# Adv Alegbra HW Solutions 87 - f eld. Solution. False. (v)...

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87 5 are even; moreover, the quotients obtained after dividing each by 2 have the same parity. It follows that bb 0 R . This example can be generalized by replacing 5 by any integer D with D 1 mod 4; the ring R is a special case of the ring of integers in a quadratic number f eld of the form Q ( D ) . 3.16 Prove that the set of all C -functions is a subring of F ( R ) . Solution. Clearly the constant function 1 is a C -function; if f and g are C -functions, then f g is obviously a C -function, while fg is a C -function, by Exercise 1.42. Therefore, the set of all C -functions is a subring of F ( R ) . 3.17 True or false with reasons. (i) Every f eld is a domain. Solution. True. (ii) There is a f nite f eld with more than 10 100 elements. Solution. True. (iii) If R is a domain, then there is a unique f eld containing R . Solution. False. (iv) Every commutative ring is a subring of some
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Unformatted text preview: f eld. Solution. False. (v) The subset R = Q [ i ] = { a + bi : a , b ∈ Q } is a sub f eld of C . Solution. True. (vi) The prime f eld of Q [ i ] = { a + bi : a , b ∈ Q } is Q . Solution. True. (vii) If R = Q [ √ 2 ] , then Frac ( R ) = R . Solution. False. 3.18 (i) If R is a commutative ring, de f ne the circle operation a ◦ b by a ◦ b = a + b − ab . Prove that the circle operation is associative and that 0 ◦ a = a for all a ∈ R . Solution. It is easy to see that 0 ◦ a = a for all a ∈ R : ◦ a = + a − · a = a . Let us show that ◦ is associative. ( a ◦ b ) ◦ c = ( a + b − ab ) ◦ c = a + b − ab + c − ( a + b − ab ) c = a + b + c − ab − ac − bc + ( ab ) c ;...
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## This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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