Let $K$ be a subfield of $A$, $C$ the centre of $A$, and put $K' = K \cap C$; then $K'$ is a subfield of $A$. Let $f$ be the unique homomorphism of $\mathbf{Z}$ into $A$ and $p$ its kernel. Every subring of $A$, in particular $K'$, contains $f(\mathbf{Z})$; hence the ideal $p$ is prime (I, p. 116-117). If $p = (0)$, the homomorphism $f$ of $\mathbf{Z}$ into $K'$ is injective; hence it extends (I, p. 116) to an isomorphism $\bar{f}$ of $\mathbf{Q}$ onto a subfield $P$ of $K'$. If $p \neq (0)$, there is a strictly positive integer $p$ such that $p = (p)$ (I, p. 111); if we had $p = ab$ with $a > 1$ and $b > 1$, this would mean $a \notin p, b \notin p$ and $ab \in p$ in contradiction with the fact that $p$ is prime. The number $p$ is therefore prime and by passage to quotients $f$ defines an isomorphism of $F_p = \mathbf{Z}/p$ onto a subfield $P$ of $K'$. In both cases $P$ is a subfield of $A$ contained in the centre $C$ of $A$, and it is a prime field. Let $L$ be a subfield of $A$; then $P \cap L$ is a subfield of $P$, and since $P$ is prime, we have $P \cap L = P$, whence $P \subset L$. If $P'$ is a subfield of $A$ and is a prime field, then by what has been said, $P \subset P'$, whence $P = P'$ because $P'$ is a prime field.

b) Let $p$ be a prime number. For $A$ to be of characteristic $p$ it is necessary and sufficient that $p \cdot x = 0$ for all $x \in A$.

Let $f$ be the unique homomorphism of $\mathbf{Z}$ into $A$; we have $n \cdot x = f(n)x$ for every integer $n$ and every $x$ in $A$. For $A$ to be of characteristic 0, it is necessary and sufficient that $f$ should extend to a homomorphism of $\mathbf{Q}$ into $A$, that is, $f(n)$ should be invertible in $A$ for every $n \neq 0$ (I, p. 113); this proves $a$. Similarly for $A$ to be of characteristic $p$ it is necessary and sufficient that $f$ should annihilate $p\mathbf{Z}$, that is, $f(p) = 0$, or also that $p \cdot x = 0$ for all $x \in A$; this proves $b$.

Let us take for $A$ a not necessarily commutative field. The centre of $A$ is a (commutative) field; therefore the characteristic and the prime subfield of $A$ are defined.

(1)

(2)

for $a, b$ in $A$.

Formula (2) follows from the commutativity of $A$. To prove (1) we use the binomial formula $(a + b)^p = a^p + b^p + \sum_{i=1}^{p-1} \binom{p}{i} \cdot a^i b^{p-i}$; since $p \cdot x = 0$ for all $x \in A$, it suffices to prove the following lemma:

We argue by induction on $i$, the case $i = 1$ being immediate from the formula $\binom{p}{1} = p$. Suppose that $2 \leq i \leq p - 1$ and that $\binom{i}{i-1}$ is divisible by $p$. Then the integer $i \binom{p}{i} = (p - i + 1) \binom{p}{i-1}$ belongs to the prime ideal $p\mathbf{Z}$ of $\mathbf{Z}$; since $i \notin p\mathbf{Z}$, we have $\binom{p}{i} \in p\mathbf{Z}$ and the lemma follows.

Let $A$ be a commutative ring of characteristic $p$ and $f$ an integer $\geq 0$. From Th. 2 we deduce by induction on $f$ that the mapping $a \mapsto a^{p^f}$ is an endomorphism of the ring $A$. In particular we have the relation

for any $a_1, \ldots, a_n$ in $A$. The mapping $a \mapsto a^p$ is sometimes called the Frobenius endomorphism of $A$. Taking $A = \mathbf{F}_p$ and $a_i = 1$, we obtain from (3) the relation:

For each subset $S$ of $A$ we denote by $S^{p^f}$ the set of elements of $A$ of the form $x^{p^f}$ with $x \in S$^1. In particular, if $K$ is a subring of $A$, the set $K^{p^f}$ is a subring of $A$. If $K$ is a subring of $A$ and $S$ a subset of $A$ we denote by $K[S]$ the subring of $A$ generated by $K \cup S$; when $A$ is a field, we denote by $K(S)$ the field of fractions of $K[S]$, that is, the subfield of $A$ generated by $K \cup S$.

a) We have $K[S]^{p^f} = K^{p^f}[S^{p^f}]$, and if $A$ is a field, $K(S)^{p^f} = K^{p^f}(S^{p^f})$.

b) If the $K$-module $K[S]$ is generated by the family $(a_i)_{i \in I}$ of elements of $A$, then the $K$-module $K[S^{p^f}]$ is generated by the family $(a_i^{p^f})_{i \in I'}$

Since $K[S]$ is the subring of $A$ generated by $K \cup S$, its image $KISlp^f$ under the endomorphism $\pi : a \mapsto a^{p^f}$ of the ring $A$ is the subring of $A$ generated by the image $K^{p^f} \cup S^{p^f}$ of $K \cup S$ under $\pi$, whence $K[S]^{p^f} = K^{p^f}[S^{p^f}]$. The case of fields is treated similarly; this proves a).

It is clear that the family $(a_i^{p^f})_{i \in I}$ generates the $K^{p^f}$-module $K[S]^{p^f}$. The $K$-module $K[S^{p^f}]$ is generated by products of the form $x_1^{p^f} \cdots x_n^{p^f} = (x, \ldots x_n)^{p^f}$ with $x_1, \ldots, x_n$ arbitrary in $S$, hence also by the set $K[S]^{p^f}$. Assertion b) follows directly from this.

^1 Of course the set $S^{p^f}$ should not be confused with the set product of $p^f$ sets equal to $S$, nor with the set of products of $p^f$ elements belonging to $S$.

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 $A$.

For each integer $f \in \mathbf{Z}$ denote by $\pi_f$ the automorphism $b \mapsto b^{p^f}$ of $B$. The sequence of subrings $\pi_{-f}(A)$ of $B$ (for $f \geq 0$) is ascending and its union $\mathbf{A}^{p^{-\infty}}$ is thus a subring of $B$. We have $\pi_1(\mathbf{A}^{p^{-\infty}}) = \bigcup_{f \geq 0} \pi_{-(f-1)}(A) = \mathbf{A}^{p^{-m}}$, hence $\mathbf{A}^{p^{-\infty}}$ is a perfect subring of $B$. Finally let $B_0$ be a perfect subring of $B$ containing $A$; for every integer $f \geq 0$ we have $\pi_{-f}(A) \subset \pi_{-f}(B_0) = B_0$, whence $\mathbf{A}^{p^{-\infty}} \subset B_0$.

If $v$ is a homomorphism of $A$ into a perfect ring $B'$ of characteristic $p$, then for every integer $f \geq 0$ we can define a homomorphism $h_f$ of $\pi_{-f}(A)$ into $B'$ by $h_f(\pi_{-f}(a)) = v(a)^{p^{-f}}$ for all $a \in A$. We see at once that $h_{f+1}$ agrees with $h_f$ on $\pi_{-f}(A)$; thus there exists a homomorphism $h$ of $\mathbf{A}^{p^{-\infty}}$ into $B'$ which induces $h_f$ on $\pi_{-f}(A)$ for all $f \geq 0$ and in particular, $h$ extends $h_0 = v$. If $h'$ is another extension of $v$ to a homomorphism of $A^{p^{-\infty}}$ into B, the equality $h' = h$ may be established as in the proof of Th. 3.

a) If K is a ring of characteristic 0, then S = K. b) If K is a ring of characteristic $p \neq 0$, then $S = K[X_i^p]_{i \in I}$; if moreover K is perfect, then $K = A^P$.

The mapping $F \mapsto dF$ of A into the module $\Omega_K(A)$ of K-differentials of A is K-linear and satisfies the relation

(III, p. 569). Therefore S is a subalgebra.

When K is of characteristic $p \neq 0$, put $T = K[X_i^p]_{i \in I}$; we thus have $T = A^P$ if K is perfect ($V$, p. 4, Prop. 2); moreover, we have $d(X_i^p) = pX_i^{p-1} \cdot dX_i = 0$ for all $i \in I$, hence the subalgebra S of A contains T. If K is of characteristic 0, we put $T = K$, and still find that $T \subset S$. It remains to show that S is contained in T.

For every finite subset J of I let $A_J$ be the subalgebra of A generated by the family $(X_j)_{j \in J}$. We have $A_\emptyset = K$ and $A = \bigcup_{J \in I} A_J$; so it suffices to prove the relation $S \cap A_J \subset T$, which we shall accomplish by induction on the cardinal of J. Thus let J be a finite subset of I such that $S \cap A_J \subset T$, let i be an element of $I - J$ and $J' = J \cup \{ i \}$. Every element F of A may be written in just one way in the form

with $F_n \in A_J$ for all $n \geq 0$, and then

Suppose that F belongs to S; the family $(dX_r)_{r \in J}$ is a basis of the A-module $\Omega_K(A)$ (III, p. 570) and $dF_n$ is a linear combination of the differentials $dX_j$ for $j \in J$ because $F_n \in A_J = K[X_j]_{j \in J}$. By (6) we then have $dF_n = 0$ and $nF_n = 0$ for each integer $n \geq 0$. By the induction hypothesis $F_n \in T$ for all $n$, since $dF_n = 0$.

a) If K is of characteristic 0, we have $nF_n = 0$ for all $n \geq 1$, whence $F_n = 0$ by Prop. 1 ($V$, p. 2); so we have $F = F_0$, whence $F \in T$.

b) If K is of characteristic $p \neq 0$, then A is an algebra over the field $\mathbf{F}_p$ and the relation $nF_n = 0$ implies $F_n = 0$ for every integer $n$ not divisible by $p$. So we have $F = \sum_{m=0}^m F_{mp} X_i^{mp}$, whence $F \in T$