## Tag `00PI`

## 10.122. 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 $\mathop{\rm Spec}(S)$ where the map is quasi-finite is

open(Theorem 10.122.13). 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 $\mathop{\rm Spec}(S')$ (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 10.122.15). 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.Lemma 10.122.1. 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 $R$.

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 10.122.2. 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 $t \varphi(p) \in \text{Im}(\varphi)$. 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 10.35.6. In other words there exists an equation of the form $t^d + \sum_{i<d} (\varphi(r_i)/t^{n_i}) t^i = 0$ in $S_t$ with $r_i \in R$. This means that $t^{d + N} + \sum_{i < d} \varphi(r_i) t^{i + N - n_i} = 0$ in $S$ for some $N$ large enough. In other words $t$ is integral over $R$. $\square$Lemma 10.122.3. Let $R$ be a ring and let $\varphi : R[x] \to S$ be a ring map. Let $t \in S$. If $t$ is integral over $R[x]$, then there exists an $\ell \geq 0$ such that for every $a \in R$ the element $\varphi(a)^\ell t$ is integral over $\varphi_a : R[y] \to S$, defined by $y \mapsto \varphi(ax)$ and $r \mapsto \varphi(r)$ for $r\in R$.

Proof.Say $t^d + \sum_{i<d} \varphi(f_i)t^i = 0$ with $f_i \in R[x]$. Let $\ell$ be the maximum degree in $x$ of all the $f_i$. Multiply the equation by $\varphi(a)^\ell$ to get $\varphi(a)^\ell t^d + \sum_{i<d} \varphi(a^\ell f_i)t^i = 0$. Note that each $\varphi(a^\ell f_i)$ is in the image of $\varphi_a$. The result follows from Lemma 10.122.1. $\square$Lemma 10.122.4. 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 $t \varphi(p) \in \text{Im}(\varphi)$. 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.By Lemma 10.122.3 there exists an $\ell \geq 0$ such that the element $\varphi(a_k)^\ell t$ is integral over the map $\varphi' : R[y] \to S$, $\varphi'(y) = \varphi(a_k x)$ and $\varphi'(r) = \varphi(r)$, for $r\in R$. The polynomial $p' = a_k^{k-1} a_0 + a_k^{k-2} a_1 y + \ldots + y^k$ is monic and $t \varphi'(p') = \varphi(a_k^{k-1}) t \varphi(p) \in \text{Im}(\varphi)$. By definition of $\varphi'$ this implies there exists a $n \geq k-1$ such that $\varphi(a_k^n)t \varphi'(p') \in \text{Im}(\varphi')$. If also $n \geq \ell$, then $\varphi(a_k)^n t$ is still integral over $R[y]$. By Lemma 10.122.2 we see that $\varphi(a_k)^n t - \varphi'(q)$ is integral over $R$ for some $q \in R[y]$. Again by the simple relationship between $\varphi'$ and $\varphi$ this implies the lemma. $\square$Situation 10.122.5. Let $R$ be a ring. Let $\varphi : R[x] \to S$ be finite. Let $$ J = \{ g \in S \mid gS \subset \text{Im}(\varphi)\} $$ be the ''conductor ideal'' of $\varphi$. Assume $\varphi(R) \subset S$ integrally closed in $S$.

Lemma 10.122.6. In Situation 10.122.5. 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 $J = \{ g \in S \mid gt_i \in \text{Im}(\varphi)\text{ for all }i\}$. Note that each element $u t_i$ is integral over $R[x]$, see Lemma 10.35.3. We have $\varphi(a_0 + a_1x + \ldots + a_k x^k) u t_i \in \text{Im}(\varphi)$. By Lemma 10.122.4, 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 $\varphi(a_k^{n_i}) u t_i \in \text{Im}(\varphi)$. It follows that $m = \max\{n_1, \ldots, n_n\}$ works. $\square$Lemma 10.122.7. In Situation 10.122.5. 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 10.122.6 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 $u \varphi(a_0 + a_1x + \ldots + a_k x^k) - u \varphi(a_k) = u \varphi(a_0 + a_1x + \ldots + a_{k-1} x^{k-1}) \in \sqrt{J}$. We win by induction on $k$. $\square$This lemma suggests the following definition.

Definition 10.122.8. 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 10.122.9. 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 10.24.1 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 10.122.10. 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 10.36.11. By Lemma 10.36.14 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 10.37.7, we find a prime $\mathfrak q' \subset \mathfrak q$, lying over the prime $\mathfrak pR[x]$. Hence the fibre $\mathop{\rm Spec}(S \otimes_R \kappa(\mathfrak p))$ 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 $\mathop{\rm Spec}$, see Lemma 10.35.17. We conclude by Lemma 10.121.6 and the normal case we just discussed. $\square$

Lemma 10.122.11. 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 10.122.9 and 10.122.10 the extension $R/(R \cap \mathfrak q') \subset S/\mathfrak q'$ is not quasi-finite at the prime corresponding to $\mathfrak q$. By Lemma 10.121.6 the extension $R \to S$ is not quasi-finite at $\mathfrak q$. $\square$Lemma 10.122.12. 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 $\mathop{\rm Spec}(R)$. 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 case $b_m \in \mathfrak q$. In this case we have $b_mx \in S'$ by Lemma 10.122.1. Set $b'_{m - 1} = b_mx + b_{m - 1}$. Then $$ b'_{m - 1}x^{m - 1} + b_{m - 2}x^{m - 2} + \ldots + b_0 = 0 $$ 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 $m$.

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 10.122.1. Hence the injective map $S'_{b_m} \to S_{b_m}$ is also surjective, i.e., an isomorphism as desired. $\square$

Theorem 10.122.13 (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 10.35.3.

The case $n = 1$. We may replace $R$ by its integral closure in $S$ (Lemma 10.121.9 guarantees that $R \to S$ is still quasi-finite at $\mathfrak q$). Thus we may assume $R \subset S$ is integrally closed in $S$. 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 10.122.5. Let $J \subset S$ be the ''conductor ideal'' defined in Situation 10.122.5. Consider the diagram $$ \xymatrix{ R[x] \ar[r] & S \ar[r] & S/\sqrt{J} & R/(R \cap \sqrt{J})[x] \ar[l] \\ & R \ar[lu] \ar[r] \ar[u] & R/(R \cap \sqrt{J}) \ar[u] \ar[ru] & } $$ According to Lemma 10.122.7 the image of $x$ in the quotient $S/\sqrt{J}$ is strongly transcendental over $R/ (R \cap \sqrt{J})$. Hence by Lemma 10.122.11 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 10.121.6 we deduce that $\mathfrak q$ does not lie in $V(J) \subset \mathop{\rm Spec}(S)$. 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 $f \in R[x]$. Now let $I = \text{Ker}(R[x] \to S)$. 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 10.121.5. Let $C \subset R[x]/I$ be the integral closure of $R$. By Lemma 10.122.12 there exists an element $h \in C$, $h \not \in \varphi^{-1}(\mathfrak q)$ such that $C_h \cong (R[x]/I)_h$. We conclude that $(R[x]/I)_{fh} = S_{\varphi(fh)}$ is isomorphic to a principal localization $C_{h'}$ of $C$ for some $h' \in C$, $h' \not \in \varphi^{-1}(\mathfrak q)$. Since $\varphi(C) \subset S'$ we get $g = \varphi(h') \in S'$, $g \not \in \mathfrak q$ and moreover the injective map $S'_g \to S_g$ is also surjective because by our choice of $h'$ the map $C_{h'} \to S_g$ is surjective.

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 10.121.6 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 10.35.4. 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 10.121.2. 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 10.122.14. Let $R \to S$ be a finite type ring map. The set of points $\mathfrak q$ of $\mathop{\rm Spec}(S)$ at which $S/R$ is quasi-finite is open in $\mathop{\rm Spec}(S)$.

Proof.Let $\mathfrak q \subset S$ be a point at which the ring map is quasi-finite. By Theorem 10.122.13 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 $S''_g \cong S$ 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 10.35.4) we see that $S''$ is quasi-finite over $R$ (see Lemma 10.121.4). 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 10.122.15. 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

- $\mathop{\rm Spec}(S) \to \mathop{\rm Spec}(S')$ 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 10.122.13 to each point $\mathfrak q$ of $\mathop{\rm Spec}(S)$. Since $\mathop{\rm Spec}(S)$ is quasi-compact, see Lemma 10.16.10, 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 $\mathop{\rm Spec}(S)$ associated to $g_1, \ldots, g_n$ cover all of $\mathop{\rm Spec}(S)$).Suppose that $D(g) \subset \mathop{\rm Spec}(S')$ 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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\section{Zariski's Main Theorem}
\label{section-Zariski}
\noindent
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.
\medskip\noindent
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 $\Spec(S)$ where the map is quasi-finite
is {\it open} (Theorem \ref{theorem-main-theorem}).
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 $\Spec(S')$
(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 \ref{lemma-quasi-finite-open-integral-closure}).
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 \cite{Henselian}, and \cite{Peskine}. We also found notes
by Thierry Coquand helpful.
\begin{lemma}
\label{lemma-make-integral-trivial}
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 $R$.
\end{lemma}
\begin{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
$\varphi(a_0 a_n^{n-1}) +
\varphi(a_1 a_n^{n-2}) (\varphi(a_n)t) +
\ldots +
(\varphi(a_n) t)^n = 0$.
\end{proof}
\noindent
The following lemma is in some sense the key lemma in this
section.
\begin{lemma}
\label{lemma-make-integral-trick}
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
$t \varphi(p) \in \Im(\varphi)$. Then there
exists a $q \in R[x]$ such that $t - \varphi(q)$
is integral over $R$.
\end{lemma}
\begin{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
\ref{lemma-integral-transitive}. In other words
there exists an equation of the form
$t^d + \sum_{i<d} (\varphi(r_i)/t^{n_i}) t^i = 0$
in $S_t$ with $r_i \in R$. This means that
$t^{d + N} + \sum_{i < d} \varphi(r_i) t^{i + N - n_i} = 0$ in $S$
for some $N$ large enough. In other words
$t$ is integral over $R$.
\end{proof}
\begin{lemma}
\label{lemma-change-equation-multiply}
Let $R$ be a ring and let $\varphi : R[x] \to S$ be
a ring map. Let $t \in S$. If $t$ is integral over
$R[x]$, then there exists an $\ell \geq 0$ such that
for every $a \in R$ the element $\varphi(a)^\ell t$
is integral over $\varphi_a : R[y] \to S$, defined by
$y \mapsto \varphi(ax)$ and $r \mapsto \varphi(r)$
for $r\in R$.
\end{lemma}
\begin{proof}
Say $t^d + \sum_{i<d} \varphi(f_i)t^i = 0$
with $f_i \in R[x]$. Let $\ell$ be the maximum degree
in $x$ of all the $f_i$. Multiply the equation
by $\varphi(a)^\ell$ to get
$\varphi(a)^\ell t^d + \sum_{i<d} \varphi(a^\ell f_i)t^i = 0$.
Note that each $\varphi(a^\ell f_i)$ is in the image of
$\varphi_a$. The result follows from
Lemma \ref{lemma-make-integral-trivial}.
\end{proof}
\begin{lemma}
\label{lemma-combine-lemmas}
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 $t \varphi(p) \in \Im(\varphi)$.
Then there exists a $q \in R[x]$ and $n \geq 0$
such that $\varphi(a_k)^n t - \varphi(q) $ is integral
over $R$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-change-equation-multiply} there exists
an $\ell \geq 0$ such that
the element $\varphi(a_k)^\ell t$ is integral
over the map $\varphi' : R[y] \to S$, $\varphi'(y) =
\varphi(a_k x)$ and $\varphi'(r) = \varphi(r)$, for $r\in R$.
The polynomial $p' = a_k^{k-1} a_0 + a_k^{k-2} a_1 y
+ \ldots + y^k$ is monic and $t \varphi'(p')
= \varphi(a_k^{k-1}) t \varphi(p) \in \Im(\varphi)$.
By definition of $\varphi'$ this implies there exists
a $n \geq k-1$ such that $\varphi(a_k^n)t \varphi'(p')
\in \Im(\varphi')$. If also $n \geq \ell$, then
$\varphi(a_k)^n t$ is still integral over $R[y]$.
By Lemma \ref{lemma-make-integral-trick}
we see that $\varphi(a_k)^n t - \varphi'(q)$ is integral over $R$
for some $q \in R[y]$. Again by the simple relationship between
$\varphi'$ and $\varphi$ this implies the lemma.
\end{proof}
\begin{situation}
\label{situation-one-transcendental-element}
Let $R$ be a ring.
Let $\varphi : R[x] \to S$ be finite.
Let
$$
J = \{ g \in S \mid gS \subset \Im(\varphi)\}
$$
be the ``conductor ideal'' of $\varphi$.
Assume $\varphi(R) \subset S$ integrally closed in $S$.
\end{situation}
\begin{lemma}
\label{lemma-leading-coefficient-in-J}
In Situation \ref{situation-one-transcendental-element}.
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
$u \varphi(a_k)^m \in J$.
\end{lemma}
\begin{proof}
Assume that $S$ is generated by $t_1, \ldots, t_n$
as an $R[x]$-module. In this case
$J = \{ g \in S \mid gt_i \in \Im(\varphi)\text{ for all }i\}$.
Note that each element $u t_i$ is integral over
$R[x]$, see Lemma \ref{lemma-finite-is-integral}.
We have $\varphi(a_0 + a_1x + \ldots + a_k x^k) u t_i \in
\Im(\varphi)$. By Lemma \ref{lemma-combine-lemmas}, 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 $\varphi(a_k^{n_i}) u t_i \in \Im(\varphi)$.
It follows that $m = \max\{n_1, \ldots, n_n\}$ works.
\end{proof}
\begin{lemma}
\label{lemma-all-coefficients-in-J}
In Situation \ref{situation-one-transcendental-element}.
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 $i$.
\end{lemma}
\begin{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 \ref{lemma-leading-coefficient-in-J}
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
$u \varphi(a_0 + a_1x + \ldots + a_k x^k) - u \varphi(a_k) =
u \varphi(a_0 + a_1x + \ldots + a_{k-1} x^{k-1}) \in \sqrt{J}$.
We win by induction on $k$.
\end{proof}
\noindent
This lemma suggests the following definition.
\begin{definition}
\label{definition-strongly-transcendental}
Given an inclusion of rings $R \subset S$ and
an element $x \in S$ we say that $x$ is
{\it 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 $i$.
\end{definition}
\noindent
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 $R$.
\begin{lemma}
\label{lemma-reduced-strongly-transcendental-minimal-prime}
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 $R/\mathfrak p$.
\end{lemma}
\begin{proof}
Suppose $u(a_0 + a_1x + \ldots + a_k x^k) \in \mathfrak q$.
By Lemma \ref{lemma-minimal-prime-reduced-ring}
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 $ua_i \in \mathfrak q$.
\end{proof}
\begin{lemma}
\label{lemma-domains-transcendental-not-quasi-finite}
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 $S$.
\end{lemma}
\begin{proof}
As a first case, assume that $R$ is normal, see
Definition \ref{definition-ring-normal}.
By Lemma \ref{lemma-polynomial-ring-normal}
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 \ref{proposition-going-down-normal-integral},
we find a prime $\mathfrak q' \subset \mathfrak q$,
lying over the prime $\mathfrak pR[x]$. Hence
the fibre $\Spec(S \otimes_R \kappa(\mathfrak p))$
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.
\medskip\noindent
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 $\Spec$, see
Lemma \ref{lemma-integral-overring-surjective}.
We conclude by Lemma \ref{lemma-four-rings} and the normal case
we just discussed.
\end{proof}
\begin{lemma}
\label{lemma-reduced-strongly-transcendental-not-quasi-finite}
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 $S$.
\end{lemma}
\begin{proof}
Let $\mathfrak q \subset S$ be any prime.
Choose a minimal prime $\mathfrak q' \subset \mathfrak q$.
According to Lemmas
\ref{lemma-reduced-strongly-transcendental-minimal-prime} and
\ref{lemma-domains-transcendental-not-quasi-finite}
the extension $R/(R \cap \mathfrak q') \subset
S/\mathfrak q'$ is not quasi-finite at the prime corresponding
to $\mathfrak q$. By Lemma \ref{lemma-four-rings}
the extension $R \to S$ is not quasi-finite
at $\mathfrak q$.
\end{proof}
\begin{lemma}
\label{lemma-quasi-finite-monogenic}
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
$S'_g \cong S_g$.
\end{lemma}
\begin{proof}
Let $\mathfrak p$ be the image of $\mathfrak q$ in $\Spec(R)$.
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 $m$.
\medskip\noindent
The case $b_m \in \mathfrak q$. In this case we have $b_mx \in S'$ by
Lemma \ref{lemma-make-integral-trivial}.
Set $b'_{m - 1} = b_mx + b_{m - 1}$. Then
$$
b'_{m - 1}x^{m - 1} + b_{m - 2}x^{m - 2} + \ldots + b_0 = 0
$$
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 $m$.
\medskip\noindent
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 \ref{lemma-make-integral-trivial}.
Hence the injective map $S'_{b_m} \to S_{b_m}$ is also surjective, i.e.,
an isomorphism as desired.
\end{proof}
\begin{theorem}[Zariski's Main Theorem]
\label{theorem-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 $S'_g \cong S_g$.
\end{theorem}
\begin{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 $n$.
\medskip\noindent
The case $n = 0$ is trivial, because in this case $S' = S$,
see Lemma \ref{lemma-finite-is-integral}.
\medskip\noindent
The case $n = 1$. We may replace $R$ by its integral closure in $S$
(Lemma \ref{lemma-quasi-finite-permanence} guarantees that $R \to S$
is still quasi-finite at $\mathfrak q$). Thus we may assume
$R \subset S$ is integrally closed in $S$.
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
\ref{situation-one-transcendental-element}.
Let $J \subset S$ be the ``conductor ideal'' defined
in Situation \ref{situation-one-transcendental-element}.
Consider the diagram
$$
\xymatrix{
R[x] \ar[r] & S \ar[r] & S/\sqrt{J} & R/(R \cap \sqrt{J})[x] \ar[l]
\\
& R \ar[lu] \ar[r] \ar[u] & R/(R \cap \sqrt{J}) \ar[u] \ar[ru] &
}
$$
According to Lemma \ref{lemma-all-coefficients-in-J}
the image of $x$ in the quotient $S/\sqrt{J}$
is strongly transcendental over $R/ (R \cap \sqrt{J})$.
Hence by Lemma \ref{lemma-reduced-strongly-transcendental-not-quasi-finite}
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 \ref{lemma-four-rings} we deduce that $\mathfrak q$
does not lie in $V(J) \subset \Spec(S)$.
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 $f \in R[x]$.
Now let $I = \Ker(R[x] \to S)$. 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 \ref{lemma-quasi-finite-local}.
Let $C \subset R[x]/I$ be the integral closure of $R$. By
Lemma \ref{lemma-quasi-finite-monogenic}
there exists an element $h \in C$, $h \not \in \varphi^{-1}(\mathfrak q)$
such that $C_h \cong (R[x]/I)_h$. We conclude that
$(R[x]/I)_{fh} = S_{\varphi(fh)}$ is isomorphic to a principal
localization $C_{h'}$ of $C$ for some
$h' \in C$, $h' \not \in \varphi^{-1}(\mathfrak q)$.
Since $\varphi(C) \subset S'$ we get
$g = \varphi(h') \in S'$, $g \not \in \mathfrak q$
and moreover the injective map $S'_g \to S_g$ is also surjective
because by our choice of $h'$ the map $C_{h'} \to S_g$ is surjective.
\medskip\noindent
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 \ref{lemma-four-rings} 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 \ref{lemma-characterize-integral}.
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 \ref{lemma-isolated-point-fibre}.
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.
\end{proof}
\begin{lemma}
\label{lemma-quasi-finite-open}
Let $R \to S$ be a finite type ring map.
The set of points $\mathfrak q$ of $\Spec(S)$ at which
$S/R$ is quasi-finite is open in $\Spec(S)$.
\end{lemma}
\begin{proof}
Let $\mathfrak q \subset S$ be a point at which the ring map
is quasi-finite. By Theorem \ref{theorem-main-theorem}
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 $S''_g \cong S$
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 \ref{lemma-characterize-integral}) we see that
$S''$ is quasi-finite over $R$ (see Lemma \ref{lemma-quasi-finite}).
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 $D(g)$.
\end{proof}
\begin{lemma}
\label{lemma-quasi-finite-open-integral-closure}
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
\begin{enumerate}
\item $\Spec(S) \to \Spec(S')$ is a homeomorphism
onto an open subset,
\item if $g \in S'$ and $D(g)$ is contained in the image
of the map, then $S'_g \cong S_g$, and
\item there exists a finite $R$-algebra $S'' \subset S'$
such that (1) and (2) hold for the ring map
$S'' \to S$.
\end{enumerate}
\end{lemma}
\begin{proof}
Because $S/R$ is quasi-finite we may apply
Theorem \ref{theorem-main-theorem} to
each point $\mathfrak q$ of $\Spec(S)$.
Since $\Spec(S)$ is quasi-compact, see
Lemma \ref{lemma-quasi-compact}, 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 $\Spec(S)$ associated
to $g_1, \ldots, g_n$ cover all of $\Spec(S)$).
\medskip\noindent
Suppose that $D(g) \subset \Spec(S')$
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 \ref{lemma-cover}.
\medskip\noindent
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.
\end{proof}
```

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