# How much can we say about the number of nilpotents in a finite local commutative ring?

A commutative ring is local if it has a single maximal ideal. If the ring is finite, this implies that all elements are either units or nilpotents. Further, all finite local rings have prime power order.

Given a prime power $p^k$ and a positive integer $n < p^k$, under what conditions on $p, k, n$ does there exist a local ring $R$ with $|R| = p^k$ and $n$ nilpotents?

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what partial results have you found or worked out thus far? –  Yemon Choi Jul 11 '10 at 21:09
Well, certainly we need $n = p^a$ for $0 \leq a < k$. The cases $a = 0$ and $a = k-1$ are easy: take $R = \mathbb{F}_{p^a}$ and $R = \mathbb{Z}/p^k \mathbb{Z}$. This brings me to YC's comment: maybe you can just get every possibility by writing down a natural family of examples: to what extent have you tried this? –  Pete L. Clark Jul 11 '10 at 21:13
Interpolating between finite fields ${\mathbf F}_{p^k}$ and ${\mathbf Z}/p^k{\mathbf Z}$, consider the length $m$ Witt vectors over ${\mathbf F}_{p^r}$, $W_{m-1}({\mathbf F}_{p^r})$. (By length $m$ I mean $m$ coordinates: $(a_0,\dots,a_{m-1})$.) This is a local ring of size $p^{rm}$ and its units have nonzero $a_0$, so there are $p^{r(m-1)}$ nilpotents. If you want this to have size $p^k$ and $p^a$ nilpotents then set $rm = k$ and $r(m-1) = a$, so $r = k-a$ and $k-a$ has to divide $a$. OK, so if $0 \leq a < k$ and $k-a$ divides $a$, write $a = (k-a)d$ and use $W_d({\mathbf F}_{p^{k-a}})$. –  KConrad Jul 11 '10 at 21:58
@KConrad: In fact that exhausts all possibities. The residue field of our ring is $\mathbb{F}_{p^{k-a}}$, and the ring must have the same cardinality as $R/\mathfrak{m}\oplus\mathfrak{m}/\mathfrak{m^2}\oplus\ldots$, which has cardinality a power of $p^{k-a}$, so $k-a|k$. –  Kevin Ventullo Jul 11 '10 at 23:41
Also, $\mathbb{F}_{p^{k-a}}[x]/x^{d+1}$ works. –  Kevin Ventullo Jul 11 '10 at 23:47