Define $ Jac(R) = \cap_{\mathfrak{m}, maximal } \mathfrak{m}$ and $Nil(R) = \cap_{\mathfrak{p}, prime} \mathfrak{p}$, where $R$ is a commutative ring. Can someone give me an example of a commutative ring $R$ such that $Jac(R) \ne Nil(R)$. I have ruled out $\mathbb{Z}_{n}, \mathbb{Z}$, and any Artinian ring. Should I be looking at polynomial rings of some sort?
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1I found this duplicate by searching “Jacobson nil radical different”. You should try something similar first, next time, before posting. – rschwieb Feb 16 '20 at 18:09
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Could you explain why $Nil(R) = {0}$, where $R = F[[x]]$ for some field $F$? – user100101212 Feb 16 '20 at 19:19
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A nilpotent element in an integral domain must be zero. Otherwise, it would be a zero divisor. – rschwieb Feb 16 '20 at 19:20