Show that \(\GL_n(\CC)\) is an affine variety.
Solution
Let \[ M_n(\CC)\cong\AA^{n^2}_\CC \] have coordinate functions \[ x_{ij}, \qquad 1\leq i,j\leq n, \] and write \[ \Delta=\det(x_{ij}). \]
1. The subset\[ Y = V(t\Delta-1) \subseteq \AA^{n^2+1}_\CC \]is an affine variety.
Its coordinate ring is \[ \CC[x_{ij},t]/(t\Delta-1). \] The homomorphism \[ \CC[x_{ij},t] \longrightarrow \CC[x_{ij}]_\Delta \] defined by \[ x_{ij}\longmapsto x_{ij}, \qquad t\longmapsto\Delta^{-1} \] is surjective and has kernel \((t\Delta-1)\). Therefore \[ \CC[Y] \cong \CC[x_{ij}]_\Delta. \] The localization of the domain \(\CC[x_{ij}]\) is again a domain, so \((t\Delta-1)\) is prime. Hence \(Y\) is an affine variety.
2. Projection to the matrix coordinates gives a bijection\[ \pi:Y\longrightarrow\GL_n(\CC). \]
If \[ (A,t)\in Y, \] then \[ t\det A=1, \] so \(\det A\neq0\) and \(A\in\GL_n(\CC)\).
Conversely, if \(A\in\GL_n(\CC)\), then \[ \left(A,\frac1{\det A}\right)\in Y. \] The value of \(t\) is uniquely determined by \(A\), so \(\pi\) is bijective.
3. The bijection in step 2 is an isomorphism of varieties.
The projection \[ \pi:Y\longrightarrow M_n(\CC) \] is the restriction of a coordinate projection, hence is a morphism.
Its inverse is \[ \GL_n(\CC)\longrightarrow Y, \qquad A\longmapsto \left(A,\frac1{\det A}\right). \] Now \[ \GL_n(\CC) = D(\Delta) \subseteq \AA^{n^2}_\CC, \] and \(\Delta^{-1}\) is a regular function on the distinguished open set \(D(\Delta)\). Hence the inverse is a morphism. Therefore \[ Y\cong\GL_n(\CC). \]
4. Consequently,\[ \boxed{ \GL_n(\CC)\text{ is an affine variety} } \]with coordinate ring\[ \boxed{ \CC[\GL_n] \cong \CC[x_{ij},\Delta^{-1}]. } \]
Step 1 shows that \(Y\) is affine, and step 3 identifies \(Y\) with \(\GL_n(\CC)\). The coordinate-ring description is the isomorphism established in step 1.
5. Q.E.D.
Step 4 is the required conclusion.