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**Sample text**

3) over the unital commutative ring R we ask the question: Which elements of S(flR) are invariant under all permutation automorphisms? Certainly the elements of R. Theorem. 1Sb) where [0/2] = {KES(fIR)12K = O}( = ([0/2]nR)S(fIR». 1Sc) is an entire ring of characteristic 1= 2 then S(fIR)Gn= R. 1Sd) Proof Trivially, S(flR) = S(fIR)G 1 = R in case n = 1. The case n = 2 was discussed already in section 1. Let n > 2 and apply induction over n. 1Sb) is contained in the left-hand side. Conversely, let YES(fIR)G n.

If it is not a zero divisor, then f is said to be separable, otherwise inseparable. 4c) of a separable monic polynomial are distinct. The converse need not be true unless we restrict our considerations to entire rings. 4a) and h(t) = t m + b,t m - , + ... Sc) i= I k= 1 is a polynomial in a l , ... , am b, , ... , bm with rational integral coefficients. It is said to be the resultant off and h. 5d) which implies that the product is separable, if and only if each factor is separable and the resultant off, h is a non-zero divisor of Ro.

We remark that it is sometimes useful also to consider power sums for kE~ <0 in the corresponding quotient ring. 2. 13a) of an element ¢ of the equation ring R[t]/fR[t] of a monic equation f(x) = 0 (3. 13 b) of degree n over R we proceed as follows. la) of degree n over the polynomial ring Rn = Ro[t I' .. l4a) /=0 contained in S(g/Rn) = RO[¢I, ... l4b) /= 1 Realizing that the powers of the n-cycle C= (1,2, ... , say 1, C, ... 15a) are fixed by the permutation automorphisms of S(gjRn). ERn[t] is a monic polynomial of degree n over Rn satisfying the equation = o.