Lemma 15.109.2. Let $(A, \mathfrak m)$ be a Noetherian local ring which is not formally catenary. Then $A$ is not universally catenary.

Proof. By assumption there exists a minimal prime $\mathfrak p \subset A$ such that the spectrum of $A^\wedge /\mathfrak p A^\wedge$ is not equidimensional. After replacing $A$ by $A/\mathfrak p$ we may assume that $A$ is a domain and that the spectrum of $A^\wedge$ is not equidimensional. Let $\mathfrak q$ be a minimal prime of $A^\wedge$ such that $d = \dim (A^\wedge /\mathfrak q)$ is minimal and hence $0 < d < \dim (A)$. We prove the lemma by induction on $d$.

The case $d = 1$. In this case $\dim (A^\wedge _\mathfrak q) = 0$. Hence $A^\wedge _\mathfrak q$ is Artinian local and we see that for some $n > 0$ the ideal $J = \mathfrak q^ n$ maps to zero in $A^\wedge _\mathfrak q$. It follows that $\mathfrak m$ is the only associated prime of $J/J^2$, whence $\mathfrak m^ m$ annihilates $J/J^2$ for some $m > 0$. Thus we can use Lemma 15.108.4 to find $A \to B$ of finite type such that $B^\wedge \cong A^\wedge /J$. It follows that $\mathfrak m_ B = \sqrt{\mathfrak mB}$ is a maximal ideal with the same residue field as $\mathfrak m$ and $B^\wedge$ is the $\mathfrak m_ B$-adic completion (Algebra, Lemma 10.97.7). Then

$\dim (B_{\mathfrak m_ B}) = \dim (B^\wedge ) = 1 = d.$

Since we have the factorization $A \to B \to A^\wedge /J$ the inverse image of $\mathfrak q/J$ is a prime $\mathfrak q' \subset \mathfrak m_ B$ lying over $(0)$ in $A$. Thus, if $A$ were universally catenary, the dimension formula (Algebra, Lemma 10.113.1) would give

\begin{align*} \dim (B_{\mathfrak m_ B}) & \geq \dim ((B/\mathfrak q')_{\mathfrak m_ B}) \\ & = \dim (A) + \text{trdeg}_ A(B/\mathfrak q') - \text{trdeg}_{\kappa (\mathfrak m)}(\kappa (\mathfrak m_ B)) \\ & = \dim (A) + \text{trdeg}_ A(B/\mathfrak q') \end{align*}

This contradiction finishes the argument in case $d = 1$.

Assume $d > 1$. Let $Z \subset \mathop{\mathrm{Spec}}(A^\wedge )$ be the union of the irreducible components distinct from $V(\mathfrak q)$. Let $\mathfrak r_1, \ldots , \mathfrak r_ m \subset A^\wedge$ be the prime ideals corresponding to irreducible components of $V(\mathfrak q) \cap Z$ of dimension $> 0$. Choose $f \in \mathfrak m$, $f \not\in A \cap \mathfrak r_ j$ using prime avoidance (Algebra, Lemma 10.15.2). Then $\dim (A/fA) = \dim (A) - 1$ and there is some irreducible component of $V(\mathfrak q, f)$ of dimension $d - 1$. Thus $A/fA$ is not formally catenary and the invariant $d$ has decreased. By induction $A/fA$ is not universally catenary, hence $A$ is not universally catenary. $\square$

Comment #5380 by Bjorn Poonen on

Typo in "This contradictions" at the end of the d=1 case.

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