$\dim (A/I) \le \dim (A)$

Question

Let $A$ be a ring and $I$ be an ideal. I'm trying to prove that $\dim (A/I) \le \dim (A)$.

My attempt to proof

Suppose that $\dim (A)=n$, then there are prime ideals $\mathfrak p_0,\ldots,\mathfrak p_n$ such that $\mathfrak p_0\subsetneq\ldots\subsetneq\mathfrak p_n$.

Then if I prove chain bellow

$$\mathfrak p_0/I\subsetneq\ldots\subsetneq \mathfrak p_n/I$$

is a chain of prime ideals, we're done. The problem is I don't know why $I\subset \mathfrak p_i, 0\le i\le n$.

It should be a silly thing, I need help.

Thanks

Answer 1

Recall that the primes of $A/I$ are in inclusion-preserving one-to-one correspondence with the primes of $A$ containing $I$. For a proof of this (and related facts), see for example Zariski-Samuel's Commutative Algebra I, sections III.4-III.8.

The argument proceeds by a simple contradiction. If $\dim A/I >\dim A=n$, then there must exist a chain of primes in $A/I$ with length more than $n$. By the fact above, there exists a corresponding chain of primes in $A$ (containing $I$) of length more than $n$. This contradicts that $\dim A=n$.

The fact above, in addition to the fact that the primes of $A_{\mathfrak{p}}$ are in inclusion-preserving one-to-one correspondence with the primes of $A$ contained in $\mathfrak p$, are very good to keep in mind, especially in the context of dimension theory.