I'm working on this exercise:
Show that if X is a quasicompact scheme, then every point has a closed point in its closure. Show that every nonempty closed subset of X contains a closed point of X.
I know this question was asked a couple of times here (I'll come back to one of those questions at the end), but I was trying to do this problem by myself, and here's what I got.
Let $X$ be a quasi-compact scheme. Every point of $X$ has an open neighborhood of the form $Spec(A_i)$ for some $A_i$. These neighborhoods cover $X$, and since $X$ is quasi-compact, we can write $X=\cup_{i=1}^k Spec(A_i)$.
Now let $x\in X$ be a point. I need to show that there is a closed point $P$ of $X$ in the closure of $x$ in $X$. My $x$ lies in some $Spec(A_i)$, WLOG assume $x\in Spec(A_1)$. I know that the closure of $x$ in $Spec(A_1)$ is $V(x)$. My first question is whether the closure of $x$ in $X$ is equal to the closure of $x$ in $Spec(A_1)$ (which is $V(x)$), or is this false in general?
Let's assume for a moment that this is true. I'm looking for a closed $P\in X$. This $P$ must lie in some $Spec(A_i)$. It must have the property that $\{P\}=V(P)$ and also $P\in V(x)=\{I\in Spec(A_1):x\subset I\}$, so $i$ must be equal to $1$. But then we can just take $P$ to be a maximal ideal that contains $x$ (if $x$ isn't maximal itself, otherwise take $x=P$). But then quasi-compactness isn't used anywhere, so this argument must be false. Is it because my assumption about closures is false?
Now, there's a solution in this answer:
Notice it is enough to show that every closed subset $Z$ of $X$ has a closed point. Observe a point $p \in Z$ is closed in $Z$ if and only if it is closed in $X$ so it suffices to show that $Z$ has a closed point. But $Z$ is also a quasicompact scheme so we reduce to the case of showing that a quasicompact sheme $X$ has a closed point. For this, say $X = U_1 \cup \dots \cup U_n$ is an irredundant decomposition of $X$ as a union of open affines. We can then pick a point $p \in U_1$ that is closed in $U_1$ and such that $p \notin U_j$ for $j \neq 1$. Because $p \in (U_2 \cup \dots \cup U_n)^c$ the closure is also in $(U_2 \cup \dots \cup U_n)^c$. It is then easy to check that the closure of $p$ in $X$ is $p$.
But there are several things I don't understand.
Why is it enough to show that every closed subset $Z$ of $X$ has a closed point? And doesn't the second sentence in the quoted text say essentially the same as the first sentence says? Or am I misunderstanding something?
Why does an irredundant decomposition of $X$ exist? I only know that a closed subset of a Noetherian topological space has such a decomposition, but does $X$ have to be a Noetherian topologcal space? There's some discussion of this in the comments of the cited question, but there's no any comprehensive explanation.
The last point, "It is then easy to check that the closure of $p$ in $X$ is $p$", is I guess something like my very first question above (whether the closure in $X$ is equal to the closure in $Spec(A_1)$). I still don't see why this is true in this particular case. The closure of $p$ is the intersection of all closed sets in $X$ that contain $p$. At this point I realized that I may be misunderstanding what the topology on $X$ is. It's supposed to be an abstract topological space, but at the same time we know what the closed sets in its open cover look like. Does it mean that any closed subset of $X$ is also of the form $V(I)=\{Q\in U_i:I\subset Q\}$ for some $i$?
