Comparation of dimensions of rings Let $k$ be a field, $A$ be an integral domain, $B \subset A$, and $A, B$ are both finitely generated $k$ algebra. Let $p \subset B$ be a prime ideal. Suppose there exists prime ideals $q \subset A$, such that $q \cap B=p$, and $q$ is the minimal such ideal in the sense of inclusion.  Then, is it true $ dim A_q \leq dim B_p$ ?
For the geometric meaning, it comes from the exercise of Chapter 2, 3.22(a) of Hartshorne, where:           Let $ f: Spec(A) \to Spec(B) $ be a dominant morphism, $p \in Spec(B), Y'=$ {$ \bar{ p }$} (the closure of {$p$}) and $Z$ be an irreducible component of $f^{-1}(Y')$, whose generic point $q$ maps to $p$, then show that $ codim(Z,X) \leq codim(Y',Y)$.
I guess, everything translates faithfully to the above algebra fact except "$f$ dominant " is weakend by " $ B \to A$ is injective".  
 A: Dear Li, first of all I think that when you write "... such that $q \cap B=p$, and $q$ is the minimal such ideal in the sense of inclusion", you mean "... and $q$ is a minimal ideal...".
The answer to your question is given, I think, by the following more general result
Theorem Let $\phi: B\to A$ be a morphism of noetherian rings and ${\frak q} \subset A$ a prime ideal with inverse image ${\frak p}\subset B$. Then we have the following formula 
$$dim A_{\frak q}\leq dim B_{\frak p} +dim (A_{\frak q}\otimes_B \kappa(\frak p))                                  $$
Notice that there is no mention of injectivity for $\phi$ , nor of a field nor of finite generation of $A$ or $B$.
Now, if  $\frak q$ happens to be to be -as in your case- the generic point of one of the irreducible components of the fibre at $\frak p$ of the morphism $Spec(\phi): Spec(A) \to Spec(B)$, then the local ring of the fiber at $\frak q$, namely  $A_{\frak q}/{\frak p}A_{\frak q} =     A_{\frak q}\otimes_B \kappa(\frak p)$, is zero-dimensional (see reminder below) and you get the formula you wished.
Reminder The local ring of the generic point of an irreducible scheme is a ring having only one prime ideal (its nilpotent radical) and thus has dimension zero. If the scheme is also reduced, the local ring of its generic point is a field.
Bibliography
 Matsumura, Commutative Algebra, Theorem 19, page 79
 Matsumura, Commutative Ring Theory, Theorem 15.1, page 116
An example In view of Li's comment it might be of interest to some users to see an example.
Let $k$ be a field   ,    $B=k[t]$ and $A=k[t,X,Y]/(t-XY)=k[t,x,y]$. Let $\phi:B\to A$ be the inclusion. Then the fibre of ${\frak p} =(t)\in Spec(B)$ is the subscheme 
$F=V(t) \subset  Spec(A)$. Please note that, even though $A$ and $B$ are domains, the fibre has two irreducible components with generic points ${\frak q}=(t,x)$ and ${\frak q}'=(t,y)$. The potentially confusing fact is that to the "physical" point $Q={\frak q}$ (say) are associated two local ring. On the one hand the local ring of $Q$ in the scheme $Spec(A)$, which is $ \mathcal O_{Spec(A), Q}=A_{{\frak q}}$ $=A_{(t,x)}$ . And on the other the local ring of $Q$ in the scheme $F$, which is $\mathcal O_{F,Q}=A_{{\frak q}}/tA_{{\frak q}}=k(y)$, a field as expected.
