Lemma 15.38.2. Let $k$ be a field and let $(A, \mathfrak m, K)$ be a Noetherian local $k$-algebra. If $k \to A$ is formally smooth for the $\mathfrak m$-adic topology, then $A$ is a regular local ring.

Proof. Let $k_0 \subset k$ be the prime field. Then $k_0$ is perfect, hence $k / k_0$ is separable, hence formally smooth by Algebra, Lemma 10.158.7. By Lemmas 15.37.2 and 15.37.7 we see that $k_0 \to A$ is formally smooth for the $\mathfrak m$-adic topology on $A$. Hence we may assume $k = \mathbf{Q}$ or $k = \mathbf{F}_ p$.

By Algebra, Lemmas 10.97.3 and 10.110.9 it suffices to prove the completion $A^\wedge$ is regular. By Lemma 15.37.4 we may replace $A$ by $A^\wedge$. Thus we may assume that $A$ is a Noetherian complete local ring. By the Cohen structure theorem (Algebra, Theorem 10.160.8) there exist a map $K \to A$. As $k$ is the prime field we see that $K \to A$ is a $k$-algebra map.

Let $x_1, \ldots , x_ n \in \mathfrak m$ be elements whose images form a basis of $\mathfrak m/\mathfrak m^2$. Set $T = K[[X_1, \ldots , X_ n]]$. Note that

$A/\mathfrak m^2 \cong K[x_1, \ldots , x_ n]/(x_ ix_ j)$

and

$T/\mathfrak m_ T^2 \cong K[X_1, \ldots , X_ n]/(X_ iX_ j).$

Let $A/\mathfrak m^2 \to T/m_ T^2$ be the local $K$-algebra isomorphism given by mapping the class of $x_ i$ to the class of $X_ i$. Denote $f_1 : A \to T/\mathfrak m_ T^2$ the composition of this isomorphism with the quotient map $A \to A/\mathfrak m^2$. The assumption that $k \to A$ is formally smooth in the $\mathfrak m$-adic topology means we can lift $f_1$ to a map $f_2 : A \to T/\mathfrak {m}_ T^3$, then to a map $f_3 : A \to T/\mathfrak {m}_ T^4$, and so on, for all $n \geq 1$. Warning: the maps $f_ n$ are continuous $k$-algebra maps and may not be $K$-algebra maps. We get an induced map $f : A \to T = \mathop{\mathrm{lim}}\nolimits T/\mathfrak m_ T^ n$ of local $k$-algebras. By our choice of $f_1$, the map $f$ induces an isomorphism $\mathfrak m/\mathfrak m^2 \to \mathfrak m_ T/\mathfrak m_ T^2$ hence each $f_ n$ is surjective and we conclude $f$ is surjective as $A$ is complete. This implies $\dim (A) \geq \dim (T) = n$. Hence $A$ is regular by definition. (It also follows that $f$ is an isomorphism.) $\square$

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