Definition 10.37.1. A domain $R$ is called *normal* if it is integrally closed in its field of fractions.

## 10.37 Normal rings

We first introduce the notion of a normal domain, and then we introduce the (very general) notion of a normal ring.

Lemma 10.37.2. Let $R \to S$ be a ring map. If $S$ is a normal domain, then the integral closure of $R$ in $S$ is a normal domain.

**Proof.**
Omitted.
$\square$

The following notion is occasionally useful when studying normality.

Definition 10.37.3. Let $R$ be a domain.

An element $g$ of the fraction field of $R$ is called

*almost integral over $R$*if there exists an element $r \in R$, $r\not= 0$ such that $rg^ n \in R$ for all $n \geq 0$.The domain $R$ is called

*completely normal*if every almost integral element of the fraction field of $R$ is contained in $R$.

The following lemma shows that a Noetherian domain is normal if and only if it is completely normal.

Lemma 10.37.4. Let $R$ be a domain with fraction field $K$. If $u, v \in K$ are almost integral over $R$, then so are $u + v$ and $uv$. Any element $g \in K$ which is integral over $R$ is almost integral over $R$. If $R$ is Noetherian then the converse holds as well.

**Proof.**
If $ru^ n \in R$ for all $n \geq 0$ and $v^ nr' \in R$ for all $n \geq 0$, then $(uv)^ nrr'$ and $(u + v)^ nrr'$ are in $R$ for all $n \geq 0$. Hence the first assertion. Suppose $g \in K$ is integral over $R$. In this case there exists an $d > 0$ such that the ring $R[g]$ is generated by $1, g, \ldots , g^ d$ as an $R$-module. Let $r \in R$ be a common denominator of the elements $1, g, \ldots , g^ d \in K$. It is follows that $rR[g] \subset R$, and hence $g$ is almost integral over $R$.

Suppose $R$ is Noetherian and $g \in K$ is almost integral over $R$. Let $r \in R$, $r\not= 0$ be as in the definition. Then $R[g] \subset \frac{1}{r}R$ as an $R$-module. Since $R$ is Noetherian this implies that $R[g]$ is finite over $R$. Hence $g$ is integral over $R$, see Lemma 10.36.3. $\square$

Lemma 10.37.5. Any localization of a normal domain is normal.

**Proof.**
Let $R$ be a normal domain, and let $S \subset R$ be a multiplicative subset. Suppose $g$ is an element of the fraction field of $R$ which is integral over $S^{-1}R$. Let $P = x^ d + \sum _{j < d} a_ j x^ j$ be a polynomial with $a_ i \in S^{-1}R$ such that $P(g) = 0$. Choose $s \in S$ such that $sa_ i \in R$ for all $i$. Then $sg$ satisfies the monic polynomial $x^ d + \sum _{j < d} s^{d-j}a_ j x^ j$ which has coefficients $s^{d-j}a_ j$ in $R$. Hence $sg \in R$ because $R$ is normal. Hence $g \in S^{-1}R$.
$\square$

Lemma 10.37.6. A principal ideal domain is normal.

**Proof.**
Let $R$ be a principal ideal domain. Let $g = a/b$ be an element of the fraction field of $R$ integral over $R$. Because $R$ is a principal ideal domain we may divide out a common factor of $a$ and $b$ and assume $(a, b) = R$. In this case, any equation $(a/b)^ n + r_{n-1} (a/b)^{n-1} + \ldots + r_0 = 0$ with $r_ i \in R$ would imply $a^ n \in (b)$. This contradicts $(a, b) = R$ unless $b$ is a unit in $R$.
$\square$

Lemma 10.37.7. Let $R$ be a domain with fraction field $K$. Suppose $f = \sum \alpha _ i x^ i$ is an element of $K[x]$.

If $f$ is integral over $R[x]$ then all $\alpha _ i$ are integral over $R$, and

If $f$ is almost integral over $R[x]$ then all $\alpha _ i$ are almost integral over $R$.

**Proof.**
We first prove the second statement. Write $f = \alpha _0 + \alpha _1 x + \ldots + \alpha _ r x^ r$ with $\alpha _ r \not= 0$. By assumption there exists $h = b_0 + b_1 x + \ldots + b_ s x^ s \in R[x]$, $b_ s \not= 0$ such that $f^ n h \in R[x]$ for all $n \geq 0$. This implies that $b_ s \alpha _ r^ n \in R$ for all $n \geq 0$. Hence $\alpha _ r$ is almost integral over $R$. Since the set of almost integral elements form a subring (Lemma 10.37.4) we deduce that $f - \alpha _ r x^ r = \alpha _0 + \alpha _1 x + \ldots + \alpha _{r - 1} x^{r - 1}$ is almost integral over $R[x]$. By induction on $r$ we win.

In order to prove the first statement we will use absolute Noetherian reduction. Namely, write $\alpha _ i = a_ i / b_ i$ and let $P(t) = t^ d + \sum _{j < d} f_ j t^ j$ be a polynomial with coefficients $f_ j \in R[x]$ such that $P(f) = 0$. Let $f_ j = \sum f_{ji}x^ i$. Consider the subring $R_0 \subset R$ generated by the finite list of elements $a_ i, b_ i, f_{ji}$ of $R$. It is a domain; let $K_0$ be its field of fractions. Since $R_0$ is a finite type $\mathbf{Z}$-algebra it is Noetherian, see Lemma 10.31.3. It is still the case that $f \in K_0[x]$ is integral over $R_0[x]$, because all the identities in $R$ among the elements $a_ i, b_ i, f_{ji}$ also hold in $R_0$. By Lemma 10.37.4 the element $f$ is almost integral over $R_0[x]$. By the second statement of the lemma, the elements $\alpha _ i$ are almost integral over $R_0$. And since $R_0$ is Noetherian, they are integral over $R_0$, see Lemma 10.37.4. Of course, then they are integral over $R$. $\square$

Lemma 10.37.8. Let $R$ be a normal domain. Then $R[x]$ is a normal domain.

**Proof.**
The result is true if $R$ is a field $K$ because $K[x]$ is a euclidean domain and hence a principal ideal domain and hence normal by Lemma 10.37.6. Let $g$ be an element of the fraction field of $R[x]$ which is integral over $R[x]$. Because $g$ is integral over $K[x]$ where $K$ is the fraction field of $R$ we may write $g = \alpha _ d x^ d + \alpha _{d-1}x^{d-1} + \ldots + \alpha _0$ with $\alpha _ i \in K$. By Lemma 10.37.7 the elements $\alpha _ i$ are integral over $R$ and hence are in $R$.
$\square$

Lemma 10.37.9. Let $R$ be a Noetherian normal domain. Then $R[[x]]$ is a Noetherian normal domain.

**Proof.**
The power series ring is Noetherian by Lemma 10.31.2. Let $f, g \in R[[x]]$ be nonzero elements such that $w = f/g$ is integral over $R[[x]]$. Let $K$ be the fraction field of $R$. Since the ring of Laurent series $K((x)) = K[[x]][1/x]$ is a field, we can write $w = a_ n x^ n + a_{n + 1} x^{n + 1} + \ldots )$ for some $n \in \mathbf{Z}$, $a_ i \in K$, and $a_ n \not= 0$. By Lemma 10.37.4 we see there exists a nonzero element $h = b_ m x^ m + b_{m + 1} x^{m + 1} + \ldots $ in $R[[x]]$ with $b_ m \not= 0$ such that $w^ e h \in R[[x]]$ for all $e \geq 1$. We conclude that $n \geq 0$ and that $b_ m a_ n^ e \in R$ for all $e \geq 1$. Since $R$ is Noetherian this implies that $a_ n \in R$ by the same lemma. Now, if $a_ n, a_{n + 1}, \ldots , a_{N - 1} \in R$, then we can apply the same argument to $w - a_ n x^ n - \ldots - a_{N - 1} x^{N - 1} = a_ N x^ N + \ldots $. In this way we see that all $a_ i \in R$ and the lemma is proved.
$\square$

Lemma 10.37.10. Let $R$ be a domain. The following are equivalent:

The domain $R$ is a normal domain,

for every prime $\mathfrak p \subset R$ the local ring $R_{\mathfrak p}$ is a normal domain, and

for every maximal ideal $\mathfrak m$ the ring $R_{\mathfrak m}$ is a normal domain.

**Proof.**
This follows easily from the fact that for any domain $R$ we have

inside the fraction field of $R$. Namely, if $g$ is an element of the right hand side then the ideal $I = \{ x \in R \mid xg \in R\} $ is not contained in any maximal ideal $\mathfrak m$, whence $I = R$. $\square$

Lemma 10.37.10 shows that the following definition is compatible with Definition 10.37.1. (It is the definition from EGA – see [IV, 5.13.5 and 0, 4.1.4, EGA].)

Definition 10.37.11. A ring $R$ is called *normal* if for every prime $\mathfrak p \subset R$ the localization $R_{\mathfrak p}$ is a normal domain (see Definition 10.37.1).

Note that a normal ring is a reduced ring, as $R$ is a subring of the product of its localizations at all primes (see for example Lemma 10.23.1).

Lemma 10.37.12. A normal ring is integrally closed in its total ring of fractions.

**Proof.**
Let $R$ be a normal ring. Let $x \in Q(R)$ be an element of the total ring of fractions of $R$ integral over $R$. Set $I = \{ f \in R, fx \in R\} $. Let $\mathfrak p \subset R$ be a prime. As $R \subset R_{\mathfrak p}$ is flat we see that $R_{\mathfrak p} \subset Q(R) \otimes _ R R_{\mathfrak p}$. As $R_{\mathfrak p}$ is a normal domain we see that $x \otimes 1$ is an element of $R_{\mathfrak p}$. Hence we can find $a, f \in R$, $f \not\in \mathfrak p$ such that $x \otimes 1 = a \otimes 1/f$. This means that $fx - a$ maps to zero in $Q(R) \otimes _ R R_{\mathfrak p} = Q(R)_{\mathfrak p}$, which in turn means that there exists an $f' \in R$, $f' \not\in \mathfrak p$ such that $f'fx = f'a$ in $R$. In other words, $ff' \in I$. Thus $I$ is an ideal which isn't contained in any of the prime ideals of $R$, i.e., $I = R$ and $x \in R$.
$\square$

Lemma 10.37.13. A localization of a normal ring is a normal ring.

**Proof.**
Omitted.
$\square$

Lemma 10.37.14. Let $R$ be a normal ring. Then $R[x]$ is a normal ring.

**Proof.**
Let $\mathfrak q$ be a prime of $R[x]$. Set $\mathfrak p = R \cap \mathfrak q$. Then we see that $R_{\mathfrak p}[x]$ is a normal domain by Lemma 10.37.8. Hence $(R[x])_{\mathfrak q}$ is a normal domain by Lemma 10.37.5.
$\square$

Lemma 10.37.15. A finite product of normal rings is normal.

**Proof.**
It suffices to show that the product of two normal rings, say $R$ and $S$, is normal. By Lemma 10.21.3 the prime ideals of $R\times S$ are of the form $\mathfrak {p}\times S$ and $R\times \mathfrak {q}$, where $\mathfrak {p}$ and $\mathfrak {q}$ are primes of $R$ and $S$ respectively. Localization yields $(R\times S)_{\mathfrak {p}\times S}=R_{\mathfrak {p}}$ which is a normal domain by assumption. Similarly for $S$.
$\square$

Lemma 10.37.16. Let $R$ be a ring. Assume $R$ is reduced and has finitely many minimal primes. Then the following are equivalent:

$R$ is a normal ring,

$R$ is integrally closed in its total ring of fractions, and

$R$ is a finite product of normal domains.

**Proof.**
The implications (1) $\Rightarrow $ (2) and (3) $\Rightarrow $ (1) hold in general, see Lemmas 10.37.12 and 10.37.15.

Let $\mathfrak p_1, \ldots , \mathfrak p_ n$ be the minimal primes of $R$. By Lemmas 10.25.2 and 10.25.4 we have $Q(R) = R_{\mathfrak p_1} \times \ldots \times R_{\mathfrak p_ n}$, and by Lemma 10.25.1 each factor is a field. Denote $e_ i = (0, \ldots , 0, 1, 0, \ldots , 0)$ the $i$th idempotent of $Q(R)$.

If $R$ is integrally closed in $Q(R)$, then it contains in particular the idempotents $e_ i$, and we see that $R$ is a product of $n$ domains (see Sections 10.22 and 10.24). Each factor is of the form $R/\mathfrak p_ i$ with field of fractions $R_{\mathfrak p_ i}$. By Lemma 10.36.10 each map $R/\mathfrak p_ i \to R_{\mathfrak p_ i}$ is integrally closed. Hence $R$ is a finite product of normal domains. $\square$

Lemma 10.37.17. Let $(R_ i, \varphi _{ii'})$ be a directed system (Categories, Definition 10.8.1) of rings. If each $R_ i$ is a normal ring so is $R = \mathop{\mathrm{colim}}\nolimits _ i R_ i$.

**Proof.**
Let $\mathfrak p \subset R$ be a prime ideal. Set $\mathfrak p_ i = R_ i \cap \mathfrak p$ (usual abuse of notation). Then we see that $R_{\mathfrak p} = \mathop{\mathrm{colim}}\nolimits _ i (R_ i)_{\mathfrak p_ i}$. Since each $(R_ i)_{\mathfrak p_ i}$ is a normal domain we reduce to proving the statement of the lemma for normal domains. If $a, b \in R$ and $a/b$ satisfies a monic polynomial $P(T) \in R[T]$, then we can find a (sufficiently large) $i \in I$ such that $a, b$ come from objects $a_ i, b_ i$ over $R_ i$, $P$ comes from a monic polynomial $P_ i\in R_ i[T]$ and $P_ i(a_ i/b_ i)=0$. Since $R_ i$ is normal we see $a_ i/b_ i \in R_ i$ and hence also $a/b \in R$.
$\square$

## 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.

## Comments (3)

Comment #5460 by Sandeep on

Comment #5461 by Johan on

Comment #5464 by Sandeep on