Well, it is a hard time for me before coming back to any topic about algebraic geometry. I have to admit that I did not understand well about those very abstract constructions, such as the definition of prevariety, or the diagonal mapping in Chapter V. It makes me fear Algebraic Geometry for a long time, before reaching to any understanding about projective varieties. Fortunately, my course in Algebraic Geometry forces me to play with these notions, or I will die soon, and this motivation let me have a further reading on Gathmann's note, Chapter VI, VII about projective varieties. It actually completes my understanding on Chapter V.
What makes these things are so abstract? The answer is projective varieties. Though the projective varieties are compact under the ground field $\mathbb{C}$ with usual Euclidean topology, and it is also compact under Zarisky topology. However, any affine variety is also compact in Zarisky topology, then the statement that any projective variety is compact is quite trivial. So, we must make difference between projective varieties and affine varieties. And the main difference is that any projective variety is complete.
We recall that an affine variety $(Y,\mathscr{O}_Y)$ is an isomorphic image as ring spaces of an affine variety as usual sense. A prevariety is defined as a ring space with finite open covers as affine varieties. And a prevariety is a variety if it is closed under the diagonal mapping, i.e. $\Delta_X=\{(x,x)|x\in X\}$ is closed in $X\times X$. We also recall that $X\times X$ is not Cartesian product, it is product of prevariety by category construction. And in Chapter VII, it is proved that the projection $\mathbb{P}^n\times Y\rightarrow Y$ is closed, for any variety $Y$. That means, it transfers a closed subset of $\mathbb{P}^n\times Y$ to a closed subset of $Y$. And the projective variety is complete in this sense, i.e. a variety $X$ is complete if for any variety $Y$, the projection map $\pi: X\times Y\rightarrow Y$ is closed. Note that any subvariety of a complete variety is also complete. By this, we can point out that $\mathbb{A}^n$ is not complete. Let us look at an example
Example 1. Consider $\mathbb{A}^1\times \mathbb{A}^1$, the and its subvariety $Y=\{(x,y)|xy-1=0\}$. The projection map $\pi$ will map $Y$ to $\mathbb{A}^1\setminus\{0\}$, which is not closed. Hence, $\mathbb{A}^1$ is not complete.
If we replace $\mathbb{P}^1$, instead of $\mathbb{A}^1$, we can complete the "missing point", and the projective varieties are complete in this sense. This is a remarkable property of projective varieties. By this, the author prove that with any morphism $f$ from a connected, complete variety $X$ to a variety $Y$, $f(X)$ is closed. By proving this, he uses the property of graph mentioned in Chapter V. When I read the note to this point, I did not understand why we need to generalize the definition of variety and why study the graph. Without instructors, it is a big loss.
Then he next prove the amazing result, that any global map (regular function) on a connected complete variety is just constant. In particular, any morphism from $\mathbb{P}^n$ to $\mathbb{A}^1$ is just constant. It is again, a big difference with affine case, because for affine case, we know that the ring of regular functions of an affine variety is isomorphic to the coordinate ring, and hence, it consists of much more than only constant functions. Let me explain how the proof work. A regular function for variety $X$ is defined as a morphism from $X$ to $\mathbb{A}^1$, and we can extend the domain to $\mathbb{P}^1=\mathbb{A}^1\cup\{\infty\}$. And hence, by above result $X$ is complete implies that $f(X)$ is closed in $\mathbb{P}^1$, and $f(X)$ is a proper subset of $\mathbb{P}^1$, since the point $\infty$ does not have inverse image. However, proper closed subset of $\mathbb{P}^1$ consists of finitely many points. And $X$ is connected implies that $f(X)$ is also connected. This yields $f(X)$ is just a point! Very beautiful proof.
Besides, some non-trivial morphisms between varieties are constructed, which are Segre's embedding and Veronese's embedding. This reminds me the spring course in Hanoi on Algebraic Geometry, at that time, I did not understand why we need these morphisms, but now it appears to be useful in the context. The Segre's embedding is used to prove any projective variety is variety, and a result from Chapter V implies that it is sufficient for us to prove for $\mathbb{P}^n$ itself. And by Lemma 4.6, it is sufficient to prove this for affine cover of $\mathbb{P}^n$. We embed $\Delta_{\mathbb{P}^n}=\{(x_0:...:x_n),(\lambda x_0:...\lambda x_n)|\lambda\in k^\times\}$ to a larger projective space $\mathbb{P}^N$, where $N=(n+1)^2-1$, which coordinate $(z_{i,j})$, where $z_{i,j}=x_i(\lambda x_j)$. This defines a closed subset of $\mathbb{P}^N$, and we are done!
The Veronese's embedding is to linearize the polynomial mapping, and its application is to prove that if $f$ is a homogeneous polynomial of $k[x_0,...,x_n]$ of degree $d$ then $\mathbb{P}^n\setminus V(f)$ is an affine variety. Let's us look at closer at the proof. Let $n,d>0$, and $N={n+d\choose n}-1$, and $f_0,...,f_N\in k[x_0,...,x_n]$ be homogeneous monomials of degree $d$, then the map $F:\mathbb{P}^n\rightarrow \mathbb{P}^N$ defined by $F(x)=(f_0(x):...:f_N(x))$ is an embedding from $\mathbb{P}^n$ to $\mathbb{P}^N$, and $\mathbb{P}^n\cong F(\mathbb{P}^n)$. Also, any subvariety of $\mathbb{P}^n$ can be sent to subvariety of $F(\mathbb{P}^n)$, with all terms now is of degree $d$. Now, let $f$ as above, we consider $V(f)$ via the Veronese's embedding, it is now a linear polynomial in $F(\mathbb{P}^n)$. And what we do now is easy, just transform $f$ to the form $f=x_0$ via the projective automorphism, and $\mathbb{P}^n\setminus V(f)$ is now is actually $\mathbb{A}^n$ via projective automorphisms. Hence, it must be an affine variety.
These lines above cannot describe all the content and the beauty of Chapter VI, VII on Gathmann's note, but it may be helpful if someone finds it difficult to digest very abstract construction on Chapter IV, V. Go ahead, and continue reading!
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