Smooth over an algebraically closed field equals regular

Let \(k\) be a field and \(X\) a scheme locally of finite type over \(k\). If \(X \to \Spec k\) is smooth of relative dimension \(n\) then \(X\) is regular of dimension \(n\). The converse holds when \(k\) is perfect, in particular when \(k = \kbar\), and fails otherwise.

(Jacobian criterion)

Let \(k\) be an algebraically closed field, \(X = V(f_1, \ldots, f_r) \subseteq \AA^m_k\), and \(x \in X\) a closed point with \(\dim \OO_{X,x} = n\). Let \(J(x) = (\partial f_i / \partial x_j (x))\) be the Jacobian matrix at \(x\).

1. \(\dim_k \mfm_x / \mfm_x^2 = \dim_k \Omega_{X/k} \otimes \kappa(x) = m - \operatorname{rank} J(x)\).

2. \(X\) is regular at \(x\) if and only if \(\operatorname{rank} J(x) = m - n\).

[@Har10a, Theorem I.5.1, Proposition II.8.7]

(Smoothness over a field)

Let \(k\) be a field and \(X\) a scheme locally of finite type over \(k\), of pure dimension \(n\). The following are equivalent.

1. \(X\) is smooth over \(k\).

2. \(\Omega_{X/k}\) is locally free of rank \(n\).

3. \(X\) is covered by open subschemes isomorphic to open subschemes of \(\Spec k[x_1, \ldots, x_m]/(f_1, \ldots, f_r)\) whose Jacobian matrix has rank \(m - n\) at every point.

4. \(X_{\bar{k}} = X \times_k \Spec \bar{k}\) is regular.

If \(k\) is perfect, these are also equivalent to: \(X\) is regular.

[@Har10a, Theorem II.8.15, Theorem III.10.2]