Theorem 3

— Let $A$ be a commutative algebra of finite degree over $K$. For $A$ to be etale it is necessary and sufficient that the module $\Omega_K(A)$ of $K$-differentials of $A$ should be reduced to $0$.

A) Let $L$ be an algebraic closure of $K$ ($V$, p. 23, Th. 2). For $A$ to be etale it is necessary and sufficient that the algebra $A_{(L)}$ over $L$ should be diagonalizable ($V$, p. 30, Prop. 2). Further, the $A$-module $\Omega_L(A_{(L)})$ is isomorphic to $\Omega_K(A) \otimes_A A_{(L)}$ (III, p. 572, Prop. 20), hence to $\Omega_K(A) \otimes_K L$, by the associativity of the tensor product; therefore $\Omega_K(A) = 0$ is equivalent to $\Omega_L(A_{(L)}) = 0$. To prove Th. 3 it is therefore enough to consider the case where $K$ is algebraically closed and to show that the algebra $A$ is diagonalizable if and only if $\Omega_K(A) = 0$.

B) Suppose that $A$ is diagonalizable; then ($V$, p. 29, Prop. 1), the vector space $A$ is generated by the idempotents of $A$. The assertion $\Omega_K(A) = 0$ is thus a consequence of the following lemma: