Let \(k\) be the ground field. Find an everywhere regular differential \(n\dash\)form on \[ \AA^n_k. \]
Solution
Let \[ x_1,\ldots,x_n \] be the standard coordinate functions on \(\AA^n_k\).
1. The cotangent sheaf of affine \(n\)-space is free:\[ \Omega^1_{\AA^n_k/k} \cong \bigoplus_{i=1}^n \mco_{\AA^n_k}\,dx_i. \]
The coordinate ring of affine space is \[ A=k[x_1,\ldots,x_n]. \] The module of Kähler differentials of a polynomial algebra is free: \[ \Omega^1_{A/k} \cong \bigoplus_{i=1}^n A\,dx_i. \] Sheafifying this free module on \[ \AA^n_k=\Spec A \] gives the displayed trivialization of the cotangent sheaf.
2. The top exterior power is a free rank-one sheaf generated by\[ \omega \coloneqq dx_1\wedge\cdots\wedge dx_n. \]
Taking the \(n\)th exterior power of the free rank-\(n\) sheaf in step 1 gives \[ \Omega^n_{\AA^n_k/k} \coloneqq \bigwedge^n \Omega^1_{\AA^n_k/k} \cong \mco_{\AA^n_k}. \] Under this isomorphism the standard generator is \[ dx_1\wedge\cdots\wedge dx_n. \] Thus \(\omega\) is a global section of the top differential sheaf.
3. The form\[ \boxed{ \omega=dx_1\wedge\cdots\wedge dx_n } \]is everywhere regular and nowhere vanishing.
Step 2 identifies \(\omega\) with the element \[ 1\in\mco_{\AA^n_k} \] under a trivialization of the line bundle \[ \Omega^n_{\AA^n_k/k}. \] Hence its germ at every point generates the corresponding stalk. In particular, it is regular at every point and vanishes nowhere.
4. Equivalently, the canonical bundle of affine space is trivial:\[ \boxed{ K_{\AA^n_k} \cong \mco_{\AA^n_k}. } \]
The canonical bundle of a smooth \(n\)-fold \(X\) is the determinant of the cotangent bundle: \[ K_X=\det T^*(X). \] For affine \(n\)-space this determinant is precisely \[ \Omega^n_{\AA^n_k/k}, \] which step 2 trivializes by \(\omega\).
5. Q.E.D.
Step 3 gives the requested everywhere regular differential \(n\)-form, and step 4 records its canonical-bundle interpretation.