Proposition 6

Let k be a field and A a noetherian k-algebra.

a) If A is absolutely regular (resp. absolutely normal, resp. separable), the same is true of S$^{-1}$A for every multiplicative subset S of A.

b) If $A_m$ is absolutely regular (resp. absolutely normal, resp. separable), for every maximal ideal m of A, then A is absolutely regular (resp. absolutely normal, resp. separable).

a) This follows from the fact that the ring $(S^{-1}A)_{(k')}$ is isomorphic to a ring of fractions of the ring $A_{(k')}$ for every extension $k'$ of k.

b) Suppose that $A_m$ is absolutely regular (resp. absolutely normal, resp. separable) for every maximal ideal m of A. Let $k'$ be a radicial extension of k of finite degree and let $m'$ be a maximal ideal of $A_{(k')}$. It remains to verify that the local ring $(A_{(k')})_{m'}$ is regular (resp. normal, resp. reduced).

The canonical homomorphism $A \to A_{(k')}$ makes $A_{(k')}$ into a finite A-algebra. The ideal $m'$ lies over a maximal ideal m of A (V, § 2, No. 1, prop. 1), and $(A_{(k')})_{m'}$ is isomorphic to a ring of fractions of the regular (resp. normal, resp. reduced) ring $(A_m)_{(k')}$, and is therefore regular (resp. normal, resp. reduced).