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    Ring and ideal proofs

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    Let R be a ring....Prove that radical I is an ideal.....Prove that there is a prime ideal P containing I....

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    Denote the radical of I by rad(I).

    Clearly rad(I) is not empty since it contains I (and I is not empty).

    If a, b are in rad(I), then for some integers n, m >= 1 we have that both a^n and b^m are in I.

    Expanding using the binomial theorem we obtain (using commutativity of R)

    (a + b)^{n + m - 1} = sum_{i = 0 to n + m - 1} (n + m - 1 choose i) a^i b^{n + m - 1 - i}

    But for each i, we have either i >= n or n + m - 1 - i >= m (since n - i - 1 is at least 0), in which case either a^i is in I or b^{n + m - i - 1} is in I; in any case, their product is still in I, and since I is an ideal, the above sum is in I, hence rad(I) itself is closed under sums.

    Since a^n is in I, and ...

    Solution Summary

    This provides an example of several proofs regarding rings and ideals, including prime and radical.