The Stacks project

Proof. Omitted. $\square$

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. $\square$

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 $g \in S^{-1}R$. $\square$

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 $R$. $\square$

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 $R$. $\square$

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 $R$. $\square$

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. $\square$

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 $I = R$. $\square$

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 $R \to R_{\mathfrak p}$ 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 $x \in R$. $\square$

Proof. Omitted. $\square$

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. $\square$

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 $S$. $\square$

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. $\square$

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 $a/b \in R$. $\square$