I'm currently studying Mumford's Geometric Invariant Theory. Unfortunately, I'm stuck understanding a detail in Theorem 1.1.

(Partial) Claim of Theorem 1.1

Let $X = \operatorname{Spec} R$ be an affine scheme over a characteristic-zero field $k$ and consider a reductive group action $G \curvearrowright X$. Then there is a categorical quotient of $X$ by $G$, induced by the inclusion $\phi\colon R_0 \hookrightarrow R$ ($R_0$ being the ring of invariants).

Part where I'm stuck

Now, he wants to use Remark 6 on page 8 to derive the claim.

Remark 6 on page 8

Let $G$ be a group scheme acting on a scheme $X$ (via $\sigma\colon G\times X \to X$). Furthermore, let $\phi\colon X \to Y$ be a morphism of schemes. The following conditions together imply that $(Y, \phi)$ is a categorical quotient of $G\curvearrowright X$:

  1. $\phi\circ\sigma = \phi\circ p_2$ (where $p_2$ is the projection to the second component of the cartesian product)
  2. $\phi$ induces an injective morphism of schemes $\mathcal{O}_y \hookrightarrow \phi_*\mathcal{O}_X$ that has the subsheaf of invariants as its image.
  3. If $W$ is an invariant closed subset of $X$, then $\phi(W)$ is closed in $Y$; if $(W_i)$ is a family of invariant closed subsets of $X$, then $\phi\left(\bigcap_i W_i\right) = \bigcap_i \phi(W_i)$.

Mumford's proof sketch for condition (3)

I understand how Mumford proves the first two conditions, but I don't get the third one. Mumford shows that for closed invariant sets $(W_i)$,

$$ \overline{\phi\left(\bigcap_i W_i\right)} = \bigcap_i \overline{\phi(W_i)}. $$

Now, it suffices to prove that $\phi(W_1)$ is closed.

Mumford then prompts me to apply the above equation to the case where $W_1$ is arbitrary and $W_2$ is the preimage of a closed point of $Y$, and claims that this implies that $\phi(W_1)$ is closed.

This last claim is the one I don't understand.

My thoughts

  • In the situation where $W_2$ is the preimage of a closed point $y \in Y$, both sides of the equation we already have are subsets of $\phi(\phi^{-1}(y))$. If this set is empty, the equation just states $\varnothing = \varnothing$, not very helpful.
  • Since the left hand side is either $\overline{\{y\}}$ or $\overline{\varnothing}$, the closure on the left-hand side is irrelevant in the situation of the previous bullet point.
  • Let's assume that $\phi$ is surjective. Then the right-hand side is $\{y\}$ if and only if $y\in \overline{\phi(W_1)}$ and $\varnothing$ otherwise. So if $y\in \overline{\phi(W_1)}$, the equation implies $y\in \bigcap_i \phi(W_i) \subseteq \phi(W_1)$. $\phi(W_1)$ contains all closed points of its closure. However, what does this help, and is $\phi$ really surjective?

Hints for how to expand Mumford's proof sketch are highly appreciated.


The idea is actually rather simple. Let $W$ be a closed $G$-invariant subset of $X$, and $y$ a closed point that is not in $\phi(W)$. Note that $\phi^{-1}(y)$ is also closed and $G$-invariant. We already know that $$ \overline{\phi(W\cap\phi^{-1}(y))}=\overline{\phi(W)}\cap\{y\}. $$ But the LHS is empty which means that $y\notin\overline{\phi(W)}$. This tells us that $\phi(W)$ must be closed. Otherwise, we may find a closed point $y\in\overline{\phi(W)}\setminus\phi(W)$ that is not contained in $\overline{\phi(W)}$ by the argument above which is absurd. See the proof of Theorem 6.1 in Dolgachev's book "Lectures on Invariant Theory" for more details.

FYI, the morphism $\phi$ is submersive which means that it is surjective and the induced topology on its image is the quotient topology. One could show that any categorical quotient satisfying the 3 conditions in remark 6 is submersive.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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