## 10.118 Around Krull-Akizuki

One application of Krull-Akizuki is to show that there are plenty of discrete valuation rings. More generally in this section we show how to construct discrete valuation rings dominating Noetherian local rings.

First we show how to dominate a Noetherian local domain by a $1$-dimensional Noetherian local domain by blowing up the maximal ideal.

Lemma 10.118.1. Let $R$ be a local Noetherian domain with fraction field $K$. Assume $R$ is not a field. Then there exist $R \subset R' \subset K$ with

$R'$ local Noetherian of dimension $1$,

$R \to R'$ a local ring map, i.e., $R'$ dominates $R$, and

$R \to R'$ essentially of finite type.

**Proof.**
Choose any valuation ring $A \subset K$ dominating $R$ (which exist by Lemma 10.49.2). Denote $v$ the corresponding valuation. Let $x_1, \ldots , x_ r$ be a minimal set of generators of the maximal ideal $\mathfrak m$ of $R$. We may and do assume that $v(x_ r) = \min \{ v(x_1), \ldots , v(x_ r)\} $. Consider the ring

\[ S = R[x_1/x_ r, x_2/x_ r, \ldots , x_{r - 1}/x_ r] \subset K. \]

Note that $\mathfrak mS = x_ rS$ is a principal ideal. Note that $S \subset A$ and that $v(x_ r) > 0$, hence we see that $x_ rS \not= S$. Choose a minimal prime $\mathfrak q$ over $x_ rS$. Then $\text{height}(\mathfrak q) = 1$ by Lemma 10.59.10 and $\mathfrak q$ lies over $\mathfrak m$. Hence we see that $R' = S_{\mathfrak q}$ is a solution.
$\square$

reference
Lemma 10.118.2 (Kollár). Let $(R, \mathfrak m)$ be a local Noetherian ring. Then exactly one of the following holds:

$(R, \mathfrak m)$ is Artinian,

$(R, \mathfrak m)$ is regular of dimension $1$,

$\text{depth}(R) \geq 2$, or

there exists a finite ring map $R \to R'$ which is not an isomorphism whose kernel and cokernel are annihilated by a power of $\mathfrak m$ such that $\mathfrak m$ is not an associated prime of $R'$.

**Proof.**
Observe that $(R, \mathfrak m)$ is not Artinian if and only if $V(\mathfrak m) \subset \mathop{\mathrm{Spec}}(R)$ is nowhere dense. See Proposition 10.59.6. We assume this from now on.

Let $J \subset R$ be the largest ideal killed by a power of $\mathfrak m$. If $J \not= 0$ then $R \to R/J$ shows that $(R, \mathfrak m)$ is as in (4).

Otherwise $J = 0$. In particular $\mathfrak m$ is not an associated prime of $R$ and we see that there is a nonzerodivisor $x \in \mathfrak m$ by Lemma 10.62.18. If $\mathfrak m$ is not an associated prime of $R/xR$ then $\text{depth}(R) \geq 2$ by the same lemma. Thus we are left with the case when there is an $y \in R$, $y \not\in xR$ such that $y \mathfrak m \subset xR$.

If $y \mathfrak m \subset x \mathfrak m$ then we can consider the map $\varphi : \mathfrak m \to \mathfrak m$, $f \mapsto yf/x$ (well defined as $x$ is a nonzerodivisor). By the determinantal trick of Lemma 10.15.2 there exists a monic polynomial $P$ with coefficients in $R$ such that $P(\varphi ) = 0$. We conclude that $P(y/x) = 0$ in $R_ x$. Let $R' \subset R_ x$ be the ring generated by $R$ and $y/x$. Then $R \subset R'$ and $R'/R$ is a finite $R$-module annihilated by a power of $x$. Thus $R$ is as in (4).

Otherwise there is a $t \in \mathfrak m$ such that $y t = u x$ for some unit $u$ of $R$. After replacing $t$ by $u^{-1}t$ we get $yt = x$. In particular $y$ is a nonzerodivisor. For any $t' \in \mathfrak m$ we have $y t' = x s$ for some $s \in R$. Thus $y (t' - s t ) = x s - x s = 0$. Since $y$ is not a zero-divisor this implies that $t' = ts$ and so $\mathfrak m = (t)$. Thus $(R, \mathfrak m)$ is regular of dimension 1.
$\square$

Lemma 10.118.3. Let $R$ be a local ring with maximal ideal $\mathfrak m$. Assume $R$ is Noetherian, has dimension $1$, and that $\dim (\mathfrak m/\mathfrak m^2) > 1$. Then there exists a ring map $R \to R'$ such that

$R \to R'$ is finite,

$R \to R'$ is not an isomorphism,

the kernel and cokernel of $R \to R'$ are annihilated by a power of $\mathfrak m$, and

$\mathfrak m$ is not an associated prime of $R'$.

**Proof.**
This follows from Lemma 10.118.2 and the fact that $R$ is not Artinian, not regular, and does not have depth $\geq 2$ (the last part because the depth does not exceed the dimension by Lemma 10.71.3).
$\square$

Example 10.118.4. Consider the Noetherian local ring

\[ R = k[[x, y]]/(y^2) \]

It has dimension 1 and it is Cohen-Macaulay. An example of an extension as in Lemma 10.118.3 is the extension

\[ k[[x, y]]/(y^2) \subset k[[x, z]]/(z^2), \ \ y \mapsto xz \]

in other words it is gotten by adjoining $y/x$ to $R$. The effect of repeating the construction $n > 1$ times is to adjoin the element $y/x^ n$.

Example 10.118.5. Let $k$ be a field of characteristic $p > 0$ such that $k$ has infinite degree over its subfield $k^ p$ of $p$th powers. For example $k = \mathbf{F}_ p(t_1, t_2, t_3, \ldots )$. Consider the ring

\[ A = \left\{ \sum a_ i x^ i \in k[[x]] \text{ such that } [k^ p(a_0, a_1, a_2, \ldots ) : k^ p] < \infty \right\} \]

Then $A$ is a discrete valuation ring and its completion is $A^\wedge = k[[x]]$. Note that the induced extension of fraction fields of $A \subset k[[x]]$ is infinite purely inseparable. Choose any $f \in k[[x]]$, $f \not\in A$. Let $R = A[f] \subset k[[x]]$. Then $R$ is a Noetherian local domain of dimension $1$ whose completion $R^\wedge $ is nonreduced (think!).

Lemma 10.118.7. Let $A$ be a ring. The following are equivalent.

The ring $A$ is a discrete valuation ring.

The ring $A$ is a valuation ring and Noetherian.

The ring $A$ is a regular local ring of dimension $1$.

The ring $A$ is a Noetherian local domain with maximal ideal $\mathfrak m$ generated by a single nonzero element.

The ring $A$ is a Noetherian local normal domain of dimension $1$.

In this case if $\pi $ is a generator of the maximal ideal of $A$, then every element of $A$ can be uniquely written as $u\pi ^ n$, where $u \in A$ is a unit.

**Proof.**
The equivalence of (1) and (2) is Lemma 10.49.18. Moreover, in the proof of Lemma 10.49.18 we saw that if $A$ is a discrete valuation ring, then $A$ is a PID, hence (3). Note that a regular local ring is a domain (see Lemma 10.105.2). Using this the equivalence of (3) and (4) follows from dimension theory, see Section 10.59.

Assume (3) and let $\pi $ be a generator of the maximal ideal $\mathfrak m$. For all $n \geq 0$ we have $\dim _{A/\mathfrak m} \mathfrak m^ n/\mathfrak m^{n + 1} = 1$ because it is generated by $\pi ^ n$ (and it cannot be zero). In particular $\mathfrak m^ n = (\pi ^ n)$ and the graded ring $\bigoplus \mathfrak m^ n/\mathfrak m^{n + 1}$ is isomorphic to the polynomial ring $A/\mathfrak m[T]$. For $x \in A \setminus \{ 0\} $ define $v(x) = \max \{ n \mid x \in \mathfrak m^ n\} $. In other words $x = u \pi ^{v(x)}$ with $u \in A^*$. By the remarks above we have $v(xy) = v(x) + v(y)$ for all $x, y \in A \setminus \{ 0\} $. We extend this to the field of fractions $K$ of $A$ by setting $v(a/b) = v(a) - v(b)$ (well defined by multiplicativity shown above). Then it is clear that $A$ is the set of elements of $K$ which have valuation $\geq 0$. Hence we see that $A$ is a valuation ring by Lemma 10.49.16.

A valuation ring is a normal domain by Lemma 10.49.10. Hence we see that the equivalent conditions (1) – (3) imply (5). Assume (5). Suppose that $\mathfrak m$ cannot be generated by $1$ element to get a contradiction. Then Lemma 10.118.3 implies there is a finite ring map $A \to A'$ which is an isomorphism after inverting any nonzero element of $\mathfrak m$ but not an isomorphism. In particular we may identify $A'$ with a subset of the fraction field of $A$. Since $A \to A'$ is finite it is integral (see Lemma 10.35.3). Since $A$ is normal we get $A = A'$ a contradiction.
$\square$

Definition 10.118.8. Let $A$ be a discrete valuation ring. A *uniformizer* is an element $\pi \in A$ which generates the maximal ideal of $A$.

By Lemma 10.118.7 any two uniformizers of a discrete valuation ring are associates.

Lemma 10.118.9. Let $R$ be a domain with fraction field $K$. Let $M$ be an $R$-submodule of $K^{\oplus r}$. Assume $R$ is local Noetherian of dimension $1$. For any nonzero $x \in R$ we have $\text{length}_ R(R/xR) < \infty $ and

\[ \text{length}_ R(M/xM) \leq r \cdot \text{length}_ R(R/xR). \]

**Proof.**
If $x$ is a unit then the result is true. Hence we may assume $x \in \mathfrak m$ the maximal ideal of $R$. Since $x$ is not zero and $R$ is a domain we have $\dim (R/xR) = 0$, and hence $R/xR$ has finite length. Consider $M \subset K^{\oplus r}$ as in the lemma. We may assume that the elements of $M$ generate $K^{\oplus r}$ as a $K$-vector space after replacing $K^{\oplus r}$ by a smaller subspace if necessary.

Suppose first that $M$ is a finite $R$-module. In that case we can clear denominators and assume $M \subset R^{\oplus r}$. Since $M$ generates $K^{\oplus r}$ as a vectors space we see that $R^{\oplus r}/M$ has finite length. In particular there exists an integer $c \geq 0$ such that $x^ cR^{\oplus r} \subset M$. Note that $M \supset xM \supset x^2M \supset \ldots $ is a sequence of modules with successive quotients each isomorphic to $M/xM$. Hence we see that

\[ n \text{length}_ R(M/xM) = \text{length}_ R(M/x^ nM). \]

The same argument for $M = R^{\oplus r}$ shows that

\[ n \text{length}_ R(R^{\oplus r}/xR^{\oplus r}) = \text{length}_ R(R^{\oplus r}/x^ nR^{\oplus r}). \]

By our choice of $c$ above we see that $x^ nM$ is sandwiched between $x^ n R^{\oplus r}$ and $x^{n + c}R^{\oplus r}$. This easily gives that

\[ r(n + c) \text{length}_ R(R/xR) \geq n \text{length}_ R(M/xM) \geq r (n - c) \text{length}_ R(R/xR) \]

Hence in the finite case we actually get the result of the lemma with equality.

Suppose now that $M$ is not finite. Suppose that the length of $M/xM$ is $\geq k$ for some natural number $k$. Then we can find

\[ 0 \subset N_0 \subset N_1 \subset N_2 \subset \ldots N_ k \subset M/xM \]

with $N_ i \not= N_{i + 1}$ for $i = 0, \ldots k - 1$. Choose an element $m_ i \in M$ whose congruence class mod $xM$ falls into $N_ i$ but not into $N_{i - 1}$ for $i = 1, \ldots , k$. Consider the finite $R$-module $M' = Rm_1 + \ldots + Rm_ k \subset M$. Let $N'_ i \subset M'/xM'$ be the inverse image of $N_ i$. It is clear that $N'_ i \not=N'_{i + 1}$ by our choice of $m_ i$. Hence we see that $\text{length}_ R(M'/xM') \geq k$. By the finite case we conclude $k \leq r\text{length}_ R(R/xR)$ as desired.
$\square$

Here is a first application.

Lemma 10.118.10. Let $R \to S$ be a homomorphism of domains inducing an injection of fraction fields $K \subset L$. If $R$ is Noetherian local of dimension $1$ and $[L : K] < \infty $ then

each prime ideal $\mathfrak n_ i$ of $S$ lying over the maximal ideal $\mathfrak m$ of $R$ is maximal,

there are finitely many of these, and

$[\kappa (\mathfrak n_ i) : \kappa (\mathfrak m)] < \infty $ for each $i$.

**Proof.**
Pick $x \in \mathfrak m$ nonzero. Apply Lemma 10.118.9 to the submodule $S \subset L \cong K^{\oplus n}$ where $n = [L : K]$. Thus the ring $S/xS$ has finite length over $R$. It follows that $S/\mathfrak m S$ has finite length over $\kappa (\mathfrak m)$. In other words, $\dim _{\kappa (\mathfrak m)} S/\mathfrak m S$ is finite (Lemma 10.51.6). Thus $S/\mathfrak mS$ is Artinian (Lemma 10.52.2). The structural results on Artinian rings implies parts (1) and (2), see for example Lemma 10.52.6. Part (3) is implied by the finiteness established above.
$\square$

Lemma 10.118.11. Let $R$ be a domain with fraction field $K$. Let $M$ be an $R$-submodule of $K^{\oplus r}$. Assume $R$ is Noetherian of dimension $1$. For any nonzero $x \in R$ we have $\text{length}_ R(M/xM) < \infty $.

**Proof.**
Since $R$ has dimension $1$ we see that $x$ is contained in finitely many primes $\mathfrak m_ i$, $i = 1, \ldots , n$, each maximal. Since $R$ is Noetherian we see that $R/xR$ is Artinian, see Proposition 10.59.6. Hence $R/xR$ is a quotient of $\prod R/\mathfrak m_ i^{e_ i}$ for certain $e_ i$ because that $\mathfrak m_1^{e_1} \ldots \mathfrak m_ n^{e_ n} \subset (x)$ for suitably large $e_ i$ as $R/xR$ is Artinian (see Section 10.52). Hence $M/xM$ similarly decomposes as a product $\prod (M/xM)_{\mathfrak m_ i} = \prod M/(\mathfrak m_ i^{e_ i}, x)M$ of its localizations at the $\mathfrak m_ i$. By Lemma 10.118.9 applied to $M_{\mathfrak m_ i}$ over $R_{\mathfrak m_ i}$ we see each $M_{\mathfrak m_ i}/xM_{\mathfrak m_ i} = (M/xM)_{\mathfrak m_ i}$ has finite length over $R_{\mathfrak m_ i}$. It easily follows that $M/xM$ has finite length over $R$.
$\square$

Lemma 10.118.12 (Krull-Akizuki). Let $R$ be a domain with fraction field $K$. Let $K \subset L$ be a finite extension of fields. Assume $R$ is Noetherian and $\dim (R) = 1$. In this case any ring $A$ with $R \subset A \subset L$ is Noetherian.

**Proof.**
To begin we may assume that $L$ is the fraction field of $A$ by replacing $L$ by the fraction field of $A$ if necessary. Let $I \subset A$ be a nonzero ideal. Clearly $I$ generates $L$ as a $K$-vector space. Hence we see that $I \cap R \not= (0)$. Pick any nonzero $x \in I \cap R$. Then we get $I/xA \subset A/xA$. By Lemma 10.118.11 the $R$-module $A/xA$ has finite length as an $R$-module. Hence $I/xA$ has finite length as an $R$-module. Hence $I$ is finitely generated as an ideal in $A$.
$\square$

Lemma 10.118.13. Let $R$ be a Noetherian local domain with fraction field $K$. Assume that $R$ is not a field. Let $K \subset L$ be a finitely generated field extension. Then there exists discrete valuation ring $A$ with fraction field $L$ which dominates $R$.

**Proof.**
If $L$ is not finite over $K$ choose a transcendence basis $x_1, \ldots , x_ r$ of $L$ over $K$ and replace $R$ by $R[x_1, \ldots , x_ r]$ localized at the maximal ideal generated by $\mathfrak m_ R$ and $x_1, \ldots , x_ r$. Thus we may assume $K \subset L$ finite.

By Lemma 10.118.1 we may assume $\dim (R) = 1$.

Let $A \subset L$ be the integral closure of $R$ in $L$. By Lemma 10.118.12 this is Noetherian. By Lemma 10.35.17 there is a prime ideal $\mathfrak q \subset A$ lying over the maximal ideal of $R$. By Lemma 10.118.7 the ring $A_{\mathfrak q}$ is a discrete valuation ring dominating $R$ as desired.
$\square$

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