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.


1 Answer 1


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.

  • $\begingroup$ Thanks, that was helpful indeed! $\endgroup$
    – bloxx
    Feb 7, 2022 at 12:08

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