# Why intermediate rings of Krull have dimension $\leq 1$?

Let $R$ be a Noetherian domain of Krull dimension $\leq 1$ with the quotient field $K$. Let $T$ be an intermediate ring between $R$ and $K$. Then, why is $T$ of Krull dimension $\leq 1$?

(Reference: Kaplansky, Commutative Rings, p.61.)

The proof given in the book shows that $T$ is a Noetherian domain and $T/I$ is a Noetherian $R$-module for each non-zero ideal $I$ of $T$.

However, I do not get how to conclude $\dim(T)\leq 1$ from the facts given above. How?

This would be true if we can show that $T/I$ is an Artinian ring for each nonzero ideal $I$, but I do not know how this can be proved.

• $R/I\cap R\subset T/I$ is a finite ring extension. Note that $I\ne(0)$ shows that $I\cap R\ne(0)$. Since $\dim R/I\cap R=0$ you get $\dim T/I=0$. – user26857 Sep 11 '17 at 18:38
• @user26857 Oh right. Thank you! – Rubertos Sep 11 '17 at 18:47