Show that $\dim(A) \leq \dim(B)$. Let $A \subseteq B$ be rings, $B$ integral over $A$. Show that $\dim(A) \leq \dim(B)$. 
My idea: Let  $\dim(A) = m $ and $\dim(B) = n$. There is a chain of prime ideals in $A$:
$$P_0 \varsubsetneq P_1 \varsubsetneq ... \varsubsetneq P_m.$$ 
Then there are prime ideals $Q_0,Q_1,...,Q_m$ in $B$ such that $P_i=Q_i \cap A$.
So 
$$Q_0 \cap A \varsubsetneq Q_1 \cap A \varsubsetneq ... \varsubsetneq Q_m \cap A$$
is a chain of prime ideals in $A$.
I need to find a prime chain in $B$ with length greater than or equal to $m$.I tried to take the chain $Q_0 \varsubsetneq Q_1  \varsubsetneq ... \varsubsetneq Q_m $, but I can not prove the "$\subseteq$", I think this is not valid. I'm trying to use extension too.
 A: As noted in the comments, this follows from the Cohen-Seidenberg Theorems, specifically the Lying Over and Going-Up Theorems.  (These are Theorem 5.10 and Theorem 5.11 (p. 62) in Atiyah-Macdonald.)
You don't actually need to assume that the dimensions of $A$ and $B$ are finite.
Suppose $\DeclareMathOperator{\p}{\mathfrak{p}} \DeclareMathOperator{\q}{\mathfrak{q}} \p_0 \subsetneq \p_1 \subsetneq \cdots \subsetneq \p_m$ is a chain of prime ideals of $A$.  By Lying Over, there exists a prime ideal $\q_0$ of $B$ such that $\q_0 \cap A = \p_0$.  By Going-Up, this can be extended to a chain $\q_0 \subseteq \q_1 \subseteq \cdots \subseteq \q_m$ of prime ideals of $B$ such that $\q_i \cap A = \p_i$ for $i = 0, \ldots, m$.  Moreover, since
$$
\q_i \cap A = \p_i \neq \p_{i+1} = \q_{i+1} \cap A
$$
then $\q_i \neq \q_{i+1}$ for each $i$, so we have produced a chain $\q_0 \subsetneq \q_1 \subsetneq \cdots \subsetneq \q_m$ of prime ideals of length $m$.  Thus for each chain of prime ideals of $A$, we have exhibited a chain of prime ideals of $B$ that has length greater than or equal to the first chain.  Since dimension is defined as the supremum of lengths of such chains, this shows that $\dim(A) \leq \dim(B)$.
The Incomparability Theorem (Corollary 5.9) shows that the reverse inequality holds as well, so in fact $\dim(A) = \dim(B)$, which shows that integral extensions preserve dimension.  (This is proved and expressed in geometric terms in Proposition 9.2 and Corollary 9.3 (p. 227) of Eisenbud's Commutative Algebra.)
