Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading the first several pages of GIT by Mumford, and I need some clarification of one requirement in the definition of geometric quotient (c.f. Definition 0.4, GIT):

Suppose a group scheme $G/S$ acts on scheme $X/S$ by $\sigma$, where $G,X$ are schemes over $S$. If a pair $(Y, \phi)$ consisting of a scheme $Y$ over $S$ and an $S-$morphism $\phi: X \to Y$ is a geometric quotient, then one requirement is " the fundamental sheaf $\mathcal{O}_Y$ is the subsheaf of $\phi_*(\mathcal{O}_X)$ consisting of invariant functions" i.e.

If $f \in \Gamma(U,\phi_*(\mathcal{O}_X))= \Gamma(\phi^{-1}(U),\mathcal{O}_X)$, then $f \in \Gamma(U, \mathcal{O}_Y)$ if and only if:

$$\begin{matrix} G \times \phi^{-1}(U)&\stackrel{\sigma}{\longrightarrow}&\phi^{-1}(U)\\ \downarrow{p_2}&&\downarrow{F}\\ \phi^{-1}(U)&\stackrel{F}{\longrightarrow}&\mathbb{A}^1 \end{matrix} $$ commutes (where $F$ is the morphism defined by $f$, and $\mathbb{A}^1 = \operatorname{Spec}(\mathbb{Z}[x])$)

My questions is how to make sense of this $F$ ?

share|cite|improve this question
up vote 2 down vote accepted

A regular function $f \in \mathcal O_X(U)$ defines a morphism $F: U \rightarrow \mathbb A^1$. In the classical case of a variety over an algebraically closed field, the affine line is identified with the base field $k$ this map is just evaluation.

More generally, if $f \in \mathcal O_X(U)$ is an element of the structure sheaf over an open affine subset $U$, then there is a natural ring homomorphism $\mathbb Z[X] \rightarrow \mathcal O_X(U)$ mapping $X$ to $f$. Taking Spec gives your morphism $F: U \rightarrow \mathbb A^1$. In your case your regular function is $f \circ \phi$ defined on $\phi^{-1}U$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.