# Why does $\dim(A/P)\leq n$ imply $\dim(A)\leq n$ for an algebra $A$ over a field $K$ with $P\in \operatorname{Spec}(A)$.

This is a rather specific question stemming from the proof of lemma 5.6 in Gregor Kemper's A Course in Commutative Algebra. The lemma is as follows:

Let $A$ be an algebra over a field $K$, and let $S\subseteq A$ be a subset that generates $A$ as an algebra. Then $$\dim(A)\leq \sup\{|T| \mid T\subseteq S ~\text{is finite and algebraically independent} \}.$$

Calling the supremum on the right side $n$, Kemper says that we need to show $\text{dim}(A/P)\leq n$ for all $P\in\text{Spec}(A)$. However, I don't see why it couldn't be that $\text{dim}(A/P)<\text{dim}(A)$ for every $P\in\text{Spec}(A)$ which would lead to the proof being insufficient as I see it. The dimension used here is the Krull dimension.

• If $\dim A<\infty$, then $\dim A=\sup\{\dim A/P:P\in\operatorname{Spec}A\}$. Commented Apr 19, 2017 at 13:32

If $\dim A < \infty$, then $\dim A = \dim A/P$ for some minimal prime ideal $P$.

Proof for this: Let $\dim A=n$. By the definition of the dimension, there is a chain of prime ideals $P_0 \subset P_1 \subset \dotsc \subset P_n$ with all inclusions strict. Then $\dim A/P_0=n$, because we have the chain $0 \subset P_1/P_0 \subset \dotsc \subset P_n/P_0$, thus $\dim A/P_0 \geq n$. The other inequality holds for any quotient.

If $\dim A = \infty$, then $\sup\limits_{P} \dim A/P = \infty$, i.e. if we show $\dim A/P \leq n$ for any $P$, we have automatically shown the desired $n=\infty$.

• Thanks for the reply. I don't quite understand why the first statement is necessarily true though. Why does taking the quotient with a certain minimal $P$ necessarily not change the dimension? Commented Apr 19, 2017 at 11:02
• Just take a chain of prime ideals with length equal to the dimension (such a chain exists if the dimension is finite). The first prime ideal of this chain has the desired property (and is of course minimal).
– MooS
Commented Apr 19, 2017 at 11:12
• So we'd get a chain $\{0\}\subsetneqq P_1\subsetneqq ...\subsetneqq P_{n-1}$ which is the longest possible chain. Then $A/P_1$ would have a chain $\{0\} = P_1/P_1\subsetneqq P_2/P_1\subsetneqq ... \subsetneqq P_{n-1}/P_1$, right? i.e. the chain would now be shorter by 1, wouldn't it? Commented Apr 19, 2017 at 11:55
• Sorry, that should extend to $P_n$, not $P_{n-1}$ for dimension $n$. Commented Apr 19, 2017 at 12:03
• Wait, is it as simple as a field being a Noetherian topological space and hence the dimension is equal to the dimension of one of the closed irreducible subsets of the field? Commented Apr 19, 2017 at 12:12