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Showing that a set is an ideal

  1. May 5, 2017 #1
    1. The problem statement, all variables and given/known data
    Show that the collection of all nilpotent elements of a commutative ring ##R## is an ideal.

    2. Relevant equations


    3. The attempt at a solution
    Showing that something is an ideal is somewhat straightforward, but I am a little confused as to what explicitly I have to show. If we denote ##N## as the set in question, then I know that we have to show that ##aN \subseteq N## and ##Nb \subseteq## for all ##a,b \in R##. But what else do I have to show? Do I have to show that N is an additive subgroup?
     
  2. jcsd
  3. May 5, 2017 #2

    andrewkirk

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    Science Advisor
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    Yes
     
  4. May 5, 2017 #3
    Consider ##(a + b)^{m+n}##, where ##a,b \in N##. In the binomial expansion, each summand contains a term ##a^{i}b^{m+n−i}##. Now
    either ##i \ge m## so that ##a^i## = 0 or ##m+ n − i \ge n## so that ##b^{m+n−i} = 0##. Thus each summand of ##(a + b)^{m+n}##
    is zero, so ##(a + b)^{m+n} ##= 0 and ##N## is closed under addition. Also, since ##0^1 = 0##, ##0 \in N##.
    Also ##(−a)^m## is either ##a^m## or ##−a^m##, so ##(−a)^m = 0## and ##-a \in N##.

    Does this show that ##N## is a additive subgroup of ##R##?
     
  5. May 5, 2017 #4

    fresh_42

    Staff: Mentor

    I think you need the power ##n+m+1##. But basically, yes. For an ideal you also need ##r \cdot a \in N##.
     
  6. May 5, 2017 #5
    In the solution to this problem in the book, it also shows that ##N## is closed under multiplication of elements in ##N##. It shows that ##(ab)^{mn} = (a^m)^n(b^n)^m = (0)(0) = 0##, so ##ab \in N##. Isn't this unnecessary? Don't I only need to show that ##N## is an additive subgroup and that is satisfies the absorption property for ideals? Where would being internally closed under multiplication fit in?
     
  7. May 5, 2017 #6

    fresh_42

    Staff: Mentor

    Well, it does no harm and is almost obvious in a commutative ring. It shows, that ##N## carries also a ring structure. But if ##r \cdot a \in N## for all ##r \in R\, , \,a \in N##, doesn't this imply ##a\cdot b \in N## for all ##a,b \in N##?
     
  8. May 5, 2017 #7
    That's what I mean. It seems superfluous. I just want to make sure I know what is strictly necessary to show that some set is an ideal of a ring.
     
  9. May 5, 2017 #8

    fresh_42

    Staff: Mentor

    Then (minimal) it's (for left ideals)
    1. ##N \neq \emptyset##
    2. ##a - b \in N## for all ##a,b \in N##
    3. ##r \cdot a \in N## for all ##r\in R \, , \,a \in N##
    Of course left and right ideals don't have to be distinguished in a commutative ring, but in general they have to be. And the first condition cannot be omitted by the second (##a-a \in N)##, as in case ##N## is empty, the second condition is still true, whereas ##0 \notin N##, which is needed. So we can also write ##0 \in N## as first condition.
     
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