Isomoprhism classes of indecomposable modules over a finite-dimensional $\mathbb{k}$-algebra Assume that $A$ is a finite-dimensional $\mathbb{k}$-algebra, where $\mathbb{k}$ is an arbitrary field. Recently I read in my notes for representation theory of associative algebras that for a given $A$-module, say $U$, we have a decomposition of $A$-modules of the form $U= \oplus_{b \in \mathcal{B}} \thinspace bU$, where $\mathcal{B}$ is the set of central primitive idempotent elements of $A$. 
So my question has to do with the number of isomorphism classes of indecomposable $A$-modules. Since the above decomposition is true, then an $A$-module is indecomposable if-f there is a unique element $b \in \mathcal{B}$, such that $bU=U$, and in the latter case we say that $U$ belongs to the block $b$. Obviously, isomorphic $A$-modules should be belonging into the same block, though I'm not sure about the opposite. 
If two $A$-modules belong to the same block, are they necessarily isomorphic?
Moreover is if the above is true then we know that the number of isomorphism classes is the same as the cardinality of $\mathcal{B}$, right? If I have made some mistake and the above fails, is there any other way to approximate somehow the number of iso.classes of indecomposables for a given $A$?
 A: There are some wrong claims in what you write:
1) It is true that $U=\bigoplus_{b\in \mathcal{B}} bU$. However you some over a set of orthogonal central primitive idempotents, not over all idempotents. For example, the unit of the algebra $1$ is always a central idempotent, but it only appears in such a decomposition if the algebra is local. 
2) If a module is indecomposable, there is a unique block it belongs to. This is correct, but the converse does not hold. For example if $M$ is indecomposable, then $M\oplus M$ also belongs to a unique block, but it is obviously not indecomposable.
3) If two $A$-modules belong to the same block, they are not necessarily isomorphic. As above consider $M$ and $M\oplus M$. If the modules are indecomposable, this is true if and only if the algebra is semisimple. 
4) The number of isomorphism classes of indecomposables for a given algebra is in general quite difficult to compute if the algebra is not semisimple. "Most of the time" it will be infinite. If it is finite, it really depends on the type of algebra what it is.
A: Update Oop, I initially was thinking of irreducible modules rather than indecomposable modules. Julian's answer is comprehensive for the question actually asked. Someone has mentioned in the comments that it is nevertheless useful in its own way, so with this disclaimer above I'll leave it up.

I remember it this way:
The simple right modules of $A$ correspond exactly with those of $A/J(A)$, so we can just consider the semisimple Artinian ring $A/J(A)$.
$A/J(A)$ decomposes into a finite direct product of simple Artinian rings, each of which have exactly one type of simple right module. With not much work, you will be able to show that these are exactly the simple $A/J(A)$ modules when you lift the module action up from the simple Artinian ring to the product ring.
