1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Commutative ring theorem

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to prove this theorem Let R be a commutative ring with identity and c1,c2,...cn E (element of) R Then the set I={r1c1+r2c2+....+rncn|r1,r2,...,rn E R} is an ideal in R

    2. Relevant equations

    Well I do know I need to prove closure under subtraction, closure under multiplication and absorption r E R and a E I show ra E I and ar E I I know because R is commutative I do not have to prove both ways. and I know that just proving absorption will prove multiplication.

    3. The attempt at a solution

    so, I guess I am just confused on what the set I looks like.

    to prove closure under subraction I need to elements from I. What would these two elements look like?
  2. jcsd
  3. Dec 2, 2009 #2


    User Avatar
    Science Advisor

    The set "looks like" exactly what it says: All things of the form [itex]r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n[/itex] where [itex]c_1, c_2, \cdot\cdot\cdot, c_n[/itex] are given (fixed) members of ring R and [itex]r_1, r_2, \cdot\cdot\cdot, r_n[/itex] can be any members of the ring.

    So you want to prove:
    (1) Closed under addition: If [itex]r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n[/itex] and [itex]s_1c_1+ s_2c_2+ \cdot\cdot\cdot+ s_nc_n[/itex] are members of this set, it their sum also a member? That is can it be written in the same way: find [itex]t_1, t_2, \cdot\cdot\cdot, t_n[/itex] such that the sum of those two is [itex]t_1c_1+ t_2c_2+ \cdot\cdot\cdot+ t_nc_n[/itex].

    (2) "Absorption" of the entire ring: If [itex]r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n[/itex] is in the set, show that [itex]r(r_1c_1+ r_2c_2+ \cdot\cdot\cdot+ r_nc_n)[/itex], for r any member of the ring, is of that same form.

    Those are both just as easy as they look.
  4. Dec 2, 2009 #3
    Ok that is easy.

    So under addition, because R is a ring then r1+s1 will be in R so the (r1+s1)c1+(r2+s2)c2... is in I

    And because R is a Ring and closed under multiplication r*r1, r*r2,.... are in R so r*r1c1+r*r2c2+..... is in I

    So it is a ideal in R

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Commutative ring theorem
  1. Commutative rings (Replies: 4)