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Now p does not divide all the coefficients of g(x), since it does not divide all the coefficients of f (x). Let j be the smallest value of i for which p does not divide bi . Then p divides j−1 aj − bj c0 , since aj − bj c0 = bi cj−i and bi is divisible by p when i < j. But i=0 bj c0 is not divisible by p, since p is prime and neither bj nor c0 is divisible by p. Therefore aj is not divisible by p, and hence j = n and deg g ≥ n = deg f . Thus deg g = deg f and deg h = 0. 17). 1 Field Extensions and the Tower Law Let K be a field.
There are 60 such permutations: the identity permutation, twenty 3-cycles, twenty-four 5-cycles, and fifteen permutations that are products of two disjoint transpositions. ) Now each 3-cycle in A5 generates a Sylow 3-subgroup of order 3, and these subgroups are all conjugate to one another by the Second Sylow Theorem. It follows that any normal subgroup of A5 that contains at least one 3-cycle must contain all twenty 3-cycles, and thus its order must therefore be at least 21 (since it must also contain the identity element).
We conclude therefore that r = 0, and thus j = qn. This shows that I = nZ, as required. 3 Quotient Rings and Homomorphisms Let R be a ring and let I be an ideal of R. If we regard R as an Abelian group with respect to the operation of addition, then the ideal I is a (normal) subgroup of R, and we can therefore form a corresponding quotient group R/I whose elements are the cosets of I in R. Thus an element of R/I is of the form I + x for some x ∈ R, and I + x = I + x if and only if x − x ∈ I. If 33 x, x , y and y are elements of R satisfying I + x = I + x and I + y = I + y then (x + y) − (x + y ) = (x − x ) + (y − y ), xy − x y = xy − xy + xy − x y = x(y − y ) + (x − x )y .