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I am searching for a prime ideal of the ring $R=∏_{n=2}^{∞} {\mathbb Z}_{2^n}$ which is not maximal. In fact, since each ${\mathbb Z}_{2^n}$ is local with $\left<\bar 2\right>$ as the maximal ideal, one could deduce that the Jacobson radical of $R$ is the ideal $\left<\bar 2, \bar 2, \bar 2,...\right>$ which is easily seen to be not nil, so the Krull dimension of $R$ does not equal zero.

Thanks for any help or suggestion!

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Pick any element $f$ in Jacobson radical but not in nil radical (say, $(\bar2,\bar2,\bar2,\dots)$), the preimage in $R$ of any maximal ideal of $R_f$ will do.

It seems impossible to give a concrete construction. A maximal ideal of $R$ containing a non-maximal prime ideal mustn't be of the form $\left<(\dots, \bar1, \bar 2, \bar1, \dots)\right>$. So the existence of non-maximal ideals of $R$ is equivalent to the existence of non-principal ultrafilters (dual notion of non-principal maximal ideals) of the Boolean ring $\prod_{n=2}^{\infty}\mathbb Z_2$ (the quotient of $R$ with respect to its Jacobson radical), which is equivalent to the Axiom of Choice, that is to say, to find a such we must turn to Zorn's Lemma. (Strictly speaking, the passage from a prime ideal to a maximal ideal containing it also calls for Axiom of Choice, so the above argument don't obviate the possibility completely, but I still think it unlikely.)

For details on the filters of a Boolean ring, see Ideals and filters.

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  • $\begingroup$ Thanks for the answer, but I want an explicit form for a prime ideal, please! $\endgroup$ – karparvar Jun 25 '16 at 10:22
  • $\begingroup$ @karparvar It seems impossible, I've added the reason in my answer. $\endgroup$ – Censi LI Jun 25 '16 at 11:27

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