Q:

Poles and zeroes - $\mathcal Z$-transform

I just have a small question, something that I am unsure about.  I have a difference equation for a filter: 
$$y[n] = x[n] - x[n-1]$$
I have worked out the $\mathcal Z$-transform: 
$$\mathcal Z\left\{y[n]\right\}= Y(z) = X(z)\left(1 - z^{-1}\right)$$ 
\begin{align}
H(z) &= \frac{Y(z)}{X(z)}\\
&= 1 - z^{-1}\\
&= 1 - \frac 1z\\
&= z - \frac zz\\
&= z - 1
\end{align}
There is no denominator?  Technically, there is: $\displaystyle H(z) = \frac{z - 1}{1}$. Does this simply mean that there are no poles? 

A:

Your term manipulation is incorrect:
$$
\begin{align}
H(z)&=1-z^{-1}\\
&= \left(1-z^{-1}\right)\frac{z}{z}\\
&= \frac{z-1}{z}
\end{align}
$$
So, there is a pole at $z=0$, which you can also see from the initial equation $H(z)=1-z^{-1}$ if you insert $z=0$.

