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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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I suppose in some sense it's the Jacobson radical of $R/I$. –  JSchlather Nov 27 '12 at 5:29
Well, it's the preimage of the Jacobson radical of $R/I$ under the projection $R \to R/I$. Of course, we don't call the radical of an ideal the "nilradical of I," despite the analogous setup. –  Daniel McLaury Nov 27 '12 at 6:28
I guess if I wanted to be particularly obtuse, I could analogously drop the first three letters and call it the "obson radical," but that's just silly. –  Daniel McLaury Nov 27 '12 at 6:29
Yes, that's the sense I was alluding to. Although I might phrase it as the contraction of the Jacobson radical of R/I under the canonical projection. It seems though that Jacobson radical of an ideal, is a defined notion. –  JSchlather Nov 27 '12 at 6:46
The intersection you are interested in need not be a prime ideal. Counter-example: $I=(0)\subset R=\mathbb Z/(6)$. By the way, be very careful with the adverb clearly : it is a magnet for false statements. –  Georges Elencwajg May 18 '13 at 7:31
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up vote 2 down vote accepted

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.

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