# Answer on Abstract Algebra Question for Jasvinder Singh

Question #44360

a) Check whether S is a subring of R in the following cases:

i) R = Q, S = f a

b 2 Q b is not divisible by 3g.

ii) R is the set of all real valued functions on R and S is the set of linear combinations of the

functions fI;cosnt; sinntg where I : R!R is defined by I(x) = 1 for all x 2 R.

i) R = Q, S = f a

b 2 Q b is not divisible by 3g.

ii) R is the set of all real valued functions on R and S is the set of linear combinations of the

functions fI;cosnt; sinntg where I : R!R is defined by I(x) = 1 for all x 2 R.

Expert's answer

i) S is a subring of R since it is in fact a localization of a ring Z by the maximal ideal 3Z, and R is a field of fractions of Z. Every localization of the integral domain is a subring of its field of fractions so it is a subring.

ii) Any element of S is a finite trigonometric series, so sum o any pair of elements of S is again in S. Using trigonometric formulas for the products of sin and cos in different combinations one can see that they can be again expressed as the linear combinations of sin and cos of some other arguments. Thus any sum and a product of elements of S again belongs to S, and S becomes a subring of R.

ii) Any element of S is a finite trigonometric series, so sum o any pair of elements of S is again in S. Using trigonometric formulas for the products of sin and cos in different combinations one can see that they can be again expressed as the linear combinations of sin and cos of some other arguments. Thus any sum and a product of elements of S again belongs to S, and S becomes a subring of R.

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