The Stacks project

10.35 Finite and integral ring extensions

Trivial lemmas concerning finite and integral ring maps. We recall the definition.

Definition 10.35.1. Let $\varphi : R \to S$ be a ring map.

  1. An element $s \in S$ is integral over $R$ if there exists a monic polynomial $P(x) \in R[x]$ such that $P^\varphi (s) = 0$, where $P^\varphi (x) \in S[x]$ is the image of $P$ under $\varphi : R[x] \to S[x]$.

  2. The ring map $\varphi $ is integral if every $s \in S$ is integral over $R$.

Lemma 10.35.2. Let $\varphi : R \to S$ be a ring map. Let $y \in S$. If there exists a finite $R$-submodule $M$ of $S$ such that $1 \in M$ and $yM \subset M$, then $y$ is integral over $R$.

Proof. Let $x_1 = 1 \in M$ and $x_ i \in M$, $i = 2, \ldots , n$ be a finite set of elements generating $M$ as an $R$-module. Write $yx_ i = \sum \varphi (a_{ij}) x_ j$ for some $a_{ij} \in R$. Let $P(T) \in R[T]$ be the characteristic polynomial of the $n \times n$ matrix $A = (a_{ij})$. By Lemma 10.15.1 we see $P(A) = 0$. By construction the map $\pi : R^ n \to M$, $(a_1, \ldots , a_ n) \mapsto \sum \varphi (a_ i) x_ i$ commutes with $A : R^ n \to R^ n$ and multiplication by $y$. In a formula $\pi (Av) = y\pi (v)$. Thus $P(y) = P(y) \cdot 1 = P(y) \cdot x_1 = P(y) \cdot \pi ((1, 0, \ldots , 0)) = \pi (P(A)(1, 0, \ldots , 0)) = 0$. $\square$

Proof. Let $R \to S$ be finite. Let $y \in S$. Apply Lemma 10.35.2 to $M = S$ to see that $y$ is integral over $R$. $\square$

Lemma 10.35.4. Let $\varphi : R \to S$ be a ring map. Let $s_1, \ldots , s_ n$ be a finite set of elements of $S$. In this case $s_ i$ is integral over $R$ for all $i = 1, \ldots , n$ if and only if there exists an $R$-subalgebra $S' \subset S$ finite over $R$ containing all of the $s_ i$.

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 10.35.3. $\square$

Lemma 10.35.5. Let $R \to S$ be a ring map. The following are equivalent

  1. $R \to S$ is finite,

  2. $R \to S$ is integral and of finite type, and

  3. there exist $x_1, \ldots , x_ n \in S$ which generate $S$ as an algebra over $R$ such that each $x_ i$ is integral over $R$.

Proof. Clear from Lemma 10.35.4. $\square$

slogan

Lemma 10.35.6. Suppose that $R \to S$ and $S \to T$ are integral ring maps. Then $R \to T$ is integral.

Proof. Let $t \in T$. Let $P(x) \in S[x]$ be a monic polynomial such that $P(t) = 0$. Apply Lemma 10.35.4 to the finite set of coefficients of $P$. Hence $t$ is integral over some subalgebra $S' \subset S$ finite over $R$. Apply Lemma 10.35.4 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 10.35.3. $\square$

Lemma 10.35.7. Let $R \to S$ be a ring homomorphism. The set

\[ S' = \{ s \in S \mid s\text{ is integral over }R\} \]

is an $R$-subalgebra of $S$.

Lemma 10.35.8. Let $R_ i\to S_ i$ be ring maps $i = 1, \ldots , n$. Let $R$ and $S$ denote the product of the $R_ i$ and $S_ i$ respectively. Then an element $s = (s_1, \ldots , s_ n) \in S$ is integral over $R$ if and only if each $s_ i$ is integral over $R_ i$.

Proof. Omitted. $\square$

Definition 10.35.9. Let $R \to S$ be a ring map. The ring $S' \subset S$ of elements integral over $R$, see Lemma 10.35.7, is called the integral closure of $R$ in $S$. If $R \subset S$ we say that $R$ is integrally closed in $S$ if $R = S'$.

In particular, we see that $R \to S$ is integral if and only if the integral closure of $R$ in $S$ is all of $S$.

Lemma 10.35.10. Let $R_ i\to S_ i$ be ring maps $i = 1, \ldots , n$. Denote the integral closure of $R_ i$ in $S_ i$ by $S'_ i$. Further let $R$ and $S$ denote the product of the $R_ i$ and $S_ i$ respectively. Then the integral closure of $R$ in $S$ is the product of the $S'_ i$. In particular $R \to S$ is integrally closed if and only if each $R_ i \to S_ i$ is integrally closed.

Proof. This follows immediately from Lemma 10.35.8. $\square$

Lemma 10.35.11. Integral closure commutes with localization: If $A \to B$ is a ring map, and $S \subset A$ is a multiplicative subset, then the integral closure of $S^{-1}A$ in $S^{-1}B$ is $S^{-1}B'$, where $B' \subset B$ is the integral closure of $A$ in $B$.

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

\[ (x/f)^ d + \sum \nolimits _{i = 1, \ldots , d} a_ i/f^ i (x/f)^{d - i} = 0 \]

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

\[ (x/f)^ d + \sum \nolimits _{i = 1, \ldots , d} (a_ i/f_ i) (x/f)^{d - i} = 0 \]

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

\[ (f'f_1 \ldots f_ d x)^ d + \sum \nolimits _{i = 1, \ldots , d} f^ i(f')^ if_1^ i \ldots f_ i^{i - 1} \ldots f_ d^ i a_ i (f'f_1 \ldots f_ dx)^{d - i} = 0 \]

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. $\square$

slogan

Lemma 10.35.12. Let $\varphi : R \to S$ be a ring map. Let $x \in S$. The following are equivalent:

  1. $x$ is integral over $R$, and

  2. for every prime ideal $\mathfrak p \subset R$ the element $x \in S_{\mathfrak p}$ is integral over $R_{\mathfrak p}$.

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 10.16.10) 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 10.22.2 that $S'$ is finite over $R$. Hence $x$ is integral over $R$ by Lemma 10.35.4. $\square$

slogan

Lemma 10.35.13. Let $R \to S$ and $R \to R'$ be ring maps. Set $S' = R' \otimes _ R S$.

  1. If $R \to S$ is integral so is $R' \to S'$.

  2. If $R \to S$ is finite so is $R' \to S'$.

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. $\square$

Lemma 10.35.14. Let $R \to S$ be a ring map. Let $f_1, \ldots , f_ n \in R$ generate the unit ideal.

  1. If each $R_{f_ i} \to S_{f_ i}$ is integral, so is $R \to S$.

  2. If each $R_{f_ i} \to S_{f_ i}$ is finite, so is $R \to S$.

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. $\square$

Lemma 10.35.15. Let $A \to B \to C$ be ring maps.

  1. If $A \to C$ is integral so is $B \to C$.

  2. If $A \to C$ is finite so is $B \to C$.

Proof. Omitted. $\square$

Lemma 10.35.16. Let $A \to B \to C$ be ring maps. Let $B'$ be the integral closure of $A$ in $B$, let $C'$ be the integral closure of $B'$ in $C$. Then $C'$ is the integral closure of $A$ in $C$.

Proof. Omitted. $\square$

Lemma 10.35.17. Suppose that $R \to S$ is an integral ring extension with $R \subset S$. Then $\varphi : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective.

Proof. Let $\mathfrak p \subset R$ be a prime ideal. We have to show $\mathfrak pS_{\mathfrak p} \not= S_{\mathfrak p}$, see Lemma 10.16.9. The localization $R_{\mathfrak p} \to S_{\mathfrak p}$ is injective (as localization is exact) and integral by Lemma 10.35.11 or 10.35.13. 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 10.35.4. 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 10.19.1 we see $S' = 0$. Contradiction. $\square$

Lemma 10.35.18. Let $R$ be a ring. Let $K$ be a field. If $R \subset K$ and $K$ is integral over $R$, then $R$ is a field and $K$ is an algebraic extension. If $R \subset K$ and $K$ is finite over $R$, then $R$ is a field and $K$ is a finite algebraic extension.

Proof. Assume that $R \subset K$ is integral. By Lemma 10.35.17 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 10.24.1). The other assertions are immediate from this. $\square$

Lemma 10.35.19. Let $k$ be a field. Let $S$ be a $k$-algebra over $k$.

  1. If $S$ is a domain and finite dimensional over $k$, then $S$ is a field.

  2. If $S$ is integral over $k$ and a domain, then $S$ is a field.

  3. If $S$ is integral over $k$ then every prime of $S$ is a maximal ideal (see Lemma 10.25.5 for more consequences).

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 10.35.4 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. $\square$

Lemma 10.35.20. Suppose $R \to S$ is integral. Let $\mathfrak q, \mathfrak q' \in \mathop{\mathrm{Spec}}(S)$ be distinct primes having the same image in $\mathop{\mathrm{Spec}}(R)$. Then neither $\mathfrak q \subset \mathfrak q'$ nor $\mathfrak q' \subset \mathfrak q$.

Proof. Let $\mathfrak p \subset R$ be the image. By Remark 10.16.8 the primes $\mathfrak q, \mathfrak q'$ correspond to ideals in $S \otimes _ R \kappa (\mathfrak p)$. Thus the lemma follows from Lemma 10.35.19. $\square$

Lemma 10.35.21. Suppose $R \to S$ is finite. Then the fibres of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ are finite.

Proof. By the discussion in Remark 10.16.8 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 10.30.1. By Lemma 10.35.20 every prime of $S \otimes _ R \kappa (\mathfrak p)$ is a minimal prime. Hence by Lemma 10.30.6 there are at most finitely many. $\square$

Lemma 10.35.22. Let $R \to S$ be a ring map such that $S$ is integral over $R$. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q$ be a prime of $S$ mapping to $\mathfrak p$. Then there exists a prime $\mathfrak q'$ with $\mathfrak q \subset \mathfrak q'$ mapping to $\mathfrak p'$.

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 10.35.17 we win. $\square$

The property expressed in the lemma above is called the “going up property” for the ring map $R \to S$, see Definition 10.40.1.

Lemma 10.35.23. Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module. Then $M$ is finitely presented as an $R$-module if and only if $M$ is finitely presented as an $S$-module.

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

\[ S^{\oplus m} \longrightarrow S^{\oplus n} \longrightarrow M \longrightarrow 0 \]

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 10.35.2 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

\[ S' = R[x_1, \ldots , x_ n]/(P_1(x_1), \ldots , P_ n(x_ n)) \]

Then we see that $S$ is of finite presentation as an $S'$-algebra by Lemma 10.6.2. Since $S' \to S$ is surjective we see that $S$ is of finite presentation as an $S'$-module (use Lemma 10.6.3). Hence, arguing as in the first paragraph, it suffices to show that $S'$ is of finite presentation as an $R$-module. To see this we write $R \to S'$ as the composition

\[ R \to R[x_1]/(P_1(x_1)) \to R[x_1, x_2]/(P_1(x_1), P_2(x_2)) \to \ldots \to S' \]

of ring maps of the form $R' \to R'[x]/(x^ d + a_1 x^{d - 1} + \ldots + a_ d)$. Again arguing as in the first paragraph of the proof it is enough to show that the $i$th ring in this sequence is of finite presentation as a module over the $(i - 1)$st one. This is true because $R'[x]/(x^ d + a_1 x^{d - 1} + \ldots + a_ d)$ is free as a module over $R'$ with basis $1, x, \ldots , x^{d - 1}$. $\square$

Lemma 10.35.24. Let $R$ be a ring. Let $x, y \in R$ be nonzerodivisors. Let $R[x/y] \subset R_{xy}$ be the $R$-subalgebra generated by $x/y$, and similarly for the subalgebras $R[y/x]$ and $R[x/y, y/x]$. If $R$ is integrally closed in $R_ x$ or $R_ y$, then the sequence

\[ 0 \to R \xrightarrow {(-1, 1)} R[x/y] \oplus R[y/x] \xrightarrow {(1, 1)} R[x/y, y/x] \to 0 \]

is a short exact sequence of $R$-modules.

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

\[ \alpha = a_0 + a_1 x/y + \ldots + a_ n (x/y)^ n = b_0 + b_1 y/x + \ldots + b_ m(y/x)^ m \]

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

\[ (x/y)^ N, (x/y)^{N - 1}, \ldots , x/y, 1, y/x, \ldots , (y/x)^{N - 1}, (y/x)^ N \]

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 10.35.2. Note that $\alpha \in R_ x$, so if $R$ is integrally closed in $R_ x$ then $\alpha \in R$ as desired. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00GH. Beware of the difference between the letter 'O' and the digit '0'.