This is a comment to 23950's question. 

To me, it seems like you are asking two different questions. <br/>
(1) How does height of an ideal in a ring related to height of it image in a factor ring? <br/> (2) How does height behave under localizations?


For 1): I don't think it behaves well in general. If a ring is caterary and equidimensional than it behaves better. For example, Lemma 2, p. 250, Matsumura, Commutative Ring Theory, says 

- If an equidimensional local ring $(A,m)$ is catenary then 
ht $p_2$ = ht $p_1$ + ht $(p_2/p_1)$ for all $p_1,p_2 \in$ Spec $A$ with $p_1 \subset p_2$.

For 2): Consider, $R = k[x,y,z]$ and $I = (xz,yz)$. Notice that ht $I = 1$ since $I \subset (z)$. Localize at $z$. Then $R_z = k[x,y,z,z/1]$ and $IR_z = (xz,yz)R_z = (x,y)R_z$. Hence ht $IR_z = 2$. This is a famous example that R/I is not Cohen-Macaulay at $(x,y,z)$ since it is not equidimensional (a line passes though a plane).

--------------------------------------------------------------- added June 3rd 2012

This is to answer the question below http://mathoverflow.net/questions/97730/height-of-ideal-in-graded-ring/98695#98695 . 

Hi. Assume ht $I \neq $ ht $I'$. We can find a homogeneous prime $q$ of the same height as $I'$. By going modulo $q$ we may assume the following <br/>

 - $R$ is graded, <br/> 
 - $I$ is not zero, $I' = 0$, <br/>
 - $q = 0$. 

Now we goto the localization $R_{(0)}$ as introduced by Thomas above. Since $I$ is not zero this ring is isomorphic to $k[t,t^{-1}]$ which is a PID. Hence $IR_{(0)}$ is of height $1$ and contained in a maximal ideal Q of height $1$. Here you can see $Q' = 0$. This shows that the preimage of $Q$ in $R$ contains $I$ and its $'$ is $q$. Now use 1.5.8b in Brunz-Herzog's book.

In other words, there exists a prime ideal $Q$ in $R$ such that $q = Q'$ and $I \subset Q$. Hence ht $I -1 \le$ ht $Q -1 = $ ht $q = $ ht $I'$ where the equality ht $Q - 1 = $ ht $q$ is from 1.5.8b.