The Stacks project

Proof. We consider the collection of local subrings of $K$ as a partially ordered set using the relation of domination. Suppose that $\{ A_ i\} _{i \in I}$ is a totally ordered collection of local subrings of $K$. Then $B = \bigcup A_ i$ is a local subring which dominates all of the $A_ i$. Hence by Zorn's Lemma, it suffices to show that if $A \subset K$ is a local ring whose fraction field is not $K$, then there exists a local ring $B \subset K$, $B \not= A$ dominating $A$.

Pick $t \in K$ which is not in the fraction field of $A$. If $t$ is transcendental over $A$, then $A[t] \subset K$ and hence $A[t]_{(t, \mathfrak m)} \subset K$ is a local ring distinct from $A$ dominating $A$. Suppose $t$ is algebraic over $A$. Then for some nonzero $a \in A$ the element $at$ is integral over $A$. In this case the subring $A' \subset K$ generated by $A$ and $ta$ is finite over $A$. By Lemma 10.36.17 there exists a prime ideal $\mathfrak m' \subset A'$ lying over $\mathfrak m$. Then $A'_{\mathfrak m'}$ dominates $A$. If $A = A'_{\mathfrak m'}$, then $t$ is in the fraction field of $A$ which we assumed not to be the case. Thus $A \not= A'_{\mathfrak m'}$ as desired. $\square$

Proof. Suppose $x$ is in the field of fractions of $A$ and integral over $A$. Let $A'$ denote the subring of $K$ generated by $A$ and $x$. Since $A\subset A'$ is an integral extension, we see by Lemma 10.36.17 that there is a prime ideal $\mathfrak m' \subset A'$ lying over $\mathfrak m$. Then $A'_{\mathfrak m'}$ dominates $A$. Since $A$ is a valuation ring we conclude that $A=A'_{\mathfrak m'}$ and therefore that $x\in A$. $\square$

Proof. Assume that $x$ is not in $A$. Let $A'$ denote the subring of $K$ generated by $A$ and $x$. Since $A$ is a valuation ring we see that there is no prime of $A'$ lying over $\mathfrak m$. Since $\mathfrak m$ is maximal we see that $V(\mathfrak m A') = \emptyset $. Then $\mathfrak m A' = A'$ by Lemma 10.17.2. Hence we can write $1 = \sum _{i = 0}^ d t_ i x^ i$ with $t_ i \in \mathfrak m$. This implies that $(1 - t_0) (x^{-1})^ d - \sum t_ i (x^{-1})^{d - i} = 0$. In particular we see that $x^{-1}$ is integral over $A$, and hence $x^{-1} \in A$ by Lemma 10.50.3. $\square$

Proof. If $A$ is not $K$, then $A$ is not a field and there is a nonzero maximal ideal $\mathfrak m$. If $\mathfrak m'$ is a second maximal ideal, then choose $x, y \in A$ with $x \in \mathfrak m$, $y \not\in \mathfrak m$, $x \not\in \mathfrak m'$, and $y \in \mathfrak m'$. Then neither $x/y \in A$ nor $y/x \in A$ contradicting the assumption of the lemma. Thus we see that $A$ is a local ring. Suppose that $A'$ is a local ring contained in $K$ which dominates $A$. Let $x \in A'$. We have to show that $x \in A$. If not, then $x^{-1} \in A$, and of course $x^{-1} \in \mathfrak m_ A$. But then $x^{-1} \in \mathfrak m_{A'}$ which contradicts $x \in A'$. $\square$

Proof. It is clear that $A$ is a domain. Let $a, b \in A$. Lemma 10.50.5 tells us we have to show that either $a | b$ or $b | a$ in $A$. Choose $i$ so large that there exist $a_ i, b_ i \in A_ i$ mapping to $a, b$. Then Lemma 10.50.4 applied to $a_ i, b_ i$ in $A_ i$ implies the result for $a, b$ in $A$. $\square$

Proof. We can replace $L$ by the fraction field $F$ of $B$ and $K$ by $K \cap F$. Then the lemma follows from a combination of Lemmas 10.50.4 and 10.50.5. $\square$

Proof. By Lemma 10.50.7 the ring $A$ is a valuation ring. If $A$ is a field, then $A = K$. Then $A = K \subset B$ is an integral extension, hence there are no proper inclusions among the primes of $B$ (Lemma 10.36.20). This contradicts the assumption that $B$ is a local domain and not a field. $\square$

Proof. Use the characterization of valuation rings given in Lemma 10.50.5. $\square$

Proof. Note that $\mathfrak m_{A'} \subset C$ and $C/\mathfrak m_{A'} = A$. In particular, the fraction field of $C$ is equal to the fraction field of $A'$. We will use the criterion of Lemma 10.50.5 to prove the lemma. Let $x$ be an element of the fraction field of $C$. By the lemma we may assume $x \in A'$. If $x \in \mathfrak m_{A'}$, then we see $x \in C$. If not, then $x$ is a unit of $A'$ and we also have $x^{-1} \in A'$. Hence either $x$ or $x^{-1}$ maps to an element of $A$ by the lemma again. $\square$

Proof. Suppose $x \in K$, $x \not\in A$. Consider $B = A[x^{-1}]$. Then $x \not\in B$. Namely, if $x = a_0 + a_1x^{-1} + \ldots + a_ d x^{-d}$ then $x^{d + 1} - a_0x^ d - \ldots - a_ d = 0$ and $x$ is integral over $A$ in contradiction with the fact that $A$ is normal. Thus $x^{-1}$ is not a unit in $B$. Thus $V(x^{-1}) \subset \mathop{\mathrm{Spec}}(B)$ is not empty (Lemma 10.17.2), and we can choose a prime $\mathfrak p \subset B$ with $x^{-1} \in \mathfrak p$. Choose a valuation ring $V \subset K$ dominating $B_\mathfrak p$ (Lemma 10.50.2). Then $x \not\in V$ as $x^{-1} \in \mathfrak m_ V$.

If $A$ is local, then we claim that $x^{-1} B + \mathfrak m_ A B \not= B$. Namely, if $1 = (a_0 + a_1x^{-1} + \ldots + a_ d x^{-d})x^{-1} + a'_0 + \ldots + a'_ d x^{-d}$ with $a_ i \in A$ and $a'_ i \in \mathfrak m_ A$, then we'd get

Since $a'_0 \in \mathfrak m_ A$ we see that $1 - a'_0$ is a unit in $A$ and we conclude that $x$ would be integral over $A$, a contradiction as before. Then choose the prime $\mathfrak p \supset x^{-1} B + \mathfrak m_ A B$ we find $V$ dominating $A$. $\square$

Proof. Omitted, but follows easily from Lemma 10.50.4. Note that in case $A = K$ we obtain the zero group $\Gamma = \{ 0\} $ endowed with its unique total ordering. $\square$

Proof. Omitted. $\square$

Proof. Assume $A$ is a valuation ring and let $f_1, \ldots , f_ n \in A$. Choose $i$ such that $v(f_ i)$ is minimal among $v(f_ j)$. Then $(f_ i) = (f_1, \ldots , f_ n)$. Conversely, assume $A$ is a local domain and every finitely generated ideal of $A$ is principal. Pick $f, g \in A$ and write $(f, g) = (h)$. Then $f = ah$ and $g = bh$ and $h = cf + dg$ for some $a, b, c, d \in A$. Thus $ac + bd = 1$ and we see that either $a$ or $b$ is a unit, i.e., either $g/f$ or $f/g$ is an element of $A$. This shows $A$ is a valuation ring by Lemma 10.50.5. $\square$

Proof. Omitted. $\square$

Proof. Omitted. $\square$

Proof. Suppose $A$ is a discrete valuation ring with valuation $v : A \setminus \{ 0\} \to \mathbf{Z}$ normalized so that $\mathop{\mathrm{Im}}(v) = \mathbf{Z}_{\geq 0}$. By Lemma 10.50.17 the ideals of $A$ are the subsets $I_ n = \{ 0\} \cup v^{-1}(\mathbf{Z}_{\geq n})$. It is clear that any element $x \in A$ with $v(x) = n$ generates $I_ n$. Hence $A$ is a PID so certainly Noetherian.

Suppose $A$ is a Noetherian valuation ring with value group $\Gamma $. By Lemma 10.50.17 we see the ascending chain condition holds for ideals in $\Gamma $. We may assume $A$ is not a field, i.e., there is a $\gamma \in \Gamma $ with $\gamma > 0$. Applying the ascending chain condition to the subsets $\gamma + \Gamma _{\geq 0}$ with $\gamma > 0$ we see there exists a smallest element $\gamma _0$ which is bigger than $0$. Let $\gamma \in \Gamma $ be an element $\gamma > 0$. Consider the sequence of elements $\gamma $, $\gamma - \gamma _0$, $\gamma - 2\gamma _0$, etc. By the ascending chain condition these cannot all be $> 0$. Let $\gamma - n \gamma _0$ be the last one $\geq 0$. By minimality of $\gamma _0$ we see that $0 = \gamma - n \gamma _0$. Hence $\Gamma $ is a cyclic group as desired. $\square$