Let R be the commutative Q-algebra generated by x1, x2, . . . with the relations (x_n)^n= 0 for all n. Show that R does not have a largest nilpotent ideal.
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Expert's answer
2012-11-19T07:46:18-0500
It is not hard to show that each xnhas index of nilpotency exactly equal to n in R. Let I beany nilpotent ideal in R; say In = 0. Then x_n+1 is notin I, and so I + (x_n+1)R is a nilpotent ideallarger than I.
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