# Krull's Principal Ideal Theorem

Let $R$ be a noetherian ring, and let $I$ be a proper ideal in $R$. If $I$ is generated by $n$ elements, we have by Krull's Principal Ideal Theorem that the height of $I$ is at most $n$. Is it true that $\operatorname{dim}R/I \ge \operatorname{dim}R - n$?

-
By definition . –  wxu May 22 '12 at 1:03
So if $\mathfrak p$ is a minimal prime over $I$... then this prime corresponds to a prime ideal in $R / I$. I'm not quite sure why I can extend this chain to length $\operatorname{dim}R - n$... –  Paul Slevin May 22 '12 at 1:14
Sorry, I make a mistake. –  wxu May 22 '12 at 1:22
Edited to make question clearer to future viewers –  Paul Slevin May 22 '12 at 9:17

The answer is no. Let $R=\mathbb{Z}_{(2)}[X]$, and $I=(2X-1)$, where $\mathbb{Z}_{(2)}$ is the ring $\mathbb{Z}$ localizing at the prime ideal $(2)$. Then $R/I$ is a field.
Another cheap counter-example is $R=k\times \mathbb{Z}[X]$ where $k$ is any field. Let $I$ generated by $(0,1)$, i.e., $I=\{0\}\times \mathbb{Z}[X]$. Then $R/I$ is a field. However $dim(R)=2$.
It is true for Noetherian local ring $(R,\mathfrak{m})$. Suppose dimension of $R/I$ is $r$. Then we can find $x_1,\ldots,x_r$ whose image generating an ideal of definition of $R/I$, that is $\mathfrak{m}/I$ is minimal over $(x_1,\ldots,x_r)/I$. Say $y_1,\ldots,y_n$ generate $I$, then $\mathfrak{m}$ is minimal over $(x_1,\ldots,x_r,y_1,\ldots,y_n)$, so $dim(R)\leq r+n=dim(R/I)+n$.
The reason I was asking is because I wanted to show that $\mathbb Z [x,y] / \langle f \rangle$ could not have dimension 1 for any f in $\mathbb Z[x,y]$ given that I know the dimension of $\mathbb Z[x,y]$ is 3. Is there another way to see this ? –  Paul Slevin May 22 '12 at 2:01