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104.16 A non catenary Noetherian local ring

Even though there is a succesful dimension theory of Noetherian local rings there are non-catenary Noetherian local rings. An example may be found in [Appendix, Example 2, Nagata]. In fact, we will present this example in the simplest case. Namely, we will construct a local Noetherian domain $A$ of dimension $2$ which is not universally catenary. (Note that $A$ is automatically catenary, see Exercises, Exercise 105.17.3.) The existence of a Noetherian local ring which is not universally catenary implies the existence of a Noetherian local ring which is not catenary – and we spell this out at the end of this section in the particular example at hand.

Let $k$ be a field, and consider the formal power series ring $k[[x]]$ in one variable over $k$. Let

\[ z = \sum \nolimits _{i = 1}^\infty a_ i x^ i \]

be a formal power series. We assume $z$ as an element of the Laurent series field $k((x)) = k[[x]][1/x]$ is transcendental over $k(x)$. Put

\[ z_ j = x^{-j}(z - \sum \nolimits _{i = 1, \ldots , j - 1} a_ i x^ i) = \sum \nolimits _{i = j}^\infty a_ i x^{i - j} \in k[[x]]. \]

Note that $z = z_1$. Let $R$ be the subring of $k[[x]]$ generated by $x$, $z$ and all of the $z_ j$, in other words

\[ R = k[x, z_1, z_2, z_3, \ldots ] \subset k[[x]]. \]

Consider the ideals $\mathfrak m = (x)$ and $\mathfrak n = (x - 1, z_1, z_2, \ldots )$ of $R$.

We have $x(z_{j + 1} + a_ j) = z_ j$. Hence $R/\mathfrak m = k$ and $\mathfrak m$ is a maximal ideal. Moreover, any element of $R$ not in $\mathfrak m$ maps to a unit in $k[[x]]$ and hence $R_{\mathfrak m} \subset k[[x]]$. In fact it is easy to deduce that $R_{\mathfrak m}$ is a discrete valuation ring and residue field $k$.

We claim that

\[ R/(x - 1) = k[x, z_1, z_2, z_3, \ldots ]/(x - 1) \cong k[z]. \]

Namely, the relation above implies that $(x - 1)(z_{j + 1} + a_ j) = -z_{j + 1} - a_ j + z_ j$, and hence we may express the class of $z_{j + 1}$ in terms of $z_ j$ in the quotient $R/(x - 1)$. Since the fraction field of $R$ has transcendence degree $2$ over $k$ by construction we see that $z$ is transcendental over $k$ in $R/(x - 1)$, whence the desired isomorphism. Hence $\mathfrak n = (x - 1, z)$ and is a maximal ideal. In fact the map

\[ k[x, x^{-1}, z]_{(x - 1, z)} \longrightarrow R_{\mathfrak n} \]

is an isomorphism (since $x^{-1}$ is invertible in $R_{\mathfrak n}$ and since $z_{j + 1} = x^{-1}z_ j - a_ j = \ldots = f_ j(x, x^{-1}, z)$). This shows that $R_{\mathfrak n}$ is a regular local ring of dimension $2$ and residue field $k$.

Let $S$ be the multiplicative subset

\[ S = (R \setminus \mathfrak m) \cap (R \setminus \mathfrak n) = R \setminus (\mathfrak m \cup \mathfrak n) \]

and set $B = S^{-1}R$. We claim that

  1. The ring $B$ is a $k$-algebra.

  2. The maximal ideals of the ring $B$ are the two ideals $\mathfrak mB$ and $\mathfrak nB$.

  3. The residue fields at these maximal ideals is $k$.

  4. We have $B_{\mathfrak mB} = R_{\mathfrak m}$ and $B_{\mathfrak nB} = R_{\mathfrak n}$ which are Noetherian regular local rings of dimensions $1$ and $2$.

  5. The ring $B$ is Noetherian.

We omit the details of the verifications.

Whenever given a $k$-algebra $B$ with the properties listed above we get an example as follows. Take $A = k + \text{rad}(B) \subset B$ with $\text{rad}(B) = \mathfrak mB + \mathfrak nB$ the Jacobson radical. It is easy to see that $B$ is finite over $A$ and hence $A$ is Noetherian by Eakin's theorem (see [Eakin], or [Appendix A1, Nagata], or insert future reference here). Also $A$ is a local domain with the same fraction field as $B$ and residue field $k$. Since the dimension of $B$ is $2$ we see that $A$ has dimension $2$ as well, by Algebra, Lemma 10.111.4.

If $A$ were universally catenary then the dimension formula, Algebra, Lemma 10.112.1 would give $\dim (B_{\mathfrak mB}) = 2$ contradiction.

Note that $B$ is generated by one element over $A$. Hence $B = A[x]/\mathfrak p$ for some prime $\mathfrak p$ of $A[x]$. Let $\mathfrak m' \subset A[x]$ be the maximal ideal corresponding to $\mathfrak mB$. Then on the one hand $\dim (A[x]_{\mathfrak m'}) = 3$ and on the other hand

\[ (0) \subset \mathfrak pA[x]_{\mathfrak m'} \subset \mathfrak m'A[x]_{\mathfrak m'} \]

is a maximal chain of primes. Hence $A[x]_{\mathfrak m'}$ is an example of a non catenary Noetherian local ring.

Comments (1)

Comment #4261 by Manuel Hoff on

It is claimed that . But with the above definition we have .

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