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    Let R be a commutative ring and let I be the set \{a\in R|a^{n_a}=0 for some n_a\in\mathbb{N}\}. Show that I is an ideal of R.
    R is a commutative ring, so ra\in I \forall r\in R, a\in I \implies I is an ideal of R.

    ra\in I since if a^{n_a}=0 then...where do I go from here?
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    (Original post by RamocitoMorales)
    R is a commutative ring, so ra\in I \forall r\in R, a\in I \implies I is an ideal of R.

    ra\in I since if a^{n_a}=0 then...where do I go from here?
    You want to show that if a is an element of I, then ra is an element of I as well, where r is any element of R. What is the defining property of elements of I? How can you use what you know about a to show that ra satisfies this defining property?
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    (Original post by Mark13)
    You want to show that if a is an element of I, then ra is an element of I as well, where r is any element of R. What is the defining property of elements of I? How can you use what you know about a to show that ra satisfies this defining property?
    Could I say:

    Consider (ra)^{n_a}=r^{n_a}a^{n_a}=r^{n_a  }\cdot 0=0\implies ra\in I
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    (Original post by RamocitoMorales)
    Could I say:

    Consider (ra)^{n_a}=r^{n_a}a^{n_a}=r^{n_a  }\cdot 0=0\implies ra\in I
    Yep that's it. You might want to say that you're using the commutativity of the ring to say that (ra)^{n_a}=r^{n_a} a^{n_a}, since for a non-commutative ring, it isn't generally the case that ra=ar.
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    Note that in Commutative Algebra, there is something called the radical of an ideal. The set you mention is just the radical of the zero ideal (0). Thus you can replicate the proof that the radical of an ideal is an ideal. Radicals play a very interesting role in Algebraic Geometry.
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    The proof that the radical of an ideal is an ideal is very standard. For instance, Chu proves this on his blog:

    http://qchu.wordpress.com/2009/11/30...orrespondence/
 
 
 
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