Let \[ f\in k[x_1,\ldots,x_n] \] be nonconstant. Why is the complement \[ \AA^n\sm V(f) \] an affine variety?
What is its coordinate ring?
Solution
Put \[ A=k[x_1,\ldots,x_n] \] and \[ D(f)=\AA^n\sm V(f). \]
1. The hypersurface\[ Y = V(tf-1) \subseteq \AA^{n+1}_{x_1,\ldots,x_n,t} \]is an affine variety with coordinate ring\[ \boxed{ \mco(Y) \cong A_f. } \]
The coordinate ring of \(Y\) is \[ \mco(Y) = A[t]/(tf-1). \] There is a homomorphism \[ \Phi:A[t]/(tf-1)\longrightarrow A_f \] which is the identity on \(A\) and sends \[ t\longmapsto f^{-1}. \] It is well-defined because \(f^{-1}f=1\), and it is surjective because \(A_f\) is generated by \(A\) and \(f^{-1}\).
Conversely, in \(A[t]/(tf-1)\) the image of \(f\) is invertible with inverse \(t\). By the universal property of localization, the quotient map \[ A\longrightarrow A[t]/(tf-1) \] extends uniquely to \[ \Psi:A_f\longrightarrow A[t]/(tf-1). \] The maps \(\Phi\) and \(\Psi\) are inverse because both fix \(A\) and interchange \(t\) with \(f^{-1}\). Hence \[ A[t]/(tf-1)\cong A_f. \]
Since \(A\) is a domain, its localization \(A_f\) is a domain. Thus \((tf-1)\) is prime and \(Y\) is an irreducible affine variety.
2. Projection onto the first \(n\) coordinates restricts to an isomorphism\[ \pi:Y\xrightarrow{\sim}D(f). \]
If \[ (a_1,\ldots,a_n,t)\in Y, \] then \[ t\,f(a_1,\ldots,a_n)=1, \] so \[ f(a_1,\ldots,a_n)\ne0. \] Hence \(\pi(Y)\subseteq D(f)\).
Conversely, if \[ a=(a_1,\ldots,a_n)\in D(f), \] then \[ \left(a_1,\ldots,a_n,\frac1{f(a)}\right)\in Y. \] Thus \(\pi\) is bijective, with inverse \[ a \longmapsto \left(a,\frac1{f(a)}\right). \] The projection is regular, and its inverse is regular on \(D(f)\) because \(f\) is nowhere zero there. Therefore \(\pi\) is an isomorphism.
3. The hypersurface complement is affine and\[ \boxed{ \mco\bigl(\AA^n\sm V(f)\bigr) \cong k[x_1,\ldots,x_n]_f. } \]
By step 2, \[ D(f)\cong Y. \] Step 1 shows that \(Y\) is affine and computes its coordinate ring as \(A_f\). Transporting this structure across the isomorphism gives the result; see also PR-WZGOQ.
4. Q.E.D.
Steps <1>1–<1>3 identify the complement with an affine hypersurface and compute its coordinate ring.