Dimension Theory from The Geometric Point of View (Gathmann's note-To be Continued-Update)

From the main theorem for projective varieties, that all projective varieties are complete, we establish further results about dimension of projective space. As applications, we amazingly prove some beautiful results in commutative algebra

1. $\dim \mathbb{A}^n=n$, or algebraically, $\dim k[x_1,...,x_n]=n$.

2. Krull's principal ideal theorem.

3. Any morphism from $\mathbb{P}^n$ to $\mathbb{P}^m$ where $n>m$ is constant.

Different from the algebraic methods, the geometric methods seem to be more natural and beautiful. We begin with the some notions about projection map, which gives us the surjective morphism with finite fibers from $X$ to some $\mathbb{P}^m$ for some projective variety $X\subset \mathbb{P}^n$. We then obtain that any homogeneous polynomial is integral over $k[x_0,...,x_n]/I(X)$, and prove the geometric going-up theorem for projection map $\pi$, from $X$ to $\mathbb{P}^{n-1}$. Assume that the $Y$ is a closed subvariety of $X$, and $\pi(X)=\pi(Y)$, then $X=Y$. This ensures the surjective morphism with finite fibers has the same dimension with the image. And hence, the statement that $\dim\mathbb{P}^n=n$ now follows.

We next turn to the very important proposition, which is the projective version of Krull's principal ideal theorem, that a proper hypersurface on a projective varieties $X$ has dimension $\dim X-1$. Continue from the projective point of view, we next prove that the dimension of any non-empty open subset of $X$ is equal to $\dim X$. The key point in the proof is the projective version of Krull's theorem. As applications, it is easy to see that $\dim \mathbb{A}^n=n$, together with the affine case for Krull's theorem. We finish this note with the proof of the beautiful statement, that any morphism from a projective space to another projective space, with smaller dimension, is always constant.

Enjoy!

Update. 

I make a clarification for Proposition 6, and then solve 3 (Application 2), which is briefly presented in the note. In Proposition 6, we first use Veronese's embedding to linearize $f=f_0$. Assume that $\dim X = m$, and $X\subset \mathbb{P}^n$ we want to construct $f_0,...,f_m$ linearly independent, and does not vanish simultaneously on $X$ so that we can find a morphism $\pi$ from $X$ to $\mathbb{P}^m$ defined by these polynomial, and then, use projective automorphism to transfer $f_i$ to $x_i$, to consider $\pi$ as a surjective morphism. From this, we prove that $\dim (X\cap V(f))=\dim X-1$.

Let $X_0=X$, and $X_i=X_i\cap V(f_i)=X\cap V(f_0)\cap...\cap V(f_i)$, we want to choose $f_{i+1}$ such that it is linearly independent to $f_0,...,f_i$.  First, $\dim X_0\le n$. If $X_{i+1}=\emptyset$, then we can choose arbitrary $f_{i+1}$ such that it is linearly independent to $f_0,...,f_i$. Note that we can choose such $f_{i+1}$ since $\dim_k(\text{k-vector space of homogeneous linear function}) = n+1\ge m+1$, hence we can choose such $f_{i+1}$. If $X_{i+1}\ne 0$, we choose $f_{i+1}$ such that it does not vanish on the whole $X_{i+1}$, and from this one can see $f_{i+1}$ is linearly independent to $f_0,...,f_i$ because all $f_j$ vanish on $X_{i+1}$, for $0\le j\le i$. In short, we can choose such $f_i$. Also, $\dim X_i>\dim X_{i-1}$, and hence, $X_{m+1}=\emptyset$, this implies $X\cap V(f_0)\cap...\cap V(f_m)=\emptyset$, or $f_i$ does not vanish simultaneously on $X$.

And $x\mapsto (f_0(x),...,f_m(x))$ defines a morphism from $X$ to $\mathbb{P}^m$. Via projective automorphisms, and due to the fact that $f_i$ are linearly independent, we can assume that $f_i=x_i$. If $X=\mathbb{P}^n$, then the morphism is just the identity map, and hence, is surjective. If $X\subsetneq \mathbb{P}^n$, then we can assume that $P=(0:...:0:1)$ does not lie on $X$, and the map $\pi$ is now a projection from $X$ to $\mathbb{P}^m$, and due to a remark before Theorem V, $\dim X=\dim \pi(X)=m$, and $\pi(X)$ is a irreducible, closed subset of $\mathbb{P}^m$, this does imply $\pi$ is surjective. And in any case, $\pi$ is the surjective map.

Now if $X_m=\emptyset$, $X\cap V(f_0)\cap...\cap V(f_{m-1})=\emptyset$, that means the point $(0:...:0:1)$ of $\mathbb{P}^m$ has no inverse image. This is the contradiction to the surjective property of $\pi$. Also, $\dim X_i>\dim X_{i-1}$, and hence, $X_{m+1}=\emptyset$, we finally obtain the chain of closed subset in $X$.

$$\emptyset \ne X_m\subsetneq ...\subsetneq X_1\subsetneq X_0=X$$

And now, we can see $\dim X_1=\dim (X\cap V(f))=m-1$. And this finishes the proof of the proposition,

We will use it to prove 3, that any morphism $\theta$ from $\mathbb{P}^n$ to $\mathbb{P}^m$, where $n>m$ is constant. Well, we want to prove first that if $\theta$ is not constant, then it is non-constant locally. Assume that for all $a\in\mathbb{P}^m$, all open neighborhood $U_a$ of $a$, such that $\theta$ can be represented as constant functions. In particular, $f$ is constant in open affine covers $\mathbb{A}^m$ of $\mathbb{P}^m$, with non-empty intersection. This implies $\theta$ is a constant function on $\mathbb{P}^m$, a contradiction to our assumption. Now, let $U_a\subset\mathbb{P}^m$ be an open subset of $\mathbb{P}^m$, neighborhood of a point $a\in\mathbb{P}^m$, where $\theta$ is not constant, then $\theta$ can be represented locally by $(\frac{g_0}{h_0}:...:\frac{g_m}{h_m})$, where $\deg g_i=\deg h_i$ are homogenous polynomial in $k[x_0,...,x_n]$ and not all $g_i$ are constant. Use the proposition above, for $X=\mathbb{P}^n$ we can see $\dim (V(g_j))=n-1$, in the case $g_j$ is not constant. This implies we have at most $m$ hypersuface of dimension $n-1$ in $\mathbb{P}^n$, and $n-1\ge m$. This yields $\cap_j V(g_j)\ne \emptyset$, and hence, the morphism is not well-defined, since it sends some points in $\mathbb{P}^n$ to the point $(0:...:0)$, which does not lie in $\mathbb{P}^n$. Or equivalently, $\theta$ is constant.