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BSc Mathematics SEM VI ATKT MATHEMATICS PAPER II ALGEBRA II OLD Question Paper - Mumbai University | munotes

ATKT Question Paper, Oct (27452).pdf
SEM VI · 1 May 2025

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Questions asked in this paper

  • (2) Figures to the right indicate marks for respective subquestions
  1. Q1 (a) any ONE State and prove the fundamental theorem of groups. (8)
    • ii. State and prove the Cayley’s theorem
    • i. Define kernel of a homomorphism f 2G 3 that it ig a of G and it is a normal subgroup of 6
    • ii. Prove that every subgroup of group G is norinal otherwise prove that A, is a normal of 6
    • iii. If H is a subgroup of H for every x prove that H is a normal subgroup of G is abelian 6
    • iv. Prove that there are groups of 6
  2. Q2 (a) Answer any ONE
    • i. Show that characteristic. of ‘an is prime. What can be said about. the characteristic of Justify. > Let f: RR! be ring homomorphism. that. (8) 8
    • (p) If J is {f(z) : I} isan
    • (q) If then = f(x) J} is an ideal of
    • i. Show integral domain is a field. ji, bea finite ring with every non zero element of R (6) or Is statement true for infinite Show: that. the homomorphism : Z Z is identity (6) that domain containing 6 elements. (6) ring of Gatissian integers is an Euclidean doamin ‘Define ideal ring. Show that an ideal M in ring (8) A is and only if R/M is field 6
    • Q. P. Code: 27452
    • (b) Answer any TWO
    • i. Show that a nonzero ideal P of a commutative ring is prime if and only if is an integral domain
    • ii. Show that the only maximal ideals in C[z] are (x= for C
    • iii. Show that an ideal J in Z is maximal if and only if = pZ for some’ prime 4, Answer any THREE A
    • (a) If H is the only subgroup of G of the given order-then is anormal. 5
    • (b) If a group G is a direct product of that G is not a cyclic group 5
    • (c) Define zero divisor and unit Show that every élement of Z, > is a zero divisor or an unit 5
    • (d) Show that if C Ig then: is an-ideal 5
    • (e) Show that the ring not isomorphic) 5
    • (f) Show that 2, 5 are 5

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