Show that \(\AA^1\) is not isomorphic to \(X = V(xy-1)\).
Solution
1. The coordinate ring of the hyperbola is\[ \boxed{ \CC[X]\cong\CC[x,x^{-1}]. } \]
One has \[ \CC[X] = \CC[x,y]/(xy-1). \] The homomorphism \[ \CC[x,y] \longrightarrow \CC[x,x^{-1}], \qquad x\longmapsto x, \qquad y\longmapsto x^{-1} \] is surjective and has kernel \((xy-1)\). Hence it induces the displayed isomorphism.
2. The only units in \(\CC[t]\) are the nonzero constants:\[ \CC[t]^\times=\CC^\times. \]
If \[ f(t)g(t)=1 \] in \(\CC[t]\), then \[ \deg f+\deg g=0. \] Thus both degrees are zero, so both polynomials are nonzero constants. Conversely every nonzero constant is a unit.
3. The coordinate ring \(\CC[X]\) has a nonconstant unit.
Under step 1, \[ x\in\CC[x,x^{-1}] \] is a unit with inverse \(x^{-1}\). It is not constant.
4. There is no \(\CC\)-algebra isomorphism\[ \CC[X]\xrightarrow{\sim}\CC[t]. \]
Suppose \[ \varphi:\CC[X]\xrightarrow{\sim}\CC[t] \] were a \(\CC\)-algebra isomorphism. By step 3, \(x\) is a unit, so \[ \varphi(x) \] is a unit in \(\CC[t]\). Step 2 gives \[ \varphi(x)=c \] for some \(c\in\CC^\times\).
Because \(\varphi\) is a \(\CC\)-algebra map, \[ \varphi(x-c)=0. \] But \[ x-c\neq0 \] in \(\CC[x,x^{-1}]\), contradicting injectivity of \(\varphi\). Therefore no such isomorphism exists.
5. Hence\[ \boxed{ \AA^1_\CC\not\cong V(xy-1). } \]
An isomorphism of affine varieties induces an isomorphism of their coordinate rings. Step 4 shows that the coordinate rings \[ \CC[t] \qquad\text{and}\qquad \CC[x,y]/(xy-1) \] are not isomorphic as \(\CC\)-algebras. Therefore the affine varieties are not isomorphic.
6. Q.E.D.
Step 5 is the required conclusion.