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judge if nilradical equals jacobson radical
1)a noetherian ring that is not a artin ring.
2)a local integral domain that is not a field.
3)a integral domain with only finite number of maximal ideals that is not a field.
4)a ring that only has finite number of prime ideals in which it's not the case that all primes are maximal.

I think 1)should be wrong but example like $\mathbf{Z}$ does not work...
and 3) is a special case of 2), I mean if 3) is wrong then 2) is also wrong..
and for the rest, I have no idea..

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    $\begingroup$ Is this your homework? It doesn't seem you are spending enough to time to think about it before asking. $\endgroup$ – Manos Dec 21 '14 at 1:29
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(1) In a Noetherian ring nilradical can be equal to Jacobson radical, even if it is not a Artin ring. $\mathbb{Z}$ is an example. In fact any Jacobson ring has this property.

(2) is not true, because in this case the nilradical is the zero ideal, but Jacobson radical is a non-zero ideal (since it is not a field). Take any Noetherian local domain which is a DVR.

(3) is not true in general because of (2).

(4) is not true in general again because of (2).

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  • $\begingroup$ yeah, it's true, but can you give me some concrete examples of (1) (2) (4) that the two radical are not equal? and for(3) is $k\times k$ a field? I think they may not be integral domain, like $(0,1)\times (1,0)$ is $(0,0)$ in $Z_2 \times Z_2$ $\endgroup$ – annimal Dec 21 '14 at 2:51
  • $\begingroup$ @annimal: Ah..!! Sorry. in (3) I missed the "integral domain" part. I'll edit my answer. and you are right $k \times k$ is not an integral domain. $\endgroup$ – Krish Dec 21 '14 at 3:01

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