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# stacks/stacks-project

Equidimensional only defined for spaces

Perhaps still a couple of places left where this is used for rings...
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aisejohan committed Jul 24, 2019
1 parent 0ee15a0 commit 16a99faeca20ec239c12d6e12138b27528208bf4
Showing with 17 additions and 16 deletions.
1. +17 −16 more-algebra.tex
 \begin{definition} \label{definition-formally-catenary} A Noetherian local ring $A$ is {\it formally catenary} if for every minimal prime $\mathfrak p \subset A$ the ring if for every minimal prime $\mathfrak p \subset A$ the spectrum of $A^\wedge/\mathfrak p A^\wedge$ is equidimensional. \end{definition} By Ratliff's result (Proposition \ref{proposition-ratliff}) we see that any quotient of $A$ is also formally catenary (because the class of universally catenary rings is stable under quotients). We conclude that $A^\wedge/\mathfrak p A^\wedge$ is equidimensional for every prime ideal $\mathfrak p$ of $A$. We conclude that the spectrum of $A^\wedge/\mathfrak p A^\wedge$ is equidimensional for every prime ideal $\mathfrak p$ of $A$. \begin{lemma} \label{lemma-not-formally-catenary} \end{lemma} \begin{proof} By assumption there exists a minimal prime $\mathfrak p \subset A$ such that $A^\wedge /\mathfrak p A^\wedge$ is not equidimensional. 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 $A^\wedge$ is not equidimensional. 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 \begin{lemma} \label{lemma-flat-under-catenary-equidimensional} Let $A \to B$ be a flat local ring map of local Noetherian rings. Assume $B$ is catenary and equidimensional. Then Assume $B$ is catenary and is $\Spec(B)$ equidimensional. Then \begin{enumerate} \item $B/\mathfrak p B$ is equidimensional for all $\mathfrak p \subset A$, \item $A$ is catenary and equidimensional. \item $\Spec(B/\mathfrak p B)$ is equidimensional for all $\mathfrak p \subset A$ and \item $A$ is catenary and $\Spec(A)$ is equidimensional. \end{enumerate} \end{lemma} \dim(A_\mathfrak p) = \dim(B_\mathfrak q) = \dim(B) - \dim(B/\mathfrak q) $$(Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total}). The second equality because B is equidimensional and catenary. The second equality because \Spec(B) is equidimensional and B is catenary. Thus \dim(B/\mathfrak q) is independent of the choice of \mathfrak q and we conclude that B/\mathfrak p B is equidimensional of and we conclude that \Spec(B/\mathfrak p B) is equidimensional of dimension \dim(B) - \dim(A_\mathfrak p). On the other hand, we have \dim(B/\mathfrak p B) = \dim(A/\mathfrak p) + \dim(B/\mathfrak m_A B) with no primes in between, then we may apply the above to the prime \mathfrak p'/\mathfrak p in A/\mathfrak p because A/\mathfrak p \to B/\mathfrak p B is flat and B/\mathfrak p B is equidimensional, to get \Spec(B/\mathfrak p B) is equidimensional, to get$$ 1 = \dim((A/\mathfrak p)_{\mathfrak p'}) = \dim(A/\mathfrak p) - \dim(A/\mathfrak p') \begin{proof} We may replace $A$ by $A/\mathfrak p$ where $\mathfrak p$ is a minimal prime of $A$, see Algebra, Lemma \ref{algebra-lemma-catenary-check-irreducible}. Thus we may assume that $A^\wedge$ is equidimensional. 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 \ref{algebra-lemma-catenary-check-local}). $$is local and flat (Algebra, Lemma \ref{algebra-lemma-completion-flat}). Hence it suffices to show that the ring on the right hand side is equidimensional and catenary, see hand side catenary with equidimensional spectrum, see Lemma \ref{lemma-flat-under-catenary-equidimensional}. It is catenary because complete local rings are universally catenary (Algebra, Remark$$ the first equality by Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total} and the second because $A^\wedge$ is equidimensional. and the second because the spectrum of $A^\wedge$ is equidimensional. This finishes the proof. \end{proof} Lemma \ref{lemma-syspar}. We can use Lemma \ref{lemma-minprimespolyhigher} because $R$ is a catenary domain, so every height one prime ideal of $R$ has dimension $d - 1$, and hence $R/(f + h)$ is has dimension $d - 1$, and hence the spectrum of $R/(f + h)$ is equidimensional. For the convenience of the reader we write out the details.

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