Factor Rings and IdealsDate: 04/22/2003 at 16:16:28 From: Kris Subject: Factor rings and ideals I have three questions and no idea where to begin. Please help. 1. Give an example to show that a factor ring of an integral domain may be a field. 2. Let R = {a + b*sqrt2|a, b in Z} and let R prime consist of all 2x2 matrices of the form |a 2b| for all a, b in Z. |b a| Show that R and R prime are isomorphic rings. 3. Let R and R prime be rings and let phi:R to R prime be a ring homomorphism such that phi[R] is not equal to {0 prime}. Show that if R has unity 1 and R prime has no 0 divisors, then phi (1) is unity for R prime. Date: 05/06/2003 at 19:56:12 From: Doctor Nitrogen Subject: Re: Factor rings and ideals Hi, Kris: (1) Look at the factor ring : = Z/pZ where p is a prime. How many elements does it have? With some effort you will be able to convince yourself that this factor ring as defined will always be a field and has no 0 divisors. (2) You will want to define a homomorphism theta: theta: R -------------> R' (your R prime) as theta( a + b*sqrt(2) ) = |a 2b| |b a| with a + b*sqrt(2) in R, a, b in Z, and the 2 X 2 matrix in R'. Now prove that this homomorphism is well-defined; that is, if p and q are both in R you must show (A) theta(p + q) = theta(p) + theta(q), (B) theta(p*q) = theta(p)*theta(q), where theta(p), theta(q), are both in R', p, q are in R. For (A) note that, for c, d in Z: theta(a + b*sqrt(2) + c + d*sqrt(2) ) = theta( (a + c) + (b + d)*sqrt(2) ) = |(a + c) 2(b + d)| |(b + d) (a + c) | = |a 2b| + |c 2d| |b a| + |d c| = theta(a + b*sqrt(2) + theta(c + d*sqrt(2) ). And also, for (B): theta( (a + b*sqrt(2))*(c + d*sqrt(2) ) ) = theta( (ac + 2bd) + (ad + bc)*sqrt(2) ) = |ac + 2bd 2(ad + bc)| |bc + ad 2bd + ac | = | a 2b | * | c 2d | | b a | * | d c | = theta(a + b*sqrt(2))*theta(c + d*sqrt(2)). To show this map is onto, suppose there is an element in R', say | r 2s | | s r | which is NOT the image of at least one element in R. Then there is no element in R that can be mapped to that element above, which is in R'. But there exist integers r, s in Z such that r + s*sqrt(2) will be in R. But this element in R must get mapped to R' by theta, a contradiction, meaning every element in R' must be the image of at least one element in R. To prove theta is 1-1, try to convince yourself the only element in R that gets mapped to the 0 element in R' is 0 + 0*sqrt(2) = 0, that is, that 0 = 0 + 0*sqrt(2) -------------> | 0 2(0) | | 0 0 | = 0_2 (the 2 X 2 0 matrix). Then since the map theta really is well defined, is both 1-1 and onto, it must be an isomorphism from R to R'. (3) You might want to let r be an element of R, and then look at phi(r) = phi(1*r) = phi(1)*phi(r). I hope this helped answer the questions you had concerning your mathematics problem. You are welcome to return to The Math Forum/Doctor Math whenever you have any math-related questions. - Doctor Nitrogen, The Math Forum http://mathforum.org/dr.math/ |
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