The Stacks project

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

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

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. $\square$

Proof. Let $\mathfrak p \subset A$ be a prime ideal. Let $\mathfrak q \subset B$ be a prime minimal over $\mathfrak pB$. Then $\mathfrak q \cap A = \mathfrak p$ by going down for $A \to B$ (Algebra, Lemma 10.39.19). Hence $A_\mathfrak p \to B_\mathfrak q$ is a flat local ring map with special fibre of dimension $0$ and hence

(Algebra, Lemma 10.112.7). The second equality because $\mathop{\mathrm{Spec}}(B)$ is equidimensional and $B$ is catenary. Thus $\dim (B/\mathfrak q)$ is independent of the choice of $\mathfrak q$ and we conclude that $\mathop{\mathrm{Spec}}(B/\mathfrak p B)$ is equidimensional of dimension $\dim (B) - \dim (A_\mathfrak p)$. On the other hand, we have $\dim (B/\mathfrak p B) = \dim (A/\mathfrak p) + \dim (B/\mathfrak m_ A B)$ and $\dim (B) = \dim (A) + \dim (B/\mathfrak m_ A B)$ by flatness (see lemma cited above) and we get

for all $\mathfrak p$ in $A$. Applying this to all minimal primes in $A$ we see that $A$ is equidimensional. If $\mathfrak p \subset \mathfrak p'$ is a strict inclusion with no primes in between, then we may apply the above to the prime $\mathfrak p'/\mathfrak p$ in $A/\mathfrak p$ because $A/\mathfrak p \to B/\mathfrak p B$ is flat and $\mathop{\mathrm{Spec}}(B/\mathfrak p B)$ is equidimensional, to get

Thus $\mathfrak p \mapsto \dim (A/\mathfrak p)$ is a dimension function and we conclude that $A$ is catenary. $\square$

Proof. We may replace $A$ by $A/\mathfrak p$ where $\mathfrak p$ is a minimal prime of $A$, see Algebra, Lemma 10.105.8. Thus we may assume that the spectrum of $A^\wedge $ is equidimensional. It suffices to show that every local ring essentially of finite type over $A$ is catenary (see for example Algebra, Lemma 10.105.6). Hence it suffices to show that $A[x_1, \ldots , x_ n]_\mathfrak m$ is catenary where $\mathfrak m \subset A[x_1, \ldots , x_ n]$ is a maximal ideal lying over $\mathfrak m_ A$, see Algebra, Lemma 10.54.5 (and Algebra, Lemmas 10.105.7 and 10.105.4). Let $\mathfrak m' \subset A^\wedge [x_1, \ldots , x_ n]$ be the unique maximal ideal lying over $\mathfrak m$. Then

is local and flat (Algebra, Lemma 10.97.2). Hence it suffices to show that the ring on the right hand side catenary with equidimensional spectrum, see Lemma 15.109.3. It is catenary because complete local rings are universally catenary (Algebra, Remark 10.160.9). Pick any minimal prime $\mathfrak q$ of $A^\wedge [x_1, \ldots , x_ n]_{\mathfrak m'}$. Then $\mathfrak q = \mathfrak p A^\wedge [x_1, \ldots , x_ n]_{\mathfrak m'}$ for some minimal prime $\mathfrak p$ of $A^\wedge $ (small detail omitted). Hence

the first equality by Algebra, Lemma 10.112.7 and the second because the spectrum of $A^\wedge $ is equidimensional. This finishes the proof. $\square$

Proof. Combine Lemmas 15.109.2 and 15.109.4. $\square$

Proof. By Lemma 15.108.8 the number of branches of $A$ and $A^\wedge $ are the same, hence Lemma 15.108.2 applies. Then for any minimal prime $\mathfrak q \subset A^ h$ we see that $A^\wedge /\mathfrak q A^\wedge $ has a unique minimal prime. Thus $A^ h$ is formally catenary (by definition) and hence universally catenary by Proposition 15.109.5. If $A$ is unibranch, then $A^ h$ has a unique minimal prime, hence $A^\wedge $ has a unique minimal prime, hence $A$ is formally catenary and we conclude in the same way. $\square$