There is another version of Zariski’s main theorem, intermediate between the baby one and the hard quasi-finite one. It basically is the big Zariski theorem but for quasi-finite and quasi-projective morphisms. The main point of the argument is a careful application of Stein factorization, discussed yesterday; I’d like to discuss that here. What one eventually has to see is that the quasi-projective hypothesis can be dropped, but this takes some work.

As before, let {f: X \rightarrow Y} be a proper morphism of noetherian schemes. We are now interested in the question of when some {x \in  X} is isolated in the fiber {f^{-1}(f(x)) \subset X}.

Proposition 9 If {f} is proper, then the set of {x \in X} isolated in their fiber {f^{-1}(f(x))} is open.

To see this, we shall use the Stein factorization. As before, we can write

\displaystyle  f = g \circ f' : X \rightarrow Z \rightarrow Y

where {g} is finite and {f'} has connected fibers. The fibers of {g} are discrete, and {f^{-1}(f(x)) =  f'^{-1}(g^{-1}(f(x)))}. The connected component of {x} in {f^{-1}(f(x))} is thus {f'^{-1}(f'(x))}. We thus find that {x} is isolated in its fiber if and only if {f'^{-1}(f'(x)) =  x}. So we may look at the corresponding set for {f'}; it suffices to prove the theorem for {f'}.

That is, we may prove the proposition under the assumption that the fibers are all connected, and furthermore that the pushforward of the structure sheaf is the structure sheaf (since that is true of {f'}). Let us now drop the notation {f'}, and assume that {f: X \rightarrow Y} itself has connected fibers and satisfies {f_*(\mathcal{O}_X) =  \mathcal{O}_Y}.

We need to show that the set of {x} such that {f^{-1}(f(x)) =  \left\{x\right\}} is open. If there is such an {x}, I claim that {(f_*(\mathcal{O}_X))_{f(x)} =  \mathcal{O}_{f(x)}}. This follows because as {U} becomes a small neighborhood of {f(x)}, then {f^{-1}(U)} becomes a small neighborhood of {x =  f^{-1}(f(x))} as {f} is closed. Thus, since {f_*(\mathcal{O}_X) = \mathcal{O}_Y}, we find that the natural map

\displaystyle  \mathcal{O}_x \simeq \mathcal{O}_{f(x)}

is an isomorphism. However, this means that there are small neighborhoods {U,V} of {x, f(x)} which are isomorphic under {f}. By choosing {V} small, we can just take {U = f^{-1}(V)}. It follows then that every point in {U} is isolated in its fiber. This proves the result.

In fact, we have shown more:

Proposition 10 Let {f} be a proper morphism of noetherian schemes satisfying {f_*(\mathcal{O}_X) = \mathcal{O}_Y}. Then if {T} is the set of {x \in X} which are isolated in their fibers, {f|_T} is an open immersion.

Indeed, we have seen that {T} as defined above is an open set in the proof of the previous proposition. We actually saw more, though, in the proof: if {x  \in T}, then there are neighborhoods {U} of {x} and {V} of {f(x)}, with {U =  f^{-1}(V)}, such that {f|_U : U \rightarrow  V} is an isomorphism. If we take the inverse {g_V: V \rightarrow U}, and do the same for every such pair {U,V} that occurs around any {x \in U}, then the various inverses {g_V} have to glue.

Finally, we can state the second version of the Zariski theorem.

Theorem 11 (Zariski) Let {f: X \rightarrow Y} be a quasiprojective morphism of noetherian schemes. Let {U} be the subset of {X} consisting of points which are isolated in their fibers. Then {f|_U} factors as a composite of an open immersion and a finite morphism.

To see this, first note that by definition of quasiprojective, {f} factors as a composite of an open immersion and a projective morphism. Consequently we can just assume {f} is projective. In this case, we have the Stein factorization {f = g \circ f'} where {g} is finite. But on the set of points which are isolated in their fibers, the above corollary shows that {f'} is actually an open immersion. So we find

\displaystyle  f|_U = g \circ f'|_U

is the requisite factorization. We can also prove:

Theorem 12 A proper, quasi-finite morphism {f: X \rightarrow Y} of noetherian schemes is finite.

Now it turns out that a quasi-finite morphism is in fact quasi-projective, but this is very nontrivial. We are not yet in a position to prove this.

However, we can at least appeal to the Stein factorization theorem to get {f = g \circ f'} where {g} is finite and {f'} is an open immersion, since every point is isolated in its fiber. However, {f} is proper, and {g} is separated, so by the cancellation property, we find that {f'} is itself proper. This means that it must be a closed immersion as well, and consequently finite. It follows that {f} is finite.

 

Advertisements