Lemma 5

Let k be an algebraically closed field, A a finitely generated k-algebra whose spectrum is connected, n an integer and (x_1, \ldots , x_n) a finite sequence of elements of A . Suppose that A is of dimension n and that the differentials dx_1, \ldots , dx_n generate the A-module \Omega_k(A) . Then the ring A is an integral domain and regular and the A-module \Omega_k(A) is free with basis (dx_1, \ldots , dx_n) .

Let m be a maximal ideal of A such that \dim(A_m) = n . One has [A/m : k] = 1 (V, § 3, No. 3, prop. 1 (iii)), hence A = m \oplus k1_A . Let p and q denote the corresponding projections. For $a$ and $b$ in $A$, one has

whence $p(ab) \equiv p(a)q(b) + q(a)p(b) \pmod{\mathfrak{m}^2}$. Consequently the mapping $\delta : A \to \mathfrak{m}/\mathfrak{m}^2$ which associates to each element $x$ of $A$ the class of $p(x)$ modulo $\mathfrak{m}^2$ is a $k$-derivation of $A$ into the $k$-vector space $\mathfrak{m}/\mathfrak{m}^2$. There therefore exists a $A$-linear mapping $\phi : \Omega_k(A) \to \mathfrak{m}/\mathfrak{m}^2$ such that $\delta(x) = \phi(dx)$ for every $x \in A$. Since $\delta$ is surjective, the $\phi(dx_i)$ generate the $A/\mathfrak{m}$-vector space $\mathfrak{m}/\mathfrak{m}^2$, and one has $[\mathfrak{m}/\mathfrak{m}^2 : A/\mathfrak{m}] \leq n = \dim(A_{\mathfrak{m}})$. It follows that $A_{\mathfrak{m}}$ is regular and that the images of the $dx_i$ form a basis of the $A/\mathfrak{m}$-vector space $A/\mathfrak{m} \otimes_A \Omega_k(A)$.

Let us now prove that the ring $A$ is an integral domain and regular. There exists a minimal prime ideal $\mathfrak{q}$ of $A$ such that $\dim(A/\mathfrak{q}) = n$. For every maximal ideal $\mathfrak{m}$ of $A$ containing $\mathfrak{q}$, one has $\dim(A_{\mathfrak{m}}) = n$ (VIII, § 2, No. 4, cor. 2 to th. 3), so that $A_{\mathfrak{m}}$ is regular by what we have just seen. In particular $A_{\mathfrak{m}}$ is an integral domain, which implies $\mathfrak{q}A_{\mathfrak{m}} = 0$. Since this holds for all the maximal ideals $\mathfrak{m}$ of $V(\mathfrak{q})$, one deduces that one has $\operatorname{Supp}(\mathfrak{q}) \cap V(\mathfrak{q}) = \varnothing$. But $\operatorname{Spec}(A)$ is connected, $V(\mathfrak{q})$ is nonempty and one has $\operatorname{Supp}(\mathfrak{q}) \cup V(\mathfrak{q}) = \operatorname{Spec}(A)$ (II, § 4, No. 4, prop. 16). One deduces that $\operatorname{Supp}(\mathfrak{q}) = \varnothing$, whence $\mathfrak{q} = 0$, which means that $A$ is an integral domain. One then has $\dim(A_{\mathfrak{m}}) = n$ for every maximal ideal $\mathfrak{m}$ of $A$; applying the first part of the proof, one deduces that $A$ is regular.

Finally, let $\sum_{i=1}^n a_i dx_i = 0$ be a linear relation between the $dx_i$ with coefficients in $A$. If the $a_i$ are not all zero, there exists an index $i$ and a maximal ideal $\mathfrak{m}$ of $A$ such that $a_i$ does not belong to $\mathfrak{m}$ (V, § 3, No. 3, Prop. 1, (iii) and (iv)); but this contradicts the fact proved above that the classes of $dx_i$ in $(A/\mathfrak{m}) \otimes_A \Omega_k(A)$ are linearly independent.