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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]
When n is itself a power of two, the multiplication operation can be nim-multiplication; alternatively, for any n, one can use multiplication of polynomials over GF(2) modulo a irreducible polynomial (as for instance for the field GF(2 8) in the description of the Advanced Encryption Standard cipher).
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). This ...
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.
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. Then, because the natural numbers less ...
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 = (+) ().
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 ...
Given an n-bit message m 0,...,m n-1, we view it as a polynomial of degree n-1 over the finite field GF(2). = + + … +We then pick a random irreducible polynomial of degree k over GF(2), and we define the fingerprint of the message m to be the remainder () after division of () by () over GF(2) which can be viewed as a polynomial of degree k − 1 or as a k-bit number.