Lemma 10.123.1. Let $\varphi : R \to S$ be a ring map. Suppose $t \in S$ satisfies the relation $\varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_ n) t^ n = 0$. Then $\varphi (a_ n)t$ is integral over $R$.

## 10.123 Zariski's Main Theorem

In this section our aim is to prove the algebraic version of Zariski's Main theorem. This theorem will be the basis of many further developments in the theory of schemes and morphisms of schemes later in the Stacks project.

Let $R \to S$ be a ring map of finite type. Our goal in this section is to show that the set of points of $\mathop{\mathrm{Spec}}(S)$ where the map is quasi-finite is *open* (Theorem 10.123.12). In fact, it will turn out that there exists a finite ring map $R \to S'$ such that in some sense the quasi-finite locus of $S/R$ is open in $\mathop{\mathrm{Spec}}(S')$ (but we will not prove this in the algebra chapter since we do not develop the language of schemes here – for the case where $R \to S$ is quasi-finite see Lemma 10.123.14). These statements are somewhat tricky to prove and we do it by a long list of lemmas concerning integral and finite extensions of rings. This material may be found in [Henselian], and [Peskine]. We also found notes by Thierry Coquand helpful.

**Proof.**
Namely, multiply the equation $\varphi (a_0) + \varphi (a_1)t + \ldots + \varphi (a_ n) t^ n = 0$ with $\varphi (a_ n)^{n-1}$ and write it as $\varphi (a_0 a_ n^{n-1}) + \varphi (a_1 a_ n^{n-2}) (\varphi (a_ n)t) + \ldots + (\varphi (a_ n) t)^ n = 0$.
$\square$

The following lemma is in some sense the key lemma in this section.

Lemma 10.123.2. Let $R$ be a ring. Let $\varphi : R[x] \to S$ be a ring map. Let $t \in S$. Assume that (a) $t$ is integral over $R[x]$, and (b) there exists a monic $p \in R[x]$ such that $t \varphi (p) \in \mathop{\mathrm{Im}}(\varphi )$. Then there exists a $q \in R[x]$ such that $t - \varphi (q)$ is integral over $R$.

**Proof.**
Write $t \varphi (p) = \varphi (r)$ for some $r \in R[x]$. Using euclidean division, write $r = qp + r'$ with $q, r' \in R[x]$ and $\deg (r') < \deg (p)$. We may replace $t$ by $t - \varphi (q)$ which is still integral over $R[x]$, so that we obtain $t \varphi (p) = \varphi (r')$. In the ring $S_ t$ we may write this as $\varphi (p) - (1/t) \varphi (r') = 0$. This implies that $\varphi (x)$ gives an element of the localization $S_ t$ which is integral over $\varphi (R)[1/t] \subset S_ t$. On the other hand, $t$ is integral over the subring $\varphi (R)[\varphi (x)] \subset S$. Combined we conclude that $t$ is integral over the subring $\varphi (R)[1/t] \subset S_ t$, see Lemma 10.36.6. In other words there exists an equation of the form

in $S_ t$ with $r_{i, j} \in R$. This means that $t^{d + N} + \sum _{i < d} \sum _{j = 0, \ldots , n_ i} \varphi (r_{i, j}) t^{i + N - j} = 0$ in $S$ for some $N$ large enough. In other words $t$ is integral over $R$. $\square$

Lemma 10.123.3. Let $R$ be a ring. Let $\varphi : R[x] \to S$ be a ring map. Let $t \in S$. Assume $t$ is integral over $R[x]$. Let $p \in R[x]$, $p = a_0 + a_1x + \ldots + a_ k x^ k$ such that $t \varphi (p) \in \mathop{\mathrm{Im}}(\varphi )$. Then there exists a $q \in R[x]$ and $n \geq 0$ such that $\varphi (a_ k)^ n t - \varphi (q)$ is integral over $R$.

**Proof.**
Let $R'$ and $S'$ be the localization of $R$ and $S$ at the element $a_ k$. Let $\varphi ' : R'[x] \to S'$ be the localization of $\varphi $. Let $t' \in S'$ be the image of $t$. Set $p' = p/a_ k \in R'[x]$. Then $t' \varphi '(p') \in \mathop{\mathrm{Im}}(\varphi ')$ since $t \varphi (p) \in \mathop{\mathrm{Im}}(\varphi )$. As $p'$ is monic, by Lemma 10.123.2 there exists a $q' \in R'[x]$ such that $t' - \varphi '(q')$ is integral over $R'$. We may choose an $n \geq 0$ and an element $q \in R[x]$ such that $a_ k^ n q'$ is the image of $q$. Then $\varphi (a_ k)^ n t - \varphi (q)$ is an element of $S$ whose image in $S'$ is integral over $R'$. By Lemma 10.36.11 there exists an $m \geq 0$ such that $\varphi (a_ k)^ m(\varphi (a_ k)^ n t - \varphi (q))$ is integral over $R$. Thus $\varphi (a_ k)^{m + n}t - \varphi (a_ k^ m q)$ is integral over $R$ as desired.
$\square$

Situation 10.123.4. Let $R$ be a ring. Let $\varphi : R[x] \to S$ be finite. Let

be the “conductor ideal” of $\varphi $. Assume $\varphi (R) \subset S$ integrally closed in $S$.

Lemma 10.123.5. In Situation 10.123.4. Suppose $u \in S$, $a_0, \ldots , a_ k \in R$, $u \varphi (a_0 + a_1x + \ldots + a_ k x^ k) \in J$. Then there exists an $m \geq 0$ such that $u \varphi (a_ k)^ m \in J$.

**Proof.**
Assume that $S$ is generated by $t_1, \ldots , t_ n$ as an $R[x]$-module. In this case $J = \{ g \in S \mid gt_ i \in \mathop{\mathrm{Im}}(\varphi )\text{ for all }i\} $. Note that each element $u t_ i$ is integral over $R[x]$, see Lemma 10.36.3. We have $\varphi (a_0 + a_1x + \ldots + a_ k x^ k) u t_ i \in \mathop{\mathrm{Im}}(\varphi )$. By Lemma 10.123.3, for each $i$ there exists an integer $n_ i$ and an element $q_ i \in R[x]$ such that $\varphi (a_ k^{n_ i}) u t_ i - \varphi (q_ i)$ is integral over $R$. By assumption this element is in $\varphi (R)$ and hence $\varphi (a_ k^{n_ i}) u t_ i \in \mathop{\mathrm{Im}}(\varphi )$. It follows that $m = \max \{ n_1, \ldots , n_ n\} $ works.
$\square$

Lemma 10.123.6. In Situation 10.123.4. Suppose $u \in S$, $a_0, \ldots , a_ k \in R$, $u \varphi (a_0 + a_1x + \ldots + a_ k x^ k) \in \sqrt{J}$. Then $u \varphi (a_ i) \in \sqrt{J}$ for all $i$.

**Proof.**
Under the assumptions of the lemma we have $u^ n \varphi (a_0 + a_1x + \ldots + a_ k x^ k)^ n \in J$ for some $n \geq 1$. By Lemma 10.123.5 we deduce $u^ n \varphi (a_ k^{nm}) \in J$ for some $m \geq 1$. Thus $u \varphi (a_ k) \in \sqrt{J}$, and so $u \varphi (a_0 + a_1x + \ldots + a_ k x^ k) - u \varphi (a_ k x^ k) = u \varphi (a_0 + a_1x + \ldots + a_{k-1} x^{k-1}) \in \sqrt{J}$. We win by induction on $k$.
$\square$

This lemma suggests the following definition.

Definition 10.123.7. Given an inclusion of rings $R \subset S$ and an element $x \in S$ we say that $x$ is *strongly transcendental over $R$* if whenever $u(a_0 + a_1 x + \ldots + a_ k x^ k) = 0$ with $u \in S$ and $a_ i \in R$, then we have $ua_ i = 0$ for all $i$.

Note that if $S$ is a domain then this is the same as saying that $x$ as an element of the fraction field of $S$ is transcendental over the fraction field of $R$.

Lemma 10.123.8. Suppose $R \subset S$ is an inclusion of reduced rings and suppose that $x \in S$ is strongly transcendental over $R$. Let $\mathfrak q \subset S$ be a minimal prime and let $\mathfrak p = R \cap \mathfrak q$. Then the image of $x$ in $S/\mathfrak q$ is strongly transcendental over the subring $R/\mathfrak p$.

**Proof.**
Suppose $u(a_0 + a_1x + \ldots + a_ k x^ k) \in \mathfrak q$. By Lemma 10.25.1 the local ring $S_{\mathfrak q}$ is a field, and hence $u(a_0 + a_1x + \ldots + a_ k x^ k) $ is zero in $S_{\mathfrak q}$. Thus $uu'(a_0 + a_1x + \ldots + a_ k x^ k) = 0$ for some $u' \in S$, $u' \not\in \mathfrak q$. Since $x$ is strongly transcendental over $R$ we get $uu'a_ i = 0$ for all $i$. This in turn implies that $ua_ i \in \mathfrak q$.
$\square$

Lemma 10.123.9. Suppose $R\subset S$ is an inclusion of domains and let $x \in S$. Assume $x$ is (strongly) transcendental over $R$ and that $S$ is finite over $R[x]$. Then $R \to S$ is not quasi-finite at any prime of $S$.

**Proof.**
As a first case, assume that $R$ is normal, see Definition 10.37.11. By Lemma 10.37.14 we see that $R[x]$ is normal. Take a prime $\mathfrak q \subset S$, and set $\mathfrak p = R \cap \mathfrak q$. Assume that the extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite. This would be the case if $R \to S$ is quasi-finite at $\mathfrak q$. Let $\mathfrak r = R[x] \cap \mathfrak q$. Then since $\kappa (\mathfrak p) \subset \kappa (\mathfrak r) \subset \kappa (\mathfrak q)$ we see that the extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak r)$ is finite too. Thus the inclusion $\mathfrak r \supset \mathfrak p R[x]$ is strict. By going down for $R[x] \subset S$, see Proposition 10.38.7, we find a prime $\mathfrak q' \subset \mathfrak q$, lying over the prime $\mathfrak pR[x]$. Hence the fibre $\mathop{\mathrm{Spec}}(S \otimes _ R \kappa (\mathfrak p))$ contains a point not equal to $\mathfrak q$, namely $\mathfrak q'$, whose closure contains $\mathfrak q$ and hence $\mathfrak q$ is not isolated in its fibre.

If $R$ is not normal, let $R \subset R' \subset K$ be the integral closure $R'$ of $R$ in its field of fractions $K$. Let $S \subset S' \subset L$ be the subring $S'$ of the field of fractions $L$ of $S$ generated by $R'$ and $S$. Note that by construction the map $S \otimes _ R R' \to S'$ is surjective. This implies that $R'[x] \subset S'$ is finite. Also, the map $S \subset S'$ induces a surjection on $\mathop{\mathrm{Spec}}$, see Lemma 10.36.17. We conclude by Lemma 10.122.6 and the normal case we just discussed. $\square$

Lemma 10.123.10. Suppose $R \subset S$ is an inclusion of reduced rings. Assume $x \in S$ be strongly transcendental over $R$, and $S$ finite over $R[x]$. Then $R \to S$ is not quasi-finite at any prime of $S$.

**Proof.**
Let $\mathfrak q \subset S$ be any prime. Choose a minimal prime $\mathfrak q' \subset \mathfrak q$. According to Lemmas 10.123.8 and 10.123.9 the extension $R/(R \cap \mathfrak q') \subset S/\mathfrak q'$ is not quasi-finite at the prime corresponding to $\mathfrak q$. By Lemma 10.122.6 the extension $R \to S$ is not quasi-finite at $\mathfrak q$.
$\square$

Lemma 10.123.11. Let $R$ be a ring. Let $S = R[x]/I$. Let $\mathfrak q \subset S$ be a prime. Assume $R \to S$ is quasi-finite at $\mathfrak q$. Let $S' \subset S$ be the integral closure of $R$ in $S$. Then there exists an element $g \in S'$, $g \not\in \mathfrak q$ such that $S'_ g \cong S_ g$.

**Proof.**
Let $\mathfrak p$ be the image of $\mathfrak q$ in $\mathop{\mathrm{Spec}}(R)$. There exists an $f \in I$, $f = a_ nx^ n + \ldots + a_0$ such that $a_ i \not\in \mathfrak p$ for some $i$. Namely, otherwise the fibre ring $S \otimes _ R \kappa (\mathfrak p)$ would be $\kappa (\mathfrak p)[x]$ and the map would not be quasi-finite at any prime lying over $\mathfrak p$. We conclude there exists a relation $b_ m x^ m + \ldots + b_0 = 0$ with $b_ j \in S'$, $j = 0, \ldots , m$ and $b_ j \not\in \mathfrak q \cap S'$ for some $j$. We prove the lemma by induction on $m$. The base case is $m = 0$ is vacuous (because the statements $b_0 = 0$ and $b_0 \not\in \mathfrak q$ are contradictory).

The case $b_ m \not\in \mathfrak q$. In this case $x$ is integral over $S'_{b_ m}$, in fact $b_ mx \in S'$ by Lemma 10.123.1. Hence the injective map $S'_{b_ m} \to S_{b_ m}$ is also surjective, i.e., an isomorphism as desired.

The case $b_ m \in \mathfrak q$. In this case we have $b_ mx \in S'$ by Lemma 10.123.1. Set $b'_{m - 1} = b_ mx + b_{m - 1}$. Then

Since $b'_{m - 1}$ is congruent to $b_{m - 1}$ modulo $S' \cap \mathfrak q$ we see that it is still the case that one of $b'_{m - 1}, b_{m - 2}, \ldots , b_0$ is not in $S' \cap \mathfrak q$. Thus we win by induction on $m$. $\square$

Theorem 10.123.12 (Zariski's Main Theorem). Let $R$ be a ring. Let $R \to S$ be a finite type $R$-algebra. Let $S' \subset S$ be the integral closure of $R$ in $S$. Let $\mathfrak q \subset S$ be a prime of $S$. If $R \to S$ is quasi-finite at $\mathfrak q$ then there exists a $g \in S'$, $g \not\in \mathfrak q$ such that $S'_ g \cong S_ g$.

**Proof.**
There exist finitely many elements $x_1, \ldots , x_ n \in S$ such that $S$ is finite over the $R$-sub algebra generated by $x_1, \ldots , x_ n$. (For example generators of $S$ over $R$.) We prove the proposition by induction on the minimal such number $n$.

The case $n = 0$ is trivial, because in this case $S' = S$, see Lemma 10.36.3.

The case $n = 1$. We may replace $R$ by its integral closure in $S$ (Lemma 10.122.9 guarantees that $R \to S$ is still quasi-finite at $\mathfrak q$). Thus we may assume $R \subset S$ is integrally closed in $S$, in other words $R = S'$. Consider the map $\varphi : R[x] \to S$, $x \mapsto x_1$. (We will see that $\varphi $ is not injective below.) By assumption $\varphi $ is finite. Hence we are in Situation 10.123.4. Let $J \subset S$ be the “conductor ideal” defined in Situation 10.123.4. Consider the diagram

According to Lemma 10.123.6 the image of $x$ in the quotient $S/\sqrt{J}$ is strongly transcendental over $R/ (R \cap \sqrt{J})$. Hence by Lemma 10.123.10 the ring map $R/ (R \cap \sqrt{J}) \to S/\sqrt{J}$ is not quasi-finite at any prime of $S/\sqrt{J}$. By Lemma 10.122.6 we deduce that $\mathfrak q$ does not lie in $V(J) \subset \mathop{\mathrm{Spec}}(S)$. Thus there exists an element $s \in J$, $s \not\in \mathfrak q$. By definition of $J$ we may write $s = \varphi (f)$ for some polynomial $f \in R[x]$. Let $I = \mathop{\mathrm{Ker}}(\varphi : R[x] \to S)$. Since $\varphi (f) \in J$ we get $(R[x]/I)_ f \cong S_{\varphi (f)}$. Also $s \not\in \mathfrak q$ means that $f \not\in \varphi ^{-1}(\mathfrak q)$. Thus $\varphi ^{-1}(\mathfrak q)$ is a prime of $R[x]/I$ at which $R \to R[x]/I$ is quasi-finite, see Lemma 10.122.5. Note that $R$ is integrally closed in $R[x]/I$ since $R$ is integrally closed in $S$. By Lemma 10.123.11 there exists an element $h \in R$, $h \not\in R \cap \mathfrak q$ such that $R_ h \cong (R[x]/I)_ h$. Thus $(R[x]/I)_{fh} = S_{\varphi (fh)}$ is isomorphic to a principal localization $R_{h'}$ of $R$ for some $h' \in R$, $h' \not\in \mathfrak q$.

The case $n > 1$. Consider the subring $R' \subset S$ which is the integral closure of $R[x_1, \ldots , x_{n-1}]$ in $S$. By Lemma 10.122.9 the extension $S/R'$ is quasi-finite at $\mathfrak q$. Also, note that $S$ is finite over $R'[x_ n]$. By the case $n = 1$ above, there exists a $g' \in R'$, $g' \not\in \mathfrak q$ such that $(R')_{g'} \cong S_{g'}$. At this point we cannot apply induction to $R \to R'$ since $R'$ may not be finite type over $R$. Since $S$ is finitely generated over $R$ we deduce in particular that $(R')_{g'}$ is finitely generated over $R$. Say the elements $g'$, and $y_1/(g')^{n_1}, \ldots , y_ N/(g')^{n_ N}$ with $y_ i \in R'$ generate $(R')_{g'}$ over $R$. Let $R''$ be the $R$-sub algebra of $R'$ generated by $x_1, \ldots , x_{n-1}, y_1, \ldots , y_ N, g'$. This has the property $(R'')_{g'} \cong S_{g'}$. Surjectivity because of how we chose $y_ i$, injectivity because $R'' \subset R'$, and localization is exact. Note that $R''$ is finite over $R[x_1, \ldots , x_{n-1}]$ because of our choice of $R'$, see Lemma 10.36.4. Let $\mathfrak q'' = R'' \cap \mathfrak q$. Since $(R'')_{\mathfrak q''} = S_{\mathfrak q}$ we see that $R \to R''$ is quasi-finite at $\mathfrak q''$, see Lemma 10.122.2. We apply our induction hypothesis to $R \to R''$, $\mathfrak q''$ and $x_1, \ldots , x_{n-1} \in R''$ and we find a subring $R''' \subset R''$ which is integral over $R$ and an element $g'' \in R'''$, $g'' \not\in \mathfrak q''$ such that $(R''')_{g''} \cong (R'')_{g''}$. Write the image of $g'$ in $(R'')_{g''}$ as $g'''/(g'')^ n$ for some $g''' \in R'''$. Set $g = g''g''' \in R'''$. Then it is clear that $g \not\in \mathfrak q$ and $(R''')_ g \cong S_ g$. Since by construction we have $R''' \subset S'$ we also have $S'_ g \cong S_ g$ as desired. $\square$

Lemma 10.123.13. Let $R \to S$ be a finite type ring map. The set of points $\mathfrak q$ of $\mathop{\mathrm{Spec}}(S)$ at which $S/R$ is quasi-finite is open in $\mathop{\mathrm{Spec}}(S)$.

**Proof.**
Let $\mathfrak q \subset S$ be a point at which the ring map is quasi-finite. By Theorem 10.123.12 there exists an integral ring extension $R \to S'$, $S' \subset S$ and an element $g \in S'$, $g\not\in \mathfrak q$ such that $S'_ g \cong S_ g$. Since $S$ and hence $S_ g$ are of finite type over $R$ we may find finitely many elements $y_1, \ldots , y_ N$ of $S'$ such that $S''_ g \cong S_ g$ where $S'' \subset S'$ is the sub $R$-algebra generated by $g, y_1, \ldots , y_ N$. Since $S''$ is finite over $R$ (see Lemma 10.36.4) we see that $S''$ is quasi-finite over $R$ (see Lemma 10.122.4). It is easy to see that this implies that $S''_ g$ is quasi-finite over $R$, for example because the property of being quasi-finite at a prime depends only on the local ring at the prime. Thus we see that $S_ g$ is quasi-finite over $R$. By the same token this implies that $R \to S$ is quasi-finite at every prime of $S$ which lies in $D(g)$.
$\square$

Lemma 10.123.14. Let $R \to S$ be a finite type ring map. Suppose that $S$ is quasi-finite over $R$. Let $S' \subset S$ be the integral closure of $R$ in $S$. Then

$\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S')$ is a homeomorphism onto an open subset,

if $g \in S'$ and $D(g)$ is contained in the image of the map, then $S'_ g \cong S_ g$, and

there exists a finite $R$-algebra $S'' \subset S'$ such that (1) and (2) hold for the ring map $S'' \to S$.

**Proof.**
Because $S/R$ is quasi-finite we may apply Theorem 10.123.12 to each point $\mathfrak q$ of $\mathop{\mathrm{Spec}}(S)$. Since $\mathop{\mathrm{Spec}}(S)$ is quasi-compact, see Lemma 10.17.10, we may choose a finite number of $g_ i \in S'$, $i = 1, \ldots , n$ such that $S'_{g_ i} = S_{g_ i}$, and such that $g_1, \ldots , g_ n$ generate the unit ideal in $S$ (in other words the standard opens of $\mathop{\mathrm{Spec}}(S)$ associated to $g_1, \ldots , g_ n$ cover all of $\mathop{\mathrm{Spec}}(S)$).

Suppose that $D(g) \subset \mathop{\mathrm{Spec}}(S')$ is contained in the image. Then $D(g) \subset \bigcup D(g_ i)$. In other words, $g_1, \ldots , g_ n$ generate the unit ideal of $S'_ g$. Note that $S'_{gg_ i} \cong S_{gg_ i}$ by our choice of $g_ i$. Hence $S'_ g \cong S_ g$ by Lemma 10.23.2.

We construct a finite algebra $S'' \subset S'$ as in (3). To do this note that each $S'_{g_ i} \cong S_{g_ i}$ is a finite type $R$-algebra. For each $i$ pick some elements $y_{ij} \in S'$ such that each $S'_{g_ i}$ is generated as $R$-algebra by $1/g_ i$ and the elements $y_{ij}$. Then set $S''$ equal to the sub $R$-algebra of $S'$ generated by all $g_ i$ and all the $y_{ij}$. Details omitted. $\square$

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