n-factor martingale representation theorem Baxter & Rennie at pag. 162 state the following theorem. 
Let $W$ be an $n$-dimensional $\mathbb Q$-Brownian motion and let $M_t=(M_1(t),...,M_n(t))$ be an $n$-dimensional $\mathbb Q$-martingale process, which has volatility matrix $(\sigma_{ij}(t))$, in that $dM_j(t) = \Sigma_i \sigma_{ij}(t) dW_i(t)$ and the matrix is non singular (for all $t$ with probability 1). Let $N_t$ be any one-dimensional $\mathbb Q$-martingale. 
Then there exists an $n$-dimensional $F$-previsible process $\phi_t = (\phi_1(t), ... \phi_n(t))$ such that $\int^T_0 (\Sigma_j \sigma_{ij}(t) d\phi_j(t))^2 dt <\infty$, and the martingale $N$ can be written as $N_t = N_0 + \Sigma_j \int_0^T \phi_j(s) dM_j(s)$. Further $\phi$ is (essentially) unique.
My question is: because each of the $M_j(t)$ is a $\mathbb Q$-martingale, why can't I chose the $\phi_j$ to write $N_t=N_0 + \int_0^T \phi_j(s) dM_j(s)$ $\forall j$? In other words, what is the idea behind uniqueness in the $n$-dimensional case ($n>1$)?
 A: The statement and associated definitions in the book seem to gloss over an important assumption: the process $N_t$ needs to be a martingale with respect to the measure $\mathbb{Q}$ and the filtration $\{\mathcal{F}_t^W\}$ generated by the $n$-dimensional Brownian motion $W(t)$, i.e. $\mathcal{F}_t^W = \sigma(W(s) : s \le t)$.  That is, for each $t$, $N_t$ needs to be $\mathcal{F}_t^W$-measurable (i.e. the process is $\mathcal{F}_t^W$-adapted), and for each $s < t$ we have $E[N_t \mid \mathcal{F}_s^W] = N_s$.
This explains why your objection doesn't apply.  I think your thought was that if you have a representation of a process $N_t$ with respect to the Brownian motion $W$, you could apply the theorem again using the first coordinate $W_1$, which is indeed a one-dimensional Brownian motion, in place of $W$, and get a different representation.  But an $\mathcal{F}_t^W$-martingale will typically not be an $\mathcal{F}_t^{W_1}$-martingale.
As a simple example, consider $N(t) = W_2(t)$.  Then $W_2(t)$ is not measurable with respect to the $\sigma$-field $\mathcal{F}^{W_1}_t$: indeed, it is independent of this $\sigma$-field.  So although $W_2(t)$ is a martingale with respect to $\mathcal{F}_t^W$, it is not an $\mathcal{F}_t^{W_1}$-martingale.
