In a finite reduced commutative ring, is every minimal prime ideal is generated by an idempotent? The number of idempotent elements and the number of minimal prime ideals are same.
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1$\begingroup$ Isn't a finite reduced ring just a finite product of finite fields? And so every ideal is generated by an idempotent? This also contradicts your claim that the number of minimal prime ideals is the same as the number of idempotent elements. $\endgroup$– Jeremy RickardCommented Nov 10, 2014 at 9:42
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1$\begingroup$ To elaborate on Jeremy's comment: the total quotient ring of any Noetherian commutative reduced ring is a finite product of fields (note that a finite ring is obviously Noetherian), and any finite commutative ring is its own total quotient ring. Then the minimal primes are in one-to-one correspondence with the field factors in question. Thus, you get an idempotent element for any string of 1's and 0's that is $n$ characters long, where $n$ is the number of field factors (i.e. the number of minimal primes). Hence, the number of idempotent elements is $2^n$ (which is never the same as $n$). $\endgroup$– Neil EpsteinCommented Nov 10, 2014 at 13:48
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