Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Product of Ideals

  1. Jun 30, 2005 #1
    Given a commutative ring R with a unit, how do you prove that the product of two ideals, I1 and I2, is also an ideal?
    The product of course is defined to be {x*y | x in I1, y in I2}, where * is the multiplication in the ring R.
    I'm having trouble proving that I1*I2 is a group under addition.

  2. jcsd
  3. Jun 30, 2005 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Are you sure?
  4. Jun 30, 2005 #3
    http://planetmath.org/encyclopedia/ProductOfIdeals.html [Broken]
    Last edited by a moderator: May 2, 2017
  5. Jun 30, 2005 #4
    Well I can see this definition doesn't match mine, but at any rate - I was asked to prove that this group:
    {x*y | x in I1, y in I2}
    is an ideal.
  6. Nov 25, 2007 #5
    I'm working on the similar problem right now, and I was going through many threads in this forum and couldn't find the answer whether the product as Chen defined it is an ideal. I wanted to find the counterexample that {x*y | x in I1, y in I2} is an ideal, but was not able to come up with anything. So, does anybody have any ideas of how to do it, or can you at list give me a hint whether or not it is an ideal? Thanks!
  7. Nov 25, 2007 #6

    Chris Hillman

    User Avatar
    Science Advisor

    Is it possible that you are confusing an abelian subgroup with a subring with an ideal? A good short textbook which should help is Herstein, Abstract Algebra. See the excellent and very readable textbook by Cox, Little, and OShea, Ideals, Varieties, and Algorithms, for much more about such constructions as [itex]IJ, \, I \cap J, \, I+J, \, I:J, \, \sqrt{I}[/itex].
    Last edited: Nov 25, 2007
  8. Nov 25, 2007 #7


    User Avatar
    Science Advisor
    Homework Helper

    I suspect that the question had meant that IJ is generated by the set {fg : f in I, g in J}, and not equal to it. Occasionally this is expressed by writing IJ = <fg : f in I, g in J>.
  9. Nov 26, 2007 #8

    Chris Hillman

    User Avatar
    Science Advisor

    Ditto morphism (this point and others are well explained in IVA).
  10. Nov 26, 2007 #9


    User Avatar
    Science Advisor
    Homework Helper

    your definition is wrong. the product of ideals is tautologically an ideal, as it is defined as the ideal generated by those products, or equivalently as all sums of them.
  11. Aug 18, 2008 #10

    I saw many people have given a wrong definition for the product of ideals. The following is the correct definition as appeared in mathematical text:

    I*J = {i[1]*j[1]+i[2]*j[2]+...+i[n]*j[n]:i[n] in I and j[n] in J where n is finite}
    Last edited: Aug 18, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook