The system of regular parameters form a basis of the module of relative differential forms

Question

The problem comes from Serre's "Groupes algébriques et corps de classes", chapter II, section 7. Let $k$ be a field, and $A$ be a $k$-algebra. $A$ is a Noetherian regular ring (every localization at prime is regular) and $\dim A = r$. Let $P$ be a prime of $A$ and $P_P = (t_1, \dots, t_r)$, where $t_1, \dots, t_r$ is a system of regular parameters.

Prove that $\Omega_{A_P/k}$, the module of relative differential forms, has a basis $\rm{d}t_1, \dots, \rm{d}t_r$.

What I tried

I know that the polynomial rings over $k$ have the above properties. But I am not sure if it holds for general regular local rings.

I also know that if $k$ is algebraically closed and $P$ is a maximal ideal of $A$. Then $A/P = k$. Actually, in this special case over varieties, $k \subset A$ and this inclusion induces $k \cong A/P$. Since $A_P$ is a Noetherian regular local ring, $G_P(A_P) = \oplus_{i \geq 0} P^i_P/P^{i+1}_P \cong k[X_1,\dots, X_r]$. This means that although every element in $P^i_P$ may have various forms, it must be like $\sum_{j_1 \leq \dots \leq j_i} c_{j_1, \dots,j_i} t_{j_1} \dots t_{j_i}$ in $P^i_P/P^{i+1}_P$, $c_{j_1, \dots,j_i} \in k$.

Answer 1

This is proved in Hartshorne's Algebraic Geometry II, 8.8. Only need to assume $k$ is perfect. Then $\Omega_{A_P/k}$ is a free $A_P$-module of rank $r$ iff $A_P$ is a regular local ring. We know that $A_P$ is a regular local ring. So $\Omega_{A_P/k}$ is a free $A_P$-module of rank $r$. (8.7) in that book also shows $dt1,…,dtr$ is a basis.