Show that every morphism \[ f:\AA^1_\CC\longrightarrow\AA^1_\CC\smz \] is constant.
Solution
Write \[ \GG_m=\AA^1_\CC\smz. \] Then \[ \mco(\AA^1_\CC)=\CC[x], \qquad \mco(\GG_m)=\CC[t,t^{-1}]. \]
1. The units of \(\CC[x]\) are exactly the nonzero constants:\[ \CC[x]^\times=\CC^\times. \]
Let \[ u,v\in\CC[x] \] satisfy \[ uv=1. \] Both polynomials are nonzero, so \[ \deg(uv)=\deg u+\deg v. \] Hence \[ 0=\deg1=\deg u+\deg v. \] Since polynomial degrees are nonnegative, \[ \deg u=\deg v=0. \] Thus every unit is a nonzero constant. Conversely every nonzero constant is a unit.
2. The comorphism of \(f\) sends the target coordinate \(t\) to a nonzeroconstant.
The morphism \(f\) corresponds to a homomorphism \[ f^*:\CC[t,t^{-1}]\longrightarrow\CC[x]. \] The element \(t\) is a unit in the source ring, so \[ f^*(t) \] must be a unit in \(\CC[x]\). By step 1, there is a \[ c\in\CC^\times \] such that \[ f^*(t)=c. \]
3. The morphism \(f\) is the constant morphism with value \(c\).
For a point \[ a\in\AA^1_\CC, \] the target coordinate evaluated at \(f(a)\) is \[ t(f(a)) = (f^*t)(a) = c. \] Thus \[ f(a)=c \] for every \(a\), so \(f\) is constant.
4. Q.E.D.
Steps <1>1–<1>3 show that every morphism from the affine line to the punctured affine line has constant image.