Questions about the sketch of the proof : every Projective variety is complete I cannot understand the following sketch of the proof: Every projective variety is complete using Nakayama Lemma.
The following lecture note on the 279 pages is one recently
that I have referred.

To show this, the lecturer defines the set
$$\mathcal{V}^{*}(I):=\left \{ (x^{*},y) \in \mathbb{P}^n \times Y : f(x^{*},y)=0 \right \}, $$
and since it is a closed set, then it is sufficient to show that $\pi(\mathcal{V}^{*}(I))$ is closed, or equivalently, $\overline{\pi(\mathcal{V}^{*}(I)) }=\pi(\mathcal{V}^{*}(I)) $...... (★).
To begin with, in my thoughts, (★) seems to hold: $\mathcal{V}^{*}(I) $ is closed, i.e consists of finite elements, and its projection is also finite on $Y$ whose topology is Zariski topology, so $\pi(\mathcal{V}^{*}(I)) $ is still closed.
However, the author also said "we must show for every $y \in Y$, there exists $x^{*} \in \mathbb{P}^n$ so that $(x^{*},y) \in \mathcal{V}^{*}(I) $".
In my thoughts, the author seems to verify that $\mathcal{V}^{*}(I)$ is a nonempty set, and to ascertain this, the author takes a maximal ideal $\mathfrak{m} \subset A $ that vanishes at $y$, and sets $J:=\mathfrak{m}B+I $.
Then he says "if $\mathcal{V}^{*}(J)$ is non-empty, we'll be done" (I understand this paragraph is an alternative way of showing that  $\mathcal{V}^{*}(I) \neq \phi$).
Here the author uses Nakayama's Lemma in order to verify $\mathcal{V}^{*}(J)\neq 0$.
I have some questions regarding the proof.

*

*I am little confused on the statement that $\mathcal{V}^{*}(I)$ is closed in $\mathbb{P}^n \times Y$. I believe that it is also closed in the Zariski topology. Since the product topology is the weakest topology (or initial topology), then the product of the Zariski topology is stronger than box topology. Is this the right idea? Anyway, if we take $\mathcal{V}^{*}(I)$ in the Zariski topology, then it is closed in $\mathbb{P}^n \times Y$ because  $\mathcal{V}^{*}(I)$ is a finite set.


*The author showed that $\mathcal{V}^{*}(J)\neq \phi$ (As I said, by guesswork, I firmly believe that this is an alternative proof of  $\mathcal{V}^{*}(I)\neq \phi $).
He proved this by contradiction: assuming that $\mathcal{V}^{*}(J)= \phi$, the author induces $\mathcal{V}^{*}(I)= \phi$. I can indirectly infer this by Nakayama's Lemma:  $J=\mathfrak{m}B+I $ implies $J=I$ where $I$ is a submodule of a finitely generated $A$-module.
Hence $$\phi=\mathcal{V}^{*}(J) \underset{Nakyama }{=} \mathcal{V}^{*}(I).$$
He then gives the contradiction "but it is impossible for a proper ideal $I \subset B$".
I do not understand why the author carried out the proof by applying Nakayama's Lemma, defining the set $J$ etc., because since $I$ is a proper ideal of $B$, the relation $\mathcal{V}^{*}(I)\neq \phi $ is always satisfied.
I have tried reasoning it out as follows:

It must be that $\mathcal{V}^{*}(I)\neq \phi$, since $\mathcal{V}^{*}(I)= \phi$ is
impossible for a proper ideal $I \subset B$. Hence,
$\mathcal{V}^{*}(I)\neq \phi$.

 A: Let's outline how the proof is supposed to go first - parts of your questions are based on misconceptions about the strategy. The idea is that we need to show that for any $Y$ and any nonempty closed $C\subset \Bbb P^n\times Y$, the image of $C$ in $Y$ under the projection is closed. So we want to consider the minimal possible situation: shrink $Y$ so that $Y=\overline{\pi(C)}$, therefore checking whether $\pi(C)$ is closed amounts to checking that $C\to \overline{\pi(C)}$ surjective. This is because $\overline{\pi(C)}$ is the smallest closed subset containing $\pi(C)$, so $\pi(C)$ is closed iff $\pi(C)=\overline{\pi(C)}$.
To show that $\pi(C)=\overline{\pi(C)}$, it's enough to show that there's a point of $C$ in the fiber $\pi^{-1}(y)$ for all $y\in Y=\overline{\pi(C)}$. Writing $C$ as $\mathcal{V}^*(I)$ for some choice of homogeneous ideal $I\subset B$, $\mathcal{V}^*(J)$ is the intersection of $\mathcal{V}^*(I)$ with the fiber $\pi^{-1}(y)$ - the strategy of the proof is to show that if $\mathcal{V}^*(J)$ is empty, then $\mathcal{V}^*(I)$ must be as well, which will be a contradiction because we took $C=\mathcal{V}^*(I)$ to be nonempty.

Aside from the issues addressed above, your post has some other misconceptions I believe should be corrected. Here goes:

To begin with, in my thoughts, (★) seems to hold: $\mathcal{V}^{*}(I) $ is closed, i.e consists of finite elements, and its projection is also finite on $Y$ whose topology is Zariski topology, so $\pi(\mathcal{V}^{*}(I)) $ is still closed.

Yes, $\mathcal{V}^*(I)$ is closed, but this business about "finite elements" does not make sense (what even are "finite elements"?). If you knew that the projection $\pi(\mathcal{V}^{*}(I))$ to $Y$ were closed, you would be done: every closed subset of $\Bbb P^n\times Y$ is of this form. You're getting ahead of yourself here.


*

*I am little confused on the statement that $\mathcal{V}^{*}(I)$ is closed in $\mathbb{P}^n \times Y$. I believe that it is also closed in the Zariski topology. Since the product topology is the weakest topology (or initial topology), then the product of the Zariski topology is stronger than box topology. Is this the right idea? Anyway, if we take $\mathcal{V}^{*}(I)$ in the Zariski topology, then it is closed in $\mathbb{P}^n \times Y$ because  $\mathcal{V}^{*}(I)$ is a finite set.


When we consider the product of varieties, we put the Zariski topology on it, not the product topology. The statement that "$\mathcal{V}^{*}(I)$ is closed in $\mathbb{P}^n \times Y$" is taking $\Bbb P^n\times Y$ to have the Zariski topology. I do not understand why you think " $\mathcal{V}^{*}(I)$ is a finite set" - this is false and you should let go of this belief. Consider  $Y=\Bbb A^2$ and $n=1$, so $B=k[t_1,t_0][x_0,x_1]$ with $x_0,x_1$ homogeneous of degree one and $t_0,t_1$ homogeneous of degree zero. Then $\mathcal{V}^*((t_0))$ is a copy of $\Bbb P^1\times\Bbb A^1$ which is not finite.
