Integral dependence over rings is transitive 
Let $A\subset B\subset C$ be commutative rings. Suppose $B$ is integral over $A$, and $C$ is integral over $B$. Then I want to show that $C$ is integral over $A$.

To be integral means that for every $b\in B$, there exists a monic polynomial with coefficients in $A$ such that $b$ is a root, i.e. $b^n+a_{n-1}b^{n-1}+\cdots+a_1b+a_0=0$ for $a_{n-1},\ldots,a_0\in A$. Likewise, for any $c\in C$ we can write $c^k+b_{k-1}c^{k-1}+\cdots+b_1c+b_0=0$. Now, how can we write $c$ as a polynomial with coefficients in $A$?
 A: This is a very standard but nontrivial result: see e.g. $\S$ 14.1 of these notes.
This occurs in the first section of a long chapter on integral extensions and is the fifth result in that section.  Especially, the proof uses the fact that if $R \subset S$ is a ring extension, then $\alpha \in S$ is integral over $R$ if and only if there is a subring $T$ with $R \subset R[\alpha] \subset T \subset S$ with $T$ finitely generated as an $R$-module.
A: Here's a simple proof for finitely generated algebras.

Lemma 1: Let $S$ be an algebra over $R$.  Then $\alpha\in S$ is integral over a $R$ iff $R[\alpha]$ is finitely generated as a module over R.
(Note that this lemma works even for $S$ not finitely generated over $R$).
Proof: 
(Forward).  If $\alpha$ is integral over $R$, then there is some monic polyomial $f\in R[x]$ s.t. $f(\alpha)=\alpha^n+f_{n-1}\alpha^{n-1}+\cdots+f_1\alpha+f_0=0$.  Thus $\alpha^n=-(f_{n-1}\alpha^{n-1}+\cdots+f_1\alpha+f_0)$, and so any powers of $\alpha$ greater than or equal to $n$ can be rewritten in terms of powers of $\alpha$ lower than $n$.  Thus $R[\alpha]$ is finitely generated as a module by $\{1,\alpha,\ldots,\alpha^{n-1}\}$.
(Reverse). If $R[\alpha]$ is finitely generated as a module over $R$, then let $\{g_1,\ldots,g_m\}\subseteq R[\alpha]$ be a finite set of generators.  We can choose some $k\in\mathbb{N}$ such that it is higher than the degree of every $g_i$.  Now consider $\alpha^k\in R[\alpha]$ is generated by these $g_i$, thus $\alpha^n=b_1 g_1+\cdots+b_m g_m$ for some $b_i\in R$.  But notice that the righthand side is all of degree lower than $k$, thus $\alpha^k-(b_1 g_1+\cdots+b_m g_m)=0$ is a monic polynomial in $R$ which is satisfied by $\alpha$, i.e. $\alpha$ is algebraic over $R$.

Lemma 2: Let $S$ be a finitely generated algebra over $R$.  $S$ is integral over a ring $R$ iff $S$ is finitely generated as a module over $R$.
Proof:  Apply Lemma 1 to all the generators of $S$.

Theorem (Integral dependence over rings is transitive):  Let $S$ be a finitely generated algebra over $R$, $T$ a finitely generated algebra over $S$.  If $S$ is integral over $R$, and $T$ is integral over $S$, then $T$ is integral over $R$.
Proof:
Applying Lemma 2, this is equivalent to showing being finitely generated as a module is transitive.  If we take $\{f_i\}$ generators of $T$ over $S$ and $\{g_i\}$ generators of $S$ over $R$ then $\{g_i f_j\}$ will form a set of generators of $T$ over $R$. The proof is left as an excercise for the reader.  
(Hint: Write down an arbitrary element of $T$ in terms of the $f_i$ and elements of $S$, and then expand the elements of $S$ in terms of the $g_i$.)
