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fields algebraic over a perfect field. Most fields that are encountered in practice are perfect. The imperfect case arises mainly in algebraic geometry in characteristic p > 0. Every imperfect field is necessarily transcendental over its prime subfield (the minimal subfield), because the latter is
Algorithm: SFF (Square-Free Factorization) Input: A monic polynomial f in F q [x] where q = p m Output: Square-free factorization of f R ← 1 # Make w be the product (without multiplicity) of all factors of f that have # multiplicity not divisible by p c ← gcd(f, f′) w ← f/c # Step 1: Identify all factors in w i ← 1 while w ≠ 1 do y ...
As every polynomial ring over a field is a unique factorization domain, every monic polynomial over a finite field may be factored in a unique way (up to the order of the factors) into a product of irreducible monic polynomials. There are efficient algorithms for testing polynomial irreducibility and factoring polynomials over finite fields.
The map x ↦ L(x) is a linear map over any field containing F q.; The set of roots of L is an F q-vector space and is closed under the q-Frobenius map.; Conversely, if U is any F q-linear subspace of some finite field containing F q, then the polynomial that vanishes exactly on U is a linearised polynomial.
In mathematics, particularly computational algebra, Berlekamp's algorithm is a well-known method for factoring polynomials over finite fields (also known as Galois fields). The algorithm consists mainly of matrix reduction and polynomial GCD computations. It was invented by Elwyn Berlekamp in 1967.
A polynomial code of length is cyclic if and only if its generator polynomial divides Since g ( x ) {\displaystyle g(x)} is the minimal polynomial with roots α c , … , α c + d − 2 , {\displaystyle \alpha ^{c},\ldots ,\alpha ^{c+d-2},} it suffices to check that each of α c , … , α c + d − 2 {\displaystyle \alpha ^{c},\ldots ,\alpha ...
The assertion "the polynomials of degree one are irreducible" is trivially true for any field. If F is algebraically closed and p(x) is an irreducible polynomial of F[x], then it has some root a and therefore p(x) is a multiple of x − a. Since p(x) is irreducible, this means that p(x) = k(x − a), for some k ∈ F \ {0} .