Suppose $A$ is a ring, what is A-algebra? When I was reading Lang's Algebraic Number Theory, it wrote
If B is integral over A and finitely generated as an A-algebra, then B is a finitely generated A-module. However, is an A-algebra automatically an A-module with an additional bilinear operation  
I feel that it would be better to add the proof for this theorem given in Lang.  
Proof. We may prove this by induction on the number of ring generators, and thus we may assume that $B = A[x]$ for some element $x$ integral over A. But we have already seen that our assertion is true in that case.  
I am still confused with this theorem.

An A-algebra is (in this context) simply a ring B which contains A as a subring.
In that situation, B can be viewed as an A-module in a hopefully obvious way.
On the other hand, if B is an A-algebra and S is a subset of B, we say that S generates B as an A-algebra if the smallest subring of B which contains A (so that it is an A-algebra) and S is B itself. Then, that B is finitely generated as an A-algebra means that there is a finite subset S of B which generates B as an A-algebra.
For example, the polynomial ring $Q[x]$ is a $Q$-algebra which is not a finitely generated $Q$-module but which is finitely generated as a $Q$-algebra: the set $S=\{x\}$ generates $Q[x]$ as a $Q$-algebra.
