If $C\supseteq A$ is finite over $A$, then any subring $A\subseteq B\subseteq C$ is finite over $A$ Let $K$ be a field and let $A, B,C$ be $K$-algebras such that:

*

*$A \subset B \subset C$

*$A\subset C$ is a finite extension.

*$B \subset C$ is a integral extension.

So $C$ is a finitely generated $A$-module (why?) and this implies (why?) $B$ is a finitely generated $A$-module.

I asked my teacher the second question and she replied that it is a result present in Atiyah-Macdonald.
However, I cannot find it in the Chapter 2 dedicated to modules (neither in paragraph 1 for modules in general, nor in paragraph 4 for finitely generated modules, nor in paragraph 9 for algebras). Can any of you help me understand this or can you tell me which result to refer to?
 A: This is not true in general. A remark first: as I noted in my comment, to say that $A\subseteq C$ is a finite extension means by definition that $C$ is a finitely-generated $A$-module, so this answers your first question. Also note that this makes condition $3$ in your question superfluous; if $C$ is a finitely generated $A$-module, then it is integral over $A$, and thus in particular is integral over $B\supseteq A$.
In any case, let $A$ be any non-Noetherian $K$-algebra, and let $I$ be any non-finitely generated ideal of $A$. (For instance, take $A=K[y_i]_{i\in\mathbb{N}}$ and $I=(y_i)_{i\in\mathbb{N}}$, although the specific choice of $A$ and $I$ is not important.) Then let $C={A[x]}\big/{(x^2)}$. Certainly $C$ is a finitely generated $A$-module, generated by the elements $\left\{\overline{1},\overline{x}\right\}$. Also, $I\overline{x}=\left\{\overline{ix}\right\}_{i\in I}$ is an ideal of $C$, and so, by the second isomorphism theorem, $$B:=A+I\overline{x}=\left\{\overline{a+ix}:a\in A,i\in I\right\}$$ is a subring of $C$. However, note that $A\cap I\overline{x}=0$, and so $B=A\oplus I\overline{x}$ as an $A$-module. Since $I\overline{x}\cong I$ is not a finitely-generated $A$-module, this means that $B$ is not a finitely generated $A$-module either, and thus we have a counterexample. (The projection of any generating set of a module onto a direct summand of the module is a generating set of the summand. In particular, if $B$ were finitely generated, then $I\overline{x}$ would be too, a contradiction.)

The desired result does hold if $A$ is Noetherian, and in that case is essentially an immediate corollary of the Artin-Tate lemma. In fact, I suspect your professor was probably thinking of this lemma in her comment to you; a proof of it may be found in proposition 7.8. of Atiyah-Macdonald, if you'd like a specific reference.
So, assume $A$ is Noetherian. Since $C$ is a finitely generated $A$-module, it is in particular a finitely generated $B$-module. Also, again since $C$ is a finitely generated $A$-module, it is in particular a finitely generated $A$-algebra. Thus, by Artin-Tate, $B$ is also a finitely generated $A$-algebra.
Now, as noted above, $C$ is integral over $A$, so likewise $B$ is integral over $A$. Then $B$ is an integral and finitely generated $A$-algebra, and thus is a finitely generated $A$-module, as desired. (Slogan: "integral and finitely generated as an algebra implies finitely generated as a module." This is a standard result and can be proved by induction, but if you need a proof of it let me know.)
