Lemma 29.17.2. Let $S$ be a locally Noetherian scheme. The following are equivalent

1. $S$ is universally catenary,

2. there exists an open covering of $S$ all of whose members are universally catenary schemes,

3. for every affine open $\mathop{\mathrm{Spec}}(R) = U \subset S$ the ring $R$ is universally catenary, and

4. there exists an affine open covering $S = \bigcup U_ i$ such that each $U_ i$ is the spectrum of a universally catenary ring.

Moreover, in this case any scheme locally of finite type over $S$ is universally catenary as well.

Proof. By Lemma 29.15.5 an open immersion is locally of finite type. A composition of morphisms locally of finite type is locally of finite type (Lemma 29.15.3). Thus it is clear that if $S$ is universally catenary then any open and any scheme locally of finite type over $S$ is universally catenary as well. This proves the final statement of the lemma and that (1) implies (2).

If $\mathop{\mathrm{Spec}}(R)$ is a universally catenary scheme, then every scheme $\mathop{\mathrm{Spec}}(A)$ with $A$ a finite type $R$-algebra is catenary. Hence all these rings $A$ are catenary by Algebra, Lemma 10.105.2. Thus $R$ is universally catenary. Combined with the remarks above we conclude that (1) implies (3), and (2) implies (4). Of course (3) implies (4) trivially.

To finish the proof we show that (4) implies (1). Assume (4) and let $X \to S$ be a morphism locally of finite type. We can find an affine open covering $X = \bigcup V_ j$ such that each $V_ j \to S$ maps into one of the $U_ i$. By Lemma 29.15.2 the induced ring map $\mathcal{O}(U_ i) \to \mathcal{O}(V_ j)$ is of finite type. Hence $\mathcal{O}(V_ j)$ is catenary. Hence $X$ is catenary by Properties, Lemma 28.11.2. $\square$

Comment #5477 by David Hansen on

Might be handy to add another equivalence here: $S$ is universally catenary iff each irreducible component of $S$ is universally catenary.

Comment #5692 by on

Thanks for the suggestion. In the algebra case we already have Lemma 10.105.8. For the schemes case I have added this fact as a separate lemma, see this commit.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).