = x (i.e. is idempotent with respect to multiplication); this is an instance of Fermat's little theorem. GF(2) is the only field with this property (Proof: if x 2 = x, then either x = 0 or ...
Over GF(3) the polynomial x 2 + 1 is irreducible but not primitive because it divides x 4 − 1: its roots generate a cyclic group of order 4, while the multiplicative group of GF(3 2) is a cyclic group of order 8. The polynomial x 2 + 2x + 2, on the other hand, is primitive. Denote one of its roots by α.
The polynomial P = x 4 + 1 is irreducible over Q but not over any finite field. On any field extension of F 2, P = (x + 1) 4. On every other finite field, at least one of −1, 2 and −2 is a square, because the product of two non-squares is a square and so we have; If =, then = (+) ().
The monic irreducible polynomial x 8 + x 4 + x 3 + x + 1 over GF(2) is not primitive. Let λ be a root of this polynomial (in the polynomial representation this would be x), that is, λ 8 + λ 4 + λ 3 + λ + 1 = 0. Now λ 51 = 1, so λ is not a primitive element of GF(2 8) and generates a multiplicative subgroup of order 51. [5]
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 elements of GF(2 n), i.e. a finite field whose order is a power of two, are usually represented as polynomials in GF(2)[X]. Multiplication of two such field elements consists of multiplication of the corresponding polynomials, followed by a reduction with respect to some irreducible polynomial which is taken from the construction of the field.