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Math Help - Are all ideals of a ring rings themselves? A question

  1. #1
    Senior Member x3bnm's Avatar
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    Are all ideals of a ring rings themselves? A question

    I've two proofs that I'm having difficulty in proving.

    Problem:

    A nonempty subset B of a ring A is called an ideal of A if B is closed with respect to addition and negatives and B absorbs products in A.

    Pinter's Abstract Algebra book on page 187-188:
    Let A be a ring, and J an ideal of A. For any element a \in A, the symbol J + a denotes the set of all sums j + a, as a remains fixed and j ranges over J. That is,

    J + a = \{j + a \colon j \in J \}

    J + a is called a coset of J in A.

    There is a way of adding and multiplying cosets which works as follows:

    (J + a) + (J + b) = J + (a + b)
    And
    (J + a)(J + b) = J + ab

    My question:

    How do I prove that:
    (J + a) + (J + b) = J + (a + b)

    And

    (J + a)(J + b) = J + ab?

    How did the author figure out that left side is equal to right side for both of these equations? Is it possible to prove both of these equations?
    Last edited by x3bnm; March 17th 2013 at 01:19 PM.
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Are all ideals of a ring rings themselves? A question

    Let j_1, j_2 \in J.

    \begin{align*}(J + a) + (J + b) =& (j_1 + a) + (j_2 + b) \\ =&(a + j_1) + (j_2 + b)\text{......[by additive commutativity of a ring]} \\ =& a + (j_1 + j_2) + b\text{.........[by associativity of addition of a ring]}\\ =& (j_1 + j_2) + a + b\text{.........[by commutativity of addition of a ring]}\end{align*}

    Because J is an ideal and ideals are subring. So by definition of subring:

    j_1 + j_2 \in J

    And so:
    \begin{align*}(J + a) + (J + b) =& (j_1 + j_2)+ a + b \\ =& J + a + b............\text{[by definition of coset of a+b]}\\ =& J + (a + b)\text{.........[By associativity of addition of a ring]}\end{align*}



    And for the second proof:

    \begin{align*}(J + a)(J + b) =& (j_1 + a)(j_2 + b) \\ =& (j_1 + a)j_2 + (j_1 + a)b \text{......[by distributive property of addition]} \\ =& j_1 j_2 + aj_2 + j_1b + ab \text{........[by distributive property of addition of a ring]}\end{align*}

    j_1 j_2 \in J\text{.............[because J is a subring and so multiplication is closed]}

    aj_2, j_1b \in J\text{.........[because J is an ideal]}
    And:

    j_1 j_2 + aj_2 + j_1b \in J\text{.........[because } j_1j_2 + aj_2 + j_1b \in J \text{ and addition is closed in a ring]}

    So (J + a)(J + b) = J + ab\text{........[by definition of coset of ab]}

    Q.E.D
    Last edited by x3bnm; March 17th 2013 at 01:54 PM.
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