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I'm looking for an example of a commutative ring $R$ with zero Krull dimension such that $R$ satisfies one of the following conditions:

(1) $R$ has no nonzero proper idempotent ideals, but has a nonzero non-nilpotent ideal.

(2) $R$ has no nonzero nilpotent ideals, but has a nonzero idempotent ideal.

As for (2), I only know that $R$ would be von Neumann regular by a classical theorem of commutative algebra.

Thanks for any help!

user26857
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karparvar
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    (2) What about the direct product of countable many copies of $\mathbb F_2$? – user26857 Oct 13 '16 at 08:19
  • @user26857 Thanks! But, any commutative Noetherian local ring with zero Krull dimension has nilpotent maximal ideal, whence all its ideals are nilpotent. If $F$ is a field and we take $F[x_2,x_3,x_4\ldots]$ and quotient it by $(x_2^2,x_3^3,x_4^4,\ldots)$ could we get a ring having no non-trivial idempotent ideals? – karparvar Oct 13 '16 at 15:46
  • (1) It seems that your proposal could be good choice. The maximal ideal $\mathfrak m=(x_2,x_3,\dots)$ of the quotient ring $R=F[X_2,X_3,\dots]/(X_2^2,X_3^3,\dots)$, has the property $\cap_{i\ge 1}\mathfrak m^i=(0)$ (why?). If $I$ is an idempotent ideal, then $I=\cap_{i\ge1}I^i\subseteq\cap_{i\ge 1}\mathfrak m^i=(0)$. – user26857 Oct 13 '16 at 21:00
  • @user26857 If we take a nonzero polynomial $f\in R$, then $f=\sum {x_2}^{u_2}{x_3}^{u_3}\dots {x_n}^{u_n}$. Now, if we put $i$ as the maximum value of $\sum {u_t}$ then $f$ would not belong to $m^{i+1}$. Is it a true argument for the proof of $\cap_{i\ge 1}\mathfrak m^i=(0)$? – karparvar Oct 14 '16 at 05:58
  • @user26857 It seems that the above argument is not valid !! Would you please elucidate the "why"? – karparvar Oct 14 '16 at 12:13
  • Why it's not valid? It seems to me that this is the right way. In fact, $\sum_{t=2}^nu_t$ is less or equal than $\sum_{t=2}^n(t-1)=n(n-1)/2$, otherwise the corresponding monomial is zero in $R$. Then, for $i>n(n-1)/2$ we have $f\notin\mathfrak m^i$. (Note that $n$ is fixed being the maximum number of variables which occur in $f$.) – user26857 Oct 14 '16 at 20:52
  • @user26857 My doubt arises from the affair that when $I$ is an ideal of a ring $R$, and $x,y\in I$, what ensures that the product $xy$ is "not" an element of, say, $I^3$? (In our problem, why $f$ does not belong to $m^i$ when $i>n(n-1)/2$?) – karparvar Oct 15 '16 at 06:56
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    Well, in our case we are working with polynomials, and basically a monomial of total degree $k$ can't be in $(X_2,X_3,\dots)^{k+1}$. – user26857 Oct 15 '16 at 07:14

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An example for part (1) may be the ring $R=F[X_2,X_3,\dots]/(X_2^2,X_3^3,\dots)$, where $F$ is a field. At the outset, note that any prime ideal of $R$ is of the form $P/J$, where $J=(X_2^2,X_3^3,\dots)$ and $P$ is a prime ideal of $F[X_2,X_3,\dots]$ containing $J$, so that $P$ contains all the $X_i$'s, whence it is equal to the maximal ideal $M=(X_2,X_3,\dots)$ of $F[X_2,X_3,\dots]$. Thus, $R$ is a local ring of zero Krull dimension with the maximal ideal $\overline M=(\overline X_2,\overline X_3,\dots)$, where the bars refer to modulo $J$. The ideal $\overline M$, having elements of arbitrary high nilpotency indices, is not nilpotent. On the other hand, since any non-zero polynomial of degree $k$ in $\overline M$ could not belong to $\overline M^{k+1}$, we deduce that $\cap_{i\ge 1}\overline M^i=(0)$. Now, if $I$ is a proper idempotent ideal of $R$, we get the conclusion that $I=\cap_{i\ge1}I^i\subseteq\cap_{i\ge 1}\overline M^i=(0)$.

karparvar
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