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A simplified version of the LLL factorization algorithm is as follows: calculate a complex (or p-adic) root α of the polynomial () to high precision, then use the Lenstra–Lenstra–Lovász lattice basis reduction algorithm to find an approximate linear relation between 1, α, α 2, α 3, . . . with integer coefficients, which might be an ...
The next odd divisor to be tested is 7. One has 77 = 7 · 11, and thus n = 2 · 3 2 · 7 · 11. This shows that 7 is prime (easy to test directly). Continue with 11, and 7 as a first divisor candidate. As 7 2 > 11, one has finished. Thus 11 is prime, and the prime factorization is; 1386 = 2 · 3 2 · 7 · 11.
The square-free factorization of a polynomial p is a factorization = where each is either 1 or a polynomial without multiple roots, and two different do not have any common root. An efficient method to compute this factorization is Yun's algorithm .
Hensel's original lemma concerns the relation between polynomial factorization over the integers and over the integers modulo a prime number p and its powers. It can be straightforwardly extended to the case where the integers are replaced by any commutative ring, and p is replaced by any maximal ideal (indeed, the maximal ideals of have the form , where p is a prime number).
In mathematics, a matrix factorization of a polynomial is a technique for factoring irreducible polynomials with matrices. David Eisenbud proved that every multivariate real-valued polynomial p without linear terms can be written as AB = pI , where A and B are square matrices and I is the identity matrix . [ 1 ]
L = p 1 p 2 …p m = q 1 q 2 …q n . Since each prime p divides L by assumption, it must also divide one of the q factors; since each q is prime as well, it must be that p = q. Iteratively dividing by the p factors shows that each p has an equal counterpart q; the two prime factorizations are identical except for their order. The unique ...
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 entry 4+2i = −i(1+i) 2 (2+i), for example, could also be written as 4+2i= (1+i) 2 (1−2i). The entries in the table resolve this ambiguity by the following convention: the factors are primes in the right complex half plane with absolute value of the real part larger than or equal to the absolute value of the imaginary part.