Corollary

— Suppose that $K$ has characteristic $p \neq 0$. For $A$ to be etale it is necessary and sufficient that $A = K[A^p]$. For every basis $(a_i)_i$, of $A$ over $K$, the family $(a_i^p)_i$ is then a basis of $A$ over $K$.

Let us choose an algebraic closure $\Omega$ of $K$. Given two $K$-homomorphisms $u$ and $v$ of $A$ into $\Omega$, if $u$ and $v$ have the same restriction to $K[A^p]$, we have

whence $u(x) = v(x)$

for all $x \in A$. We thus have the inequality $[A : K], \leq [K[A^p] : K]_s$; if $A$ is etale, we thus have

whence $A = K[A^p]$

Conversely suppose that we have $A = K[A^p]$; let $(a_i)_{i \in I}$ be a basis of $A$ over $K$. By $V$, p. 4, Prop. 2, b), the family $(a_i^p)_{i \in I}$ generates the vector $K$-space $K[A^p]$ and since $A = K[A^p]$ is of finite dimension equal to the cardinal of $I$, the family $(a_i^p)_{i \in I}$ is a basis of $A$ over $K$. Let $u$ be an element of $\Omega \otimes_K A$ such that $u^2 = 0$, whence $u^p = 0$; since $(a_i)_{i \in I}$ is a basis of $A$ over $K$, there exists a family $(\lambda_i)_{i \in I}$ of elements of $\Omega$ such that $u = \sum_{i \in I} \lambda_i \otimes a_i$, whence $u^p = \sum_{i \in I} \lambda_i^p \otimes a_i^p$. Since $(a_i^p)_{i \in I}$ is a basis of $A$ over $K$ and $u^p = 0$, we must have $\lambda_i^p = 0$, whence $\lambda_i = 0$ for all $i \in I$, and so $u = 0$. This shows the ring $\Omega \otimes_K A$ to be reduced; since $\Omega$ is perfect, the algebra $A$ over $K$ is etale by Th. 4.

For another characterization of etale algebras see $V$, p. 48, Prop. 1.