Let $B \subseteq A$ be an integral extension of integral domains where $B$ is normal (i.e. integrally closed in its own fraction field ). Let $P$ be a prime ideal of $A$ and $Q=P \cap B$ . Then how to show that $\dim A_P=\dim B_Q$ ?

I know that the extension $B \subseteq A$ satisfies both Going-Up and Going-down property (along with in comparability ). But I am not sure how to derive the dimension equality from that. Please help.


The dimension of $B_Q$ is the largest length of a chains of the primes in $B$ of the form $Q_1 \subset \cdots \subset Q_n$ such that $Q_n = Q$. Same for $A_P$ and $P$.

If you begin with a chain $Q_1 \subset \cdots \subset Q_n = Q$ in $B$, incomparability (a property of integral extensions) gives you a chain $Q_1 \cap A \subset \cdots Q_n \cap A = P$ in $A$ with the same length, with the last term equal to $P$. This shows that $\operatorname{Dim} B_Q \leq \operatorname{Dim} A_P$.

If you begin with a chain $P_1 \subset \cdots \subset P_n = P$ in $A$, and you let $Q_n =Q$, then going-down tells you that there exists a chain $Q_1 \subset \cdots \subset Q_n = Q$ of primes in $B$ with $Q_i$ lying over $P_i$. In particular, there is a chain of $B$, ending at $Q$, with the same length. So $\operatorname{Dim} A_P \leq \operatorname{Dim} B_Q$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.