Corollary 2

Let $k$ be a field, $A$ a noetherian $k$-algebra and $J$ an ideal of $A$ contained in the radical of $A$. Suppose that the $k$-algebra $A$ is formally smooth for the $J$-adic topology. Then it is absolutely regular.

Let indeed $k'$ be a finite extension of $k$ and $A'$ the $A$-algebra $A_{(k')}$; it is a question of proving that, for every maximal ideal $m'$ of $A'$, the noetherian local ring $A'_{m'}$ is regular. Now one has $JA' \subset m'$: in fact, the inverse image of $m'$ in $A$ is a maximal ideal of $A$ (V, § 2, No. 1, prop. 1), hence contains $J$. The $k'$-algebra $A'$ is formally smooth for the $JA'$-adic topology (No. 2, prop. 4, b)), and the $k'$-algebra $A'_{m'}$ is formally smooth for the $JA'_{m'}$-adic topology (No. 2, prop. 4, a)), hence also for the $m'A'_{m'}$-adic topology. Let $k_0$ be the prime subfield of $k'$. Then $A'_{m'}$ is formally smooth over $k_0$ for the $m'A'_{m'}$-adic topology (cor. to th. 1 of No. 3); since $\kappa(m')$ is separable over $k_0$, the ring $A'_{m'}$ is regular (cor. 1).