The Stacks project

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36.38 Zariski's Main Theorem

In this section we prove Zariski's main theorem as reformulated by Grothendieck. Often when we say “Zariski's main theorem” in this content we mean either of Lemma 36.38.1, Lemma 36.38.2, or Lemma 36.38.3. In most texts people refer to the last of these as Zariski's main theorem.

We have already proved the algebraic version in Algebra, Theorem 10.122.12 and we have already restated this algebraic version in the language of schemes, see Morphisms, Theorem 28.53.1. The version in this section is more subtle; to get the full result we use the étale localization techniques of Section 36.36 to reduce to the algebraic case.

Lemma 36.38.1. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is of finite type and separated. Let $S'$ be the normalization of $S$ in $X$, see Morphisms, Definition 28.51.3. Picture:

\[ \xymatrix{ X \ar[rd]_ f \ar[rr]_{f'} & & S' \ar[ld]^\nu \\ & S & } \]

Then there exists an open subscheme $U' \subset S'$ such that

  1. $(f')^{-1}(U') \to U'$ is an isomorphism, and

  2. $(f')^{-1}(U') \subset X$ is the set of points at which $f$ is quasi-finite.

Proof. By Morphisms, Lemma 28.53.2 the subset $U \subset X$ of points where $f$ is quasi-finite is open. The lemma is equivalent to

  1. $U' = f'(U) \subset S'$ is open,

  2. $U = f^{-1}(U')$, and

  3. $U \to U'$ is an isomorphism.

Let $x \in U$ be arbitrary. We claim there exists an open neighbourhood $f'(x) \in V \subset S'$ such that $(f')^{-1}V \to V$ is an isomorphism. We first prove the claim implies the lemma. Namely, then $(f')^{-1}V \cong V$ is both locally of finite type over $S$ (as an open subscheme of $X$) and for $v \in V$ the residue field extension $\kappa (v) \supset \kappa (\nu (v))$ is algebraic (as $V \subset S'$ and $S'$ is integral over $S$). Hence the fibres of $V \to S$ are discrete (Morphisms, Lemma 28.19.2) and $(f')^{-1}V \to S$ is locally quasi-finite (Morphisms, Lemma 28.19.8). This implies $(f')^{-1}V \subset U$ and $V \subset U'$. Since $x$ was arbitrary we see that (a), (b), and (c) are true.

Let $s = f(x)$. Let $(T, t) \to (S, s)$ be an elementary étale neighbourhood. Denote by a subscript ${}_ T$ the base change to $T$. Let $y = (x, t) \in X_ T$ be the unique point in the fibre $X_ t$ lying over $x$. Note that $U_ T \subset X_ T$ is the set of points where $f_ T$ is quasi-finite, see Morphisms, Lemma 28.19.13. Note that

\[ X_ T \xrightarrow {f'_ T} S'_ T \xrightarrow {\nu _ T} T \]

is the normalization of $T$ in $X_ T$, see Lemma 36.17.2. Suppose that the claim holds for $y \in U_ T \subset X_ T \to S'_ T \to T$, i.e., suppose that we can find an open neighbourhood $f'_ T(y) \in V' \subset S'_ T$ such that $(f'_ T)^{-1}V' \to V'$ is an isomorphism. The morphism $S'_ T \to S'$ is étale hence the image $V \subset S'$ of $V'$ is open. Observe that $f'(x) \in V$ as $f'_ T(y) \in V'$. Observe that

\[ \xymatrix{ (f'_ T)^{-1}V' \ar[r] \ar[d] & (f')^{-1}(V) \ar[d] \\ V' \ar[r] & V } \]

is a fibre square (as $S'_ T \times _{S'} X = X_ T$). Since the left vertical arrow is an isomorphism and $\{ V' \to V\} $ is a étale covering, we conclude that the right vertical arrow is an isomorphism by Descent, Lemma 34.20.17. In other words, the claim holds for $x \in U \subset X \to S' \to S$.

By the result of the previous paragraph we may replace $S$ by an elementary étale neighbourhood of $s = f(x)$ in order to prove the claim. Thus we may assume there is a decomposition

\[ X = V \amalg W \]

into open and closed subschemes where $V \to S$ is finite and $x \in V$, see Lemma 36.36.4. Since $X$ is a disjoint union of $V$ and $W$ over $S$ and since $V \to S$ is finite we see that the normalization of $S$ in $X$ is the morphism

\[ X = V \amalg W \longrightarrow V \amalg W' \longrightarrow S \]

where $W'$ is the normalization of $S$ in $W$, see Morphisms, Lemmas 28.51.10, 28.42.4, and 28.51.12. The claim follows and we win. $\square$

Lemma 36.38.2. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is quasi-finite and separated. Let $S'$ be the normalization of $S$ in $X$, see Morphisms, Definition 28.51.3. Picture:

\[ \xymatrix{ X \ar[rd]_ f \ar[rr]_{f'} & & S' \ar[ld]^\nu \\ & S & } \]

Then $f'$ is a quasi-compact open immersion and $\nu $ is integral. In particular $f$ is quasi-affine.

Proof. This follows from Lemma 36.38.1. Namely, by that lemma there exists an open subscheme $U' \subset S'$ such that $(f')^{-1}(U') = X$ (!) and $X \to U'$ is an isomorphism! In other words, $f'$ is an open immersion. Note that $f'$ is quasi-compact as $f$ is quasi-compact and $\nu : S' \to S$ is separated (Schemes, Lemma 25.21.14). It follows that $f$ is quasi-affine by Morphisms, Lemma 28.12.3. $\square$

reference

Lemma 36.38.3 (Zariski's Main Theorem). Let $f : X \to S$ be a morphism of schemes. Assume $f$ is quasi-finite and separated and assume that $S$ is quasi-compact and quasi-separated. Then there exists a factorization

\[ \xymatrix{ X \ar[rd]_ f \ar[rr]_ j & & T \ar[ld]^\pi \\ & S & } \]

where $j$ is a quasi-compact open immersion and $\pi $ is finite.

Proof. Let $X \to S' \to S$ be as in the conclusion of Lemma 36.38.2. By Properties, Lemma 27.22.13 we can write $\nu _*\mathcal{O}_{S'} = \mathop{\mathrm{colim}}\nolimits _{i \in I} \mathcal{A}_ i$ as a directed colimit of finite quasi-coherent $\mathcal{O}_ X$-algebras $\mathcal{A}_ i \subset \nu _*\mathcal{O}_{S'}$. Then $\pi _ i : T_ i = \underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A}_ i) \to S$ is a finite morphism for each $i$. Note that the transition morphisms $T_{i'} \to T_ i$ are affine and that $S' = \mathop{\mathrm{lim}}\nolimits T_ i$.

By Limits, Lemma 31.4.11 there exists an $i$ and a quasi-compact open $U_ i \subset T_ i$ whose inverse image in $S'$ equals $f'(X)$. For $i' \geq i$ let $U_{i'}$ be the inverse image of $U_ i$ in $T_{i'}$. Then $X \cong f'(X) = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_{i'}$, see Limits, Lemma 31.2.2. By Limits, Lemma 31.4.16 we see that $X \to U_{i'}$ is a closed immersion for some $i' \geq i$. (In fact $X \cong U_{i'}$ for sufficiently large $i'$ but we don't need this.) Hence $X \to T_{i'}$ is an immersion. By Morphisms, Lemma 28.3.2 we can factor this as $X \to T \to T_{i'}$ where the first arrow is an open immersion and the second a closed immersion. Thus we win. $\square$

Lemma 36.38.4. With notation and hypotheses as in Lemma 36.38.3. Assume moreover that $f$ is locally of finite presentation. Then we can choose the factorization such that $T$ is finite and of finite presentation over $Y$.

Proof. By Limits, Lemma 31.9.8 we can write $T = \mathop{\mathrm{lim}}\nolimits T_ i$ where all $T_ i$ are finite and of finite presentation over $Y$ and the transition morphisms $T_{i'} \to T_ i$ are closed immersions. By Limits, Lemma 31.4.11 there exists an $i$ and an open subscheme $U_ i \subset T_ i$ whose inverse image in $T$ is $X$. By Limits, Lemma 31.4.16 we see that $X \cong U_ i$ for large enough $i$. Replacing $T$ by $T_ i$ finishes the proof. $\square$


Comments (5)

Comment #1594 by kollar on

I think it could be useful to tell the reader which of the 12 lemmas you consider to be "zariski's main thm"

Comment #3041 by Rankeya on

In Lemma 36.38.5 it should probably say “Choose any open neighborhood of such that ”.

Comment #3229 by Katha on

In Lemma 36.38.13 it should say "Thus we may apply the induction hypothesis to and find an integral surjective morphism " instead of "" and "".

Comment #3230 by Katha on

sorry, I was confused, everything is right.


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