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Let $(n_1,\ldots, n_i,\ldots)$ be an infinite tuple of nonnegative integers. Is there an abstract number ring $D$ of a given characteristic $p>0$ and $I_1,\dots, I_n , \ldots$ its nonzero ideals (by assumption $D/I_i$ are finite) such that #$D/I_i=n_i$ for each $i$?

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  • $\begingroup$ Couldn't $D=\mathbb{Z}$, $I_k=k\mathbb{Z}$ for all $k>1$, and the tuple simply be $\mathbb{N}$? Or am I misunderstanding your notation? $\endgroup$
    – Alexander Gruber
    Commented Aug 15, 2012 at 20:51
  • $\begingroup$ What about characteristic $p>0$? The tuple $(n_1,n_2,\ldots)$ is given. $\endgroup$
    – user16974
    Commented Aug 15, 2012 at 20:58
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    $\begingroup$ Your question is written in the wrong order. You don't want to first introduce the ring and its ideals, and only then bring in the numbers $n_i$, but rather first introduce a sequence of positive integers $n_1, n_2,\dots$ and then ask if there is a $D$ with its ideals enumerable as $I_1, I_2,\dots$ such that $D/I_i$ has size $n_i$ for all $i$. $\endgroup$
    – KConrad
    Commented Aug 15, 2012 at 21:01

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I'm not sure what you mean by an abstract number ring; nevertheless, here's an attermpt at an answer. If $D$ is to have characteristic $p$, then each of the quotients $D/I_i$ will be a vector space over $\mathbb Z/p$, so each $n_i$ will have to be a power of $p$. Conversely, if all the $n_i$ are powers of $p$, say $n_i=p^{k_i}$, then you could take $D$ to be the polynomial ring $(\mathbb Z/p)[X]$ and take $I_i$ to be the ideal generated by $X^{k_i}$.

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  • $\begingroup$ Is it true that the $I_k$ must be necessarily be all of the distinct ideals of $D$? If so the $n_k$ would have to have appropriate multiplicities for the $\mathbb{Z}_p[x]$ example. $\endgroup$
    – Alexander Gruber
    Commented Aug 15, 2012 at 21:39
  • $\begingroup$ @lemniscate: I got the point, which answers my question in the negative. $\endgroup$
    – user16974
    Commented Aug 15, 2012 at 21:57
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    $\begingroup$ Just regarding the point 'abstract number ring' this merely means that the quotient by any ideal is finite (and some nondegeneracy condition, not a field and infinite). Pete Clark seems to lieke them math.uga.edu/~pete/aant.pdf $\endgroup$
    – user9072
    Commented Aug 15, 2012 at 22:53

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