$A\subseteq B\subseteq C$ ring extensions, $A\subseteq C$ finite/finitely-generated $\Rightarrow$ $A\subseteq B$ finite/finitely-generated? Let $A\subseteq B\subseteq C$ be commutative unital rings. Recall that the extension $A \subseteq B$ is finite / of finite type / integral, when $B$ is a finitely generated $R$-module / when $B$ is a finitely generated $A$-algebra / when $\forall b \in B$ $\exists$ monic polynomial $f \in A[x]$ with $f(b)=0$. Notation $_AB$ means "$A$-module $B$".
We know that $A\subseteq C$ is integral iff $A\subseteq B$ and $B\subseteq C$ are integral (Grillet, Abstract Algebra, 7.3.3).
Do we also have the following:


*

*$A\subseteq C$ is finite $\Leftrightarrow$ $A\subseteq B$ and $B\subseteq C$ are finite.

*$A\subseteq C$ is of finite type $\Leftrightarrow$ $A\subseteq B$ and $B\subseteq C$ are of finite type.


I'm having problems with ($A\subseteq C$ finite $\Rightarrow$ $A\subseteq B$ finite) and with ($A\subseteq C$ of finite type $\Rightarrow$ $A\subseteq B$ of finite type). If $_AB$ is a direct summand of $_AC$ (for example when $A$ is a field), i.e. $_AC= _A B\oplus _A B'$ for some submodule $B'$ of $C$, then $C=Ac_1+\cdots+Ac_n$ implies $c_i=b_i+b'_i$ for some $b_i\in B$ and $b'_i \in B$, hence $B = Ab_1 + \cdots + Ab_n$. But what if $_AB$ is not a direct summand of $_AC$? And what about the 'finite type' case?
 A: You are right to have problems because the implications you cannot prove are false!  
1)  Let $A$ be a ring and $M$ be an $A$-module.
 Define an $A$-algebra structure $A*M$ on $A\oplus M$ by decreeing that that $m\cdot m'=0$ for all $m,m'\in M$.
Now if $A$ is non-noetherian and if $M$ is a finite $A$-module with a non finitely generated sub-module $N\subset M$, you can take $B=A*N\subset C=A*M$ as an example of a non-finite subalgebra of a finite algebra .
2)  Take $A = k$, a field,  $C=k[x,y]$, the polynomial ring .
Then  $B=k[x,xy, xy^2,...,xy^n,...] $ is a subalgebra of the finitely generated $k$-algebra $C$ which is not finitely generated over $k$.  
Noetherian alert
If $A$ is noetherian, then a subalgebra of a module-finite $A$-algebra is module-finite so that no counterexample like 1) is possible.
However, since a field $k$ is a highly noetherian ring,   counterexample 2) above proves that a subalgebra $B$ of a finitely generated  $A$-algebra  need not be finitely generated even if the base ring $A$ is noetherian.
If however you add to noetherianity of $A$ the hypothesis that $C$ is module-finite over $B$, then indeed $B$ will be finitely generated over $A$: this is the Artin-Tate theorem that BR links to in a comment to George (Lowther!) 's answer.
A: Neither of the implications that you are having trouble with hold in general. As an example, consider $A=k[X_1,X_2,\ldots]$, $C=A[Y]/(Y^2)$ and $B\subseteq C$ be the (unital) algebra over $A$ generated by terms of the form $X_iY$. You should be able to show that $C$ is finitely generated both as a module and as an algebra over $A$, but $B$ is not finitely generated over $A$ in either sense.
