## 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.

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 $\varphi : R[x] \to S[x]$.

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 $P(y) = P(y) \cdot 1 = P(y) \cdot x_1 = P(y) \cdot \pi ((1, 0, \ldots , 0)) = \pi (P(A)(1, 0, \ldots , 0)) = 0$.
$\square$

Lemma 10.35.3. A finite ring extension is integral.

**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

$R \to S$ is finite,

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

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.
$\square$

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 $S$.

**Proof.**
This is clear from Lemmas 10.35.4 and 10.35.3.
$\square$

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:

$x$ is integral over $R$, and

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 $S' = R' \otimes _ R S$.

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

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.

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

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.

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

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 $k$.

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

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

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)