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Math Help - No Prime Ideal Of An Integral Domain Contains 1

  1. #1
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    No Prime Ideal Of An Integral Domain Contains 1

    Let A be a commutative integral domain with identity 1. Let X be the set of all prime ideals of A, and let V(E) be the set of all prime ideals of A containing the subset E of A.

    My task is to show that V(\{1\})=\emptyset.

    Here's my thinking on this, and what I've done:

    "To show that V(\{1\})=\emptyset, it suffices to show that an ideal of A is proper if and only if it doesn't contain 1, since by definition all prime ideals are proper ideals.

    \Rightarrow) Proof by contrapositive. Suppose 1\in I where I is an ideal of A. Since an ideal is a subgroup under addition, and further since 1 is an additive generator for A, it follows that I=A thus I is not a proper ideal.

    \Leftarrow) Suppose that 1\notin I. Since 1\in A by definition, it immediately follows that I\subset A and thus I is proper.

    Thus there is no prime ideal that contains 1, since 1 is not contained in any proper ideal."
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  2. #2
    Member HappyJoe's Avatar
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    I guess you are asking for feedback on your proof.

    For the implication in this direction ==>, you write that 1 is an additive generator for A. But there may be elements in A that cannot be written as a finite sum of 1's. Instead, remember that an ideal is closed under multiplication with elements from the ring. So since 1 is in the ideal, then for all r in A, we have also that r = r1 is in the ideal (so all elements of A are in I, hence I=A).

    The other direction is good.

    As you yourself note, this really hasn't got much to do with prime ideals. Rather, it is true that the only ideal containing 1 is the entire ring - no proper ideals contain the identity.
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