Theorem 3

— Let $A$ be a commutative ring of characteristic $p \neq 0$. There exists a perfect closure $(A, u)$ of $A$. Moreover, the kernel of $u$ is the set of all nilpotent elements of $A$ and for each $x \in A$ there exists an integer $n \geq 0$ such that $x^{p^n} \in u(A)$.

For each integer $n \geq 0$, put $A_n = A$; when $m \geq n$ we define a homomorphism $\pi_{m,n}$ of $A_n$ into $A_m$ by $\pi_{m,n}(a) = a^{p^{m-n}}$. We thus obtain a direct system of rings $(A_n, \pi_{m,n})$ (I, p. 120); let $A$ be the direct limit of this system and $u_n$ the canonical homomorphism of $A_n = A$ into $A$; we also put $u = u_0$. By construction of the direct limit the kernel $\mathfrak{n}$ of $u$ is the union of the kernels of the homomorphisms π_{n,0}: a \mapsto a^{p^n} of $A$ into $A$, thus it consists of all the nilpotent elements of $A$. The ring $A$ is commutative of characteristic $p$ by Remark 1 of V, p. 3.

The ring $A$ is the union of the ascending sequence $(u, (A))_n$ of subrings. We have $u_n(A)^{p^n} = A$; hence for each $x \in \hat{A}$ there exists an integer $n \geq 0$ such that $x^{p^n} \in u(A)$. We also have $u_n(A) = u_{n+1}(A)^p$, whence $\hat{A}^p = A$. Let $x \in A$ be such that $x^p = 0$; choose an integer $n \geq 1$ and an element $a \in A$ such that $x = u_n(a)$. Then we have $u_{n-1}(a) = u_n(a)^p = 0$; by definition of the direct limit there exists an integer $m$ such that $\pi_m^{1,n-1}(a) = 0$, that is, $a^{p^{m-n}} = 0$. We thus have $\pi_{m,n}(a) = 0$, whence $u_n(a) = 0$, that is, $x = 0$. Therefore the ring $A$ is perfect of characteristic $p$.

Let $v$ be a homomorphism of $A$ into a perfect ring $B$ of characteristic $p$. For every integer $n \geq 0$, the mapping $b \mapsto b^{p^n}$ is an automorphism of $B$ and so there exists a homomorphism $v_n$ of $A, = A$ into $B$ characterized by $v(a) = v_n(a)^{p^n}$. We then have $v_m \circ \pi_{m,n} = v_n$ for $m \geq n \geq 0$; by definition of the direct limit there exists a homomorphism $h$ of $A$ into $B$ such that $v_n = h \circ u$, for all $n \geq 0$; in particular we have $v = v_0 = h \circ u_0 = h \circ u$. Finally let $h'$ be a homomorphism of $A$ into $B$ such that $h' \circ u = v$. Let $x \in A$; as we have seen, there exist an integer $n \geq 0$ and an element $a \in A$ such that $x^{p^n} = u(a)$. Then we have

and since $B$ is perfect, we find $h(x) = h'(x)$. Thus we have $h' = h$, and this completes the proof that $(A, u)$ is a perfect closure of $A$.