Question

A least common multiple of two elements a, b ER, where R is a com- mutative ring, is an element me R such that • am and 6 m, and • if a and | , then m | 1. (a) Show that if Ris a UFD, then ab/gcd(a, b) is a least common multiple of a and b. (b) Show that if R is a PID, then k is a least common multiple of a and b if and only if (k) = (a) n(b).

Answer 1

Solution = Domain non-zero elements ine Definition : 05_UFD.+ An Integral is a unique facronization Domain (UPD) is. Every element is a product of inreducible Every ivreducible. element is prin and also know that bey 2 (it Ris Pro is UFD), if R is then any abert have d = goede a,b). theoren we then - PID 1 a Cuco le. deg.ed (9,6). Proof- ideal generated the afb. by شب 11 L= (a,b)= § uq+vbWVER} Since Ris. PID. for -xatyb , both dra and dyb hold .eفون ممي. عليه I= (d) some def and so Since ab Ela) b a and diuison then of if c is a xatyba, 9 11 So, d is a ged ca,b).

oll ab lcm (a,b) ged ca.b) ab a b 4 ab = b بان d blabla NEUFD vet and So al aold an thes such that and blor abl mb & ab/na rigedia cab), it then ab n follows that ablged (ma, nb) ablnd and lam (2,5) = ab gud.b) ab Hence d

be Proof Cuires a pale a, ber which has common multiple a least common Since, m, then alm and blm where теа & meb and thus mecanco) thus let thes ola thus last (m) C Cainco) 3 - (a) 2(b) which implies that (2) clanco). alr and (by definition of common multiple mye xcom) there cannot be ideal in principle (a) (b) that is contained in m. this there esist k which is a least common multiple of a and CK) = (@)^(b) 1 thus a net o it and only it solved