Proposition 1

Let $C$ be a $p$-ring.

a) The ring $C$ is local, with maximal ideal $pC$.

b) Suppose $p1_C$ nilpotent. Let $d$ be the smallest positive integer such that $p^d1_C = 0$. The ideals of $C$ are of the form $p^kC$ with $0 \leq k \leq d$, and one has $p^kC \neq p^lC$ when $k$ and $l$ are two distinct integers satisfying $0 \leq k \leq d, 0 \leq l \leq d$. The $C$-module $C$ is of length $d$.

c) Suppose that $p1_C$ is not nilpotent. Then $C$ is a discrete valuation ring whose residue field is of characteristic $p$, and whose field of fractions is of characteristic $0$. The ideals of the form $p^nC$, with $n \in \mathbf{N}$, are pairwise distinct; they form all the non-zero ideals of $C$. The $C$-module $C$ is not of finite length.

Assertion a) follows from prop. 19 of III, $§ 2$, no. 13.

By hypothesis, one has $\bigcap_{n \geqslant 0} p^n C = \{0\}$. Let $x \neq 0$ in $C$; there exists an integer $n \geqslant 0$ such that $x \in p^n C$, $x \notin p^{n+1} C$; therefore there exists an element $y$ of $C$ such that $x = p^n y$; since $y$ does not belong to $pC$, $y$ is invertible.

Suppose that $p 1_C$ is not nilpotent. If $x$ and $x'$ are two non-zero elements of $C$, there exist two integers $n \geqslant 0$, $n' \geqslant 0$ and two invertible elements $y$, $y'$ of $C$ such that $x = p^n y$, $x' = p^{n'} y'$. We then have $xx' = p^{n+n'} yy' \neq 0$, hence $C$ is an integral domain. Since $C$ is a local ring, but is not a field, and since the maximal ideal $m_C = pC$ of $C$ is principal, $C$ is a discrete valuation ring (VI, $§ 3$, no 6, prop. 9). The non-zero ideals of $C$ are then of the form $p^n C$ by loc. cit., prop. 8, and are pairwise distinct. In particular, the ring $C$ is not artinian, hence the $C$-module $C$ is not of finite length. The residue field $C/pC$ of $C$ is of characteristic $p$. Let $q$ be the characteristic of the field of fractions of $C$. We have $p 1_C \neq 0$, whence $p \neq q$. Moreover, if $q$ were non-zero, we should have $q 1_C = 0$ hence $C/pC$ would be of characteristic $q \neq p$, which is absurd. This proves c).

Suppose that $p 1_C$ is nilpotent. Let $d$ be the smallest positive integer such that $p^d 1_C = 0$. We have a sequence of ideals

(E)

If $k$ is an integer such that $0 \leqslant k < d$ and $p^k C = p^{k+1} C$, we deduce

contrary to the hypothesis $p^{d-1} 1_C \neq 0$. Therefore the elements of the sequence (E) are pairwise distinct. Let $a$ be an ideal of $C$ and let $k$ be the smallest positive integer such that $a \supset p^k C$. Let $x$ be a nonzero element of $a$; we have seen that $x$ is of the form $p^m u$ with $m \geqslant 0$ and $u$ invertible in $C$. Hence $p^m C \subset a$, whence $m \geqslant k$, and finally $x \in p^k C$. In conclusion, one has $a = p^k C$. The sequence (E) is then a Jordan-Hölder sequence of the $C$-module $C$, which is of length $d$.