The Stacks project

Proof. Consider the map $\varphi : M \to M$, $x \mapsto y \cdot x$. By Lemma 10.16.2 there exists a monic polynomial $P \in R[T]$ with $P(\varphi ) = 0$. In the ring $S$ we get $P(y) = P(y) \cdot 1 = P(\varphi )(1) = 0$. $\square$

Proof. Let $R \to S$ be finite. Let $y \in S$. Apply Lemma 10.36.2 to $M = S$ to see that $y$ is integral over $R$. $\square$

Proof. If each $s_ i$ is integral, then the subalgebra generated by $\varphi (R)$ and the $s_ i$ is finite over $R$. Namely, if $s_ i$ satisfies a monic equation of degree $d_ i$ over $R$, then this subalgebra is generated as an $R$-module by the elements $s_1^{e_1} \ldots s_ n^{e_ n}$ with $0 \leq e_ i \leq d_ i - 1$. Conversely, suppose given a finite $R$-subalgebra $S'$ containing all the $s_ i$. Then all of the $s_ i$ are integral by Lemma 10.36.3. $\square$

Proof. Clear from Lemma 10.36.4. $\square$

Proof. Let $t \in T$. Let $P(x) \in S[x]$ be a monic polynomial such that $P(t) = 0$. Apply Lemma 10.36.4 to the finite set of coefficients of $P$. Hence $t$ is integral over some subalgebra $S' \subset S$ finite over $R$. Apply Lemma 10.36.4 again to find a subalgebra $T' \subset T$ finite over $S'$ and containing $t$. Lemma 10.7.3 applied to $R \to S' \to T'$ shows that $T'$ is finite over $R$. The integrality of $t$ over $R$ now follows from Lemma 10.36.3. $\square$

Proof. This is clear from Lemmas 10.36.4 and 10.36.3. $\square$

Proof. Omitted. $\square$

Proof. This follows immediately from Lemma 10.36.8. $\square$

Proof. Since localization is exact we see that $S^{-1}B' \subset S^{-1}B$. Suppose $x \in B'$ and $f \in S$. Then $x^ d + \sum _{i = 1, \ldots , d} a_ i x^{d - i} = 0$ in $B$ for some $a_ i \in A$. Hence also

in $S^{-1}B$. In this way we see that $S^{-1}B'$ is contained in the integral closure of $S^{-1}A$ in $S^{-1}B$. Conversely, suppose that $x/f \in S^{-1}B$ is integral over $S^{-1}A$. Then we have

in $S^{-1}B$ for some $a_ i \in A$ and $f_ i \in S$. This means that

for a suitable $f' \in S$. Hence $f'f_1\ldots f_ dx \in B'$ and thus $x/f \in S^{-1}B'$ as desired. $\square$

Proof. It is clear that (1) implies (2). Assume (2). Consider the $R$-algebra $S' \subset S$ generated by $\varphi (R)$ and $x$. Let $\mathfrak p$ be a prime ideal of $R$. Then we know that $x^ d + \sum _{i = 1, \ldots , d} \varphi (a_ i) x^{d - i} = 0$ in $S_{\mathfrak p}$ for some $a_ i \in R_{\mathfrak p}$. Hence we see, by looking at which denominators occur, that for some $f \in R$, $f \not\in \mathfrak p$ we have $a_ i \in R_ f$ and $x^ d + \sum _{i = 1, \ldots , d} \varphi (a_ i) x^{d - i} = 0$ in $S_ f$. This implies that $S'_ f$ is finite over $R_ f$. Since $\mathfrak p$ was arbitrary and $\mathop{\mathrm{Spec}}(R)$ is quasi-compact (Lemma 10.17.10) we can find finitely many elements $f_1, \ldots , f_ n \in R$ which generate the unit ideal of $R$ such that $S'_{f_ i}$ is finite over $R_{f_ i}$. Hence we conclude from Lemma 10.23.2 that $S'$ is finite over $R$. Hence $x$ is integral over $R$ by Lemma 10.36.4. $\square$

Proof. We prove (1). Let $s_ i \in S$ be generators for $S$ over $R$. Each of these satisfies a monic polynomial equation $P_ i$ over $R$. Hence the elements $1 \otimes s_ i \in S'$ generate $S'$ over $R'$ and satisfy the corresponding polynomial $P_ i'$ over $R'$. Since these elements generate $S'$ over $R'$ we see that $S'$ is integral over $R'$. Proof of (2) omitted. $\square$

Proof. Proof of (1). Let $s \in S$. Consider the ideal $I \subset R[x]$ of polynomials $P$ such that $P(s) = 0$. Let $J \subset R$ denote the ideal (!) of leading coefficients of elements of $I$. By assumption and clearing denominators we see that $f_ i^{n_ i} \in J$ for all $i$ and certain $n_ i \geq 0$. Hence $J$ contains $1$ and we see $s$ is integral over $R$. Proof of (2) omitted. $\square$

Proof. Omitted. $\square$

Proof. Omitted. $\square$

Proof. Let $\mathfrak p \subset R$ be a prime ideal. We have to show $\mathfrak pS_{\mathfrak p} \not= S_{\mathfrak p}$, see Lemma 10.17.9. The localization $R_{\mathfrak p} \to S_{\mathfrak p}$ is injective (as localization is exact) and integral by Lemma 10.36.11 or 10.36.13. Hence we may replace $R$, $S$ by $R_{\mathfrak p}$, $S_{\mathfrak p}$ and we may assume $R$ is local with maximal ideal $\mathfrak m$ and it suffices to show that $\mathfrak mS \not= S$. Suppose $1 = \sum f_ i s_ i$ with $f_ i \in \mathfrak m$ and $s_ i \in S$ in order to get a contradiction. Let $R \subset S' \subset S$ be such that $R \to S'$ is finite and $s_ i \in S'$, see Lemma 10.36.4. The equation $1 = \sum f_ i s_ i$ implies that the finite $R$-module $S'$ satisfies $S' = \mathfrak m S'$. Hence by Nakayama's Lemma 10.20.1 we see $S' = 0$. Contradiction. $\square$

Proof. Assume that $R \subset K$ is integral. By Lemma 10.36.17 we see that $\mathop{\mathrm{Spec}}(R)$ has $1$ point. Since clearly $R$ is a domain we see that $R = R_{(0)}$ is a field (Lemma 10.25.1). The other assertions are immediate from this. $\square$

Proof. The statement on primes follows from the statement “integral $+$ domain $\Rightarrow $ field”. Let $S$ integral over $k$ and assume $S$ is a domain, Take $s \in S$. By Lemma 10.36.4 we may find a finite dimensional $k$-subalgebra $k \subset S' \subset S$ containing $s$. Hence $S$ is a field if we can prove the first statement. Assume $S$ finite dimensional over $k$ and a domain. Pick $s\in S$. Since $S$ is a domain the multiplication map $s : S \to S$ is surjective by dimension reasons. Hence there exists an element $s_1 \in S$ such that $ss_1 = 1$. So $S$ is a field. $\square$

Proof. Let $\mathfrak p \subset R$ be the image. By Remark 10.17.8 the primes $\mathfrak q, \mathfrak q'$ correspond to ideals in $S \otimes _ R \kappa (\mathfrak p)$. Thus the lemma follows from Lemma 10.36.19. $\square$

Proof. By the discussion in Remark 10.17.8 the fibres are the spectra of the rings $S \otimes _ R \kappa (\mathfrak p)$. As $R \to S$ is finite, these fibre rings are finite over $\kappa (\mathfrak p)$ hence Noetherian by Lemma 10.31.1. By Lemma 10.36.20 every prime of $S \otimes _ R \kappa (\mathfrak p)$ is a minimal prime. Hence by Lemma 10.31.6 there are at most finitely many. $\square$

Proof. We may replace $R$ by $R/\mathfrak p$ and $S$ by $S/\mathfrak q$. This reduces us to the situation of having an integral extension of domains $R \subset S$ and a prime $\mathfrak p' \subset R$. By Lemma 10.36.17 we win. $\square$

Proof. One of the implications follows from Lemma 10.6.4. To see the other assume that $M$ is finitely presented as an $S$-module. Pick a presentation

As $S$ is finite as an $R$-module, the kernel of $S^{\oplus n} \to M$ is a finite $R$-module. Thus from Lemma 10.5.3 we see that it suffices to prove that $S$ is finitely presented as an $R$-module.

Pick $y_1, \ldots , y_ n \in S$ such that $y_1, \ldots , y_ n$ generate $S$ as an $R$-module. By Lemma 10.36.2 each $y_ i$ is integral over $R$. Choose monic polynomials $P_ i(x) \in R[x]$ with $P_ i(y_ i) = 0$. Consider the ring

Then we see that $S$ is of finite presentation as an $S'$-algebra by Lemma 10.6.2. Since $S' \to S$ is surjective, the kernel $J = \mathop{\mathrm{Ker}}(S' \to S)$ is finitely generated as an ideal by Lemma 10.6.3. Hence $J$ is a finite $S'$-module (immediate from the definitions). Thus $S = \mathop{\mathrm{Coker}}(J \to S')$ is of finite presentation as an $S'$-module by Lemma 10.5.3. Hence, arguing as in the first paragraph, it suffices to show that $S'$ is of finite presentation as an $R$-module. Actually, $S'$ is free as an $R$-module with basis the monomials $x_1^{e_1} \ldots x_ n^{e_ n}$ for $0 \leq e_ i < \deg (P_ i)$. Namely, write $R \to S'$ as the composition

This shows that the $i$th ring in this sequence is free as a module over the $(i - 1)$st one with basis $1, x_ i, \ldots , x_ i^{\deg (P_ i) - 1}$. The result follows easily from this by induction. Some details omitted. $\square$

Proof. Since $x/y \cdot y/x = 1$ it is clear that the map $R[x/y] \oplus R[y/x] \to R[x/y, y/x]$ is surjective. Let $\alpha \in R[x/y] \cap R[y/x]$. To show exactness in the middle we have to prove that $\alpha \in R$. By assumption we may write

for some $n, m \geq 0$ and $a_ i, b_ j \in R$. Pick some $N > \max (n, m)$. Consider the finite $R$-submodule $M$ of $R_{xy}$ generated by the elements

We claim that $\alpha M \subset M$. Namely, it is clear that $(x/y)^ i (b_0 + b_1 y/x + \ldots + b_ m(y/x)^ m) \in M$ for $0 \leq i \leq N$ and that $(y/x)^ i (a_0 + a_1 x/y + \ldots + a_ n(x/y)^ n) \in M$ for $0 \leq i \leq N$. Hence $\alpha $ is integral over $R$ by Lemma 10.36.2. Note that $\alpha \in R_ x$, so if $R$ is integrally closed in $R_ x$ then $\alpha \in R$ as desired. $\square$