## 29.54 Zariski's Main Theorem (algebraic version)

This is the version you can prove using purely algebraic methods. Before we can prove more powerful versions (for non-affine morphisms) we need to develop more tools. See Cohomology of Schemes, Section 30.21 and More on Morphisms, Section 37.38.

Theorem 29.54.1 (Algebraic version of Zariski's Main Theorem). Let $f : Y \to X$ be an affine morphism of schemes. Assume $f$ is of finite type. Let $X'$ be the normalization of $X$ in $Y$. Picture:

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

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

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

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

**Proof.**
There is an immediate reduction to the case where $X$ and hence $Y$ are affine. Say $X = \mathop{\mathrm{Spec}}(R)$ and $Y = \mathop{\mathrm{Spec}}(A)$. Then $X' = \mathop{\mathrm{Spec}}(A')$, where $A'$ is the integral closure of $R$ in $A$, see Definitions 29.52.2 and 29.52.3. By Algebra, Theorem 10.122.12 for every $y \in Y$ at which $f$ is quasi-finite, there exists an open $U'_ y \subset X'$ such that $(f')^{-1}(U'_ y) \to U'_ y$ is an isomorphism. Set $U' = \bigcup U'_ y$ where $y \in Y$ ranges over all points where $f$ is quasi-finite. It remains to show that $f$ is quasi-finite at all points of $(f')^{-1}(U')$. If $y \in (f')^{-1}(U')$ with image $x \in X$, then we see that $Y_ x \to X'_ x$ is an isomorphism in a neighbourhood of $y$. Hence there is no point of $Y_ x$ which specializes to $y$, since this is true for $f'(y)$ in $X'_ x$, see Lemma 29.43.8. By Lemma 29.20.6 part (3) this implies $f$ is quasi-finite at $y$.
$\square$

We can use the algebraic version of Zariski's Main Theorem to show that the set of points where a morphism is quasi-finite is open.

Lemma 29.54.2. Let $f : X \to S$ be a morphism of schemes. The set of points of $X$ where $f$ is quasi-finite is an open $U \subset X$. The induced morphism $U \to S$ is locally quasi-finite.

**Proof.**
Suppose $f$ is quasi-finite at $x$. Let $x \in U = \mathop{\mathrm{Spec}}(A) \subset X$, $V = \mathop{\mathrm{Spec}}(R) \subset S$ be affine opens as in Definition 29.20.1. By either Theorem 29.54.1 above or Algebra, Lemma 10.122.13, the set of primes $\mathfrak q$ at which $R \to A$ is quasi-finite is open in $\mathop{\mathrm{Spec}}(A)$. Since these all correspond to points of $X$ where $f$ is quasi-finite we get the first statement. The second statement is obvious.
$\square$

We will improve the following lemma to general quasi-finite separated morphisms later, see More on Morphisms, Lemma 37.38.3.

Lemma 29.54.3. Let $f : Y \to X$ be a morphism of schemes. Assume

$X$ and $Y$ are affine, and

$f$ is quasi-finite.

Then there exists a diagram

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

with $Z$ affine, $\pi $ finite and $j$ an open immersion.

**Proof.**
This is Algebra, Lemma 10.122.14 reformulated in the language of schemes.
$\square$

Lemma 29.54.4. Let $f : Y \to X$ be a quasi-finite morphism of schemes. Let $T \subset Y$ be a closed nowhere dense subset of $Y$. Then $f(T) \subset X$ is a nowhere dense subset of $X$.

**Proof.**
As in the proof of Lemma 29.47.7 this reduces immediately to the case where the base $X$ is affine. In this case $Y = \bigcup _{i = 1, \ldots , n} Y_ i$ is a finite union of affine opens (as $f$ is quasi-compact). Since each $T \cap Y_ i$ is nowhere dense, and since a finite union of nowhere dense sets is nowhere dense (see Topology, Lemma 5.21.2), it suffices to prove that the image $f(T \cap Y_ i)$ is nowhere dense in $X$. This reduces us to the case where both $X$ and $Y$ are affine. At this point we apply Lemma 29.54.3 above to get a diagram

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

with $Z$ affine, $\pi $ finite and $j$ an open immersion. Set $\overline{T} = \overline{j(T)} \subset Z$. By Topology, Lemma 5.21.3 we see $\overline{T}$ is nowhere dense in $Z$. Since $f(T) \subset \pi (\overline{T})$ the lemma follows from the corresponding result in the finite case, see Lemma 29.47.7.
$\square$

## Comments (2)

Comment #377 by Tyler Lawson on

Comment #378 by Tyler Lawson on