Lemma 15.109.4. Let $A$ be a formally catenary Noetherian local ring. Then $A$ is universally catenary.

**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. $\square$

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