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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
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  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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