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For example, consider a typical case for GF(2^32), p = 2, q = 2^32, f(x) = x^32 + x^22 + x^2 + x + 1, with coefficients in GF(p) == GF(2), f(x) is primitive. However say that GF(2^32) is to be mapped to a composite field GF((2^16)^2). Let p = 2, q = 2^16, f(x) = x^32 + x^22 + x^2 + x + 1, with coefficients in GF(q) == GF(2^16) (is this the same as '''F'''<sub>''q''</sub> ?) , in this case f(x) has 16 quadratic factors, any of which can be used as the basis for a composite field. For the first algorithm note that f '(x) = 1, so it would end up in an infinite loop. For the second algorithm, the coefficients of f(x) are 0 or 1, and multiplication or xor'ing of these values is not going to produce any numbers other than 0 or 1. [[User:Rcgldr|Rcgldr]] ([[User talk:Rcgldr|talk]]) 15:41, 14 July 2020 (UTC)
:If the coefficients of a polynomial belong to a field {{mvar|F}}, all the factors of the square-free factorization have their coefficients in {{mvar|F}}. This results from the fact that the main tool of the algorithms is the polynomial greatest divisor. In your example, the input polynomial is the product of distinct irreducible quadratic factors. This implies that the results of the square-free factorization and the distinct-degree factorization are both the input polynomial. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 19:15, 14 July 2020 (UTC)
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