Trouble with operator-sum representation of a quantum operation I am reading about the operator-sum representation of quantum operations in Nielsen's and Chuang's 10th Anniversary ed of Quantum Computation and Quantum Information (N&C). I have become quite confused by some of the basic formalism presented therein. I'll get right to it.
N&C gives the definition of a quantum operation $\mathcal{E}$, for an input system $A$ in an initial state $\rho_{A}$ and environment $B$ in an initial pure state $|0_{B}\rangle\langle 0_{B}|$, as
\begin{equation}
\mathcal{E}(\rho_{A})=\text{Tr}_{B}\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)
\end{equation}
where $U$ is some transformation on the whole system. N&C then proceeds toward the operator-sum representation by expressing the partial trace $\text{Tr}_{B}$ using a finite orthonormal basis $\{|i_{B}\rangle\}$ for $B$ as
\begin{equation}
\begin{split}
\mathcal{E}(\rho_{A})&=\text{Tr}_{B}\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)
&=\sum_{i}{\langle i_{B}|\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)|i_{B}\rangle}\overset{*}{=}\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}}
\end{split}
\end{equation}
where the last equality (*) is nothing but the operation-sum representation of the operation $\mathcal{E}$. The elements $\{E_{i}\}$  of this representation are written $E_{k}\equiv \langle k_{B}|U|0_{B}\rangle$ and these are operators.
What I am unable to see is the rationale/reasoning for the equality (*) above. This is a problem to me and is what I would like to solve.
Why I am having trouble is perhaps because of the partial trace. I can't see how it is legitimate to write out the partial trace by using only the basis $\{|i_{B}\rangle\}$ as above, the dimensions do not seem to be in order as to make a meaningful expression. I would rather change to
\begin{equation}
|i_{B}\rangle\to \mathbb{I}^{A}\otimes |i_{B}\rangle
\end{equation}
whereby
\begin{equation}
\mathcal{E}(\rho^{A})=\sum_{i}{\mathbb{I}^{A}\otimes \langle i_{B}|\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)\mathbb{I}^{A}\otimes |i_{B}\rangle}
\end{equation}
as this is how I understand the partial trace. I would believe that, say, $U^{\dagger}(\mathbb{I}^{A}\otimes |k_{B}\rangle)$ is well-defined as I see dimensions of both operators (matrices) being just right. Then, if my understanding of the partial trace is correct, I would have
\begin{equation}
E_{k}=\langle k_{B}|U|0_{B}\rangle\to (\mathbb{I}^{A}\otimes \langle k_{B}|)U(\mathbb{I}^{A}\otimes |0_{B}\rangle)
\end{equation}
which is an expression where one can at least see that an $E_{k}$ is not a scalar, as opposed to how it is written by N&C.
Now, if the transformation $U$ was not present I would have no problem taking the partial trace this way. As this is not the case, I'm stuck. I have no idea as how to rearrange within the terms in the sum so that I get the elements $\{E_{i}\}$.
Any effort to help with this matter is greatly appreciated.
Edit: With the input from Norbert Schuch I have come to understand the following. The trick allowing for suitably rearranging within each term of the partial trace lies in expanding $\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|$ into products as
\begin{equation}
\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}| = \underbrace{(\rho_{A}\otimes \mathbb{I}_{B})(\mathbb{I}_{A}\otimes |0_{B}\rangle)}_{=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}}(\mathbb{I}_{A}\otimes \langle 0_{B}|)=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}(\mathbb{I}_{A}\otimes \langle 0_{B}|)\hspace{1mm},
\end{equation}
where the last equality can be seen through
\begin{equation}
\begin{split}
(\rho_{A}\otimes \mathbb{I}_{B})(\mathbb{I}_{A}\otimes |0_{B}\rangle)&=(\underbrace{\rho_{A}\mathbb{I}_{A}}_{\text{commutes}})\otimes (\underbrace{\mathbb{I}_{B}|0_{B}\rangle}_{=|0_{B}\rangle=|0_{B}\rangle\cdot 1})\\[2mm]
&=(\mathbb{I}_{A}\rho_{A})\otimes (|0_{B}\rangle\cdot 1)\\[2mm]
&=(\mathbb{I}_{A}\otimes |0_{B}\rangle)(\underbrace{\rho_{A}\otimes 1}_{=\rho_{A}})\\[2mm]
&=(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}\hspace{1mm}.
\end{split}
\end{equation}
Note the usage of the so-called mixed product property, i.e., $(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$.
Then, considering the $k$:th term in the partial trace sum, we have that
\begin{equation}
\begin{split}
& \mathbb{I}_{A}\otimes \langle k_{B}|\big(U(\rho_{A}\otimes |0_{B}\rangle\langle 0_{B}|)U^{\dagger}\big)\mathbb{I}_{A}\otimes |k_{B}\rangle\\[2mm]
=& \underbrace{(\mathbb{I}_{A}\otimes \langle k_{B}|)U(\mathbb{I}_{A}\otimes |0_{B}\rangle)}_{=E_{k}}\rho_{A}\underbrace{(\mathbb{I}_{A}\otimes \langle 0_{B}|)U^{\dagger}(\mathbb{I}_{A}\otimes |k_{B}\rangle)}_{=E^{\dagger}_{k}} \\[2mm]
=& E_{k}\rho_{A} E^{\dagger}_{k} \hspace{1mm}.
\end{split}
\end{equation}
I am pleased by Norbert's input on this matter and as far as my perspective goes I consider my question to be solved. Thank you very much.
 A: You are completely right with the first part of your question: What is meant by $|b_i\rangle_B$ is indeed $\mathbb I_A\otimes |b_i\rangle_B$. (Note that omitting identities is quite common, e.g., when writing Hamiltonians!)
Now to your question how to show
\begin{equation}
\begin{split}
\sum_{i}{\langle b_{i}|\big(U(\rho_{A}\otimes |0\rangle\langle 0|)U^{\dagger}\big)|b_{i}\rangle}\overset{*}{=}\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}}\ .
\end{split}
\end{equation}
To this end, note that 
\begin{align}
\rho_{A}\otimes |0\rangle\langle 0|
&= (\rho_A\otimes \mathbb I)(\mathbb I\otimes|0\rangle)(\mathbb I\otimes \langle 0|)
\\
&= 
(\mathbb I\otimes|0\rangle)
\rho_A
(\mathbb I\otimes \langle 0|)\ .
\end{align}
(I elaborate below why this equality holds.)
Inserting this on the LHS, we obtain
\begin{equation}
\begin{split}
\mbox{LHS}=\sum_{i}{\langle b_{i}|U
(\mathbb I\otimes|0\rangle)
\rho_A
(\mathbb I\otimes \langle 0|)
U^{\dagger}|b_{i}\rangle}
=\sum_{i}{E_{i}\rho_{A}E^{\dagger}_{i}}\ ,
\end{split}
\end{equation}
as desired.

Appendix: Why is $(\rho_A\otimes \mathbb I)(\mathbb I\otimes|0\rangle)= 
(\mathbb I\otimes|0\rangle)
\rho_A$?
First, note that $(A\otimes B)(C\otimes D)=(AC)\otimes(BD)$. Also note that we can regard $|0\rangle$ as a ($d\times 1$) matrix and $1$ as a ($1 \times 1$) matrix. We thus have
\begin{align}
(\rho_A\otimes \mathbb I)(\mathbb I\otimes|0\rangle)
&= 
(\rho_A\mathbb I)\otimes(\mathbb I|0\rangle)
\\
&=
(\mathbb I\rho_A)\otimes(|0\rangle\cdot 1)
\\
&=
(\mathbb I\otimes |0\rangle)(\rho_A\otimes 1)
\\&=
(\mathbb I\otimes|0\rangle)
\rho_A\ .
\end{align}
A: I am not convinced with the mathematics of the current answer of Norbert Schuch, that is why I wrote a more rigorous answer (to convince myself). First I want to explain my problem with the current answer. In the proof this step
\begin{equation}
\begin{split}
&=(\mathbb{I}_{A}\rho_{A})\otimes (|0_{B}\rangle\cdot 1)\\[2mm]
&=(\mathbb{I}_{A}\otimes |0_{B}\rangle)(\underbrace{\rho_{A}\otimes 1}_{=\rho_{A}})\\[2mm]
\end{split},
\end{equation}
is reasoned with the mixed product property. However when we look at this property http://www.math.uwaterloo.ca/~hwolkowi/henry/reports/kronthesisschaecke04.pdf (KRON 7). To apply it, the dimensions of the operator must fit like for the matrix multiplication. Also the last term seems strange 
\begin{equation}
\begin{split}
(\mathbb{I}_{A}\otimes |0_{B}\rangle)\rho_{A}\hspace{1mm}.
\end{split}
\end{equation}
What should this represent, when not a short form of writing the original term? Otherwise the dimensions do not fit. It seems like there is some black magic around, which I do not understand.
Therefore I propose the following reasoning using the notation in N&C:
First the operator must be in the tensor product of the principal system and environment $U\in\mathcal{H}_{\text{princ}}\otimes\mathcal{H}_{\text{env}}$. This means we decompose the operator for some bases of the corresponding Hilbert space U = $\sum_{ij} c_{ij} A_i\otimes B_j$, where $A_i$ are the bases spanning $\mathcal{H}_{\text{princ}}$ and $B_j$ are the bases spanning $\mathcal{H}_{\text{env}}$. Now we insert it 
\begin{equation}
\begin{split}
&\sum_{k}{\langle e_k |\big(U (\rho\otimes |e_0\rangle\langle e_0|)U^{\dagger}\big)|e_k\rangle}\\
&=\sum_{k}{\langle e_k |\big((\sum_{ij} c_{ij} A_i\otimes B_j) (\rho\otimes |e_0\rangle\langle e_0|)(\sum_{ij} c_{ij} A_i\otimes B_j)^{\dagger})\big)|e_k\rangle}\\
&=\sum_{k}{(I_{\mathcal{H}_{\text{princ}}} \otimes \langle e_k |)\big((\sum_{ij} c_{ij} A_i\otimes B_j) (\rho\otimes |e_0\rangle\langle e_0|)(\sum_{ij} c_{ij} A_i^{\dagger}\otimes B_j^{\dagger})\big)(I_{\mathcal{H}_{\text{princ}}} \otimes |e_k\rangle}\\
&=\sum_{k}{(\sum_{ij} c_{ij} A_i\otimes \langle e_k |B_j) (\rho\otimes |e_0\rangle\langle e_0|)(\sum_{ij} c_{ij} A_i^{\dagger}\otimes B_j^{\dagger}|e_k\rangle)}\\
&=\sum_{k}{\sum_{ijkl} c_{ij}c_{kl} A_i \rho A_k^{\dagger} \otimes \langle e_k |B_j|e_0\rangle \langle e_0|B_j^{\dagger}|e_k\rangle}\\
&=\sum_{k}{\sum_{ijkl} c_{ij}c_{kl} A_i \rho A_k^{\dagger} \langle e_k |B_j|e_0\rangle \langle e_0|B_j^{\dagger}|e_k\rangle}\\
&=\sum_{k}{(\sum_{ij}{c_{ij}A_i \langle e_k|B_j|e_0\rangle})\rho(\sum_{ij}{c_{ij}A_i^{\dagger} \langle e_0|B_j^{\dagger}|e_k\rangle})}
\end{split}
\end{equation}
The forelast step comes, because the tensor product with a scalar is the same as multiplication, which we now also apply on the term $E_k$
\begin{equation}
E_k = \langle e_k|U|e_0\rangle  = (I_{\mathcal{H}_{\text{princ}}} \otimes \langle e_k|)(\sum_{ij}c_{ij}A_i\otimes B_j) (I_{\mathcal{H}_{\text{princ}}} \otimes |e_0\rangle) \\
= \sum_{ij}c_{ij}A_i\otimes \langle e_k|B_j|e_0\rangle = \sum_{ij}c_{ij}A_i \langle e_k|B_j|e_0\rangle
\end{equation}
We can see that the above term must be
\begin{equation}\sum_{k}{(\sum_{ij}{c_{ij}A_i \langle e_k|B_j|e_0\rangle})\rho(\sum_{ij}{c_{ij}A_i^{\dagger} \langle e_k|B_j^{\dagger}|e_0\rangle})}
= \sum_{k}E_k\rho E_k^{\dagger}\end{equation}
Now some comments, not important to the answer:
I feel like this equivalence and the new notation for a extended dot product should be more explained in the N&C. It took me quite a time to get this.
