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Clearly it's a prime ideal containing the radical. Do you just call it the "Jacobson radical of I?"

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    By the way, I think it's easier to visualize the ideal corresponding to, say, a few points in the plane. Then obviously the Jacobson radical of that ideal still has to cut out just as many components and thus can't be prime.2013-05-18

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For the sake of resolving this question, it apparently is sometimes known as the "Jacobson radical of $I$." Thanks to JSchlather for supplying a reference in the comments.