# Understanding lag operators in ARMA models

I'm trying to understand the root cancellation (or sometimes pole zero cancellation of an ARMA(1,1) model as an example). I know, root cancellation occurs if the AR parameter is equal to the MA parameter multiplied by $-1$. I also understand that this is explained with the help of the lag operator.

If $a$ denotes the AR parameter and $b$ denotes the MA parameter, we can write

$$y_t = ay_{t-1} + be_{t-1} + e_t$$

as

$$y_t = \frac{1-aL}{1+bL}e_t.$$

If $a = -b$, we get

$$y_t = e_t.$$

What I do not understand is why the lag operator L is the same for the AR part and the MA part. $L$ is defined as

$$Ly_t = y_{t-1}.$$

Does this imply

$$Le_t = e_{t-1}$$

automatically?

This is simply what the lag operator does - it operates on the time series by shifting the index one period back - it is defined as $Lx_t\equiv x_{t-1}$. The lag operator behaves very much like a regular operator. For instance, $$L^kx_t=L^{k-1}Lx_{t}=L^{k-1}x_{t-1}=x_{t-k}$$ More generally, $$(aL^k+bL^m)x_t=ax_{t-k}+bx_{t-m}$$ You will thus get different results if the lag operators are additionally multiplied by different coefficients.
Zero or negative powers are also legitimate, $L^0x_t=x_t$ and $L^{-j}x_t=x_{t+j}$. Some polynomials in $L$ carry a special name. The difference operator $\Delta\equiv1-L$ is a prominent example.
Powers of functions of $L$ work perfectly analogously. E.g., the double difference operator $\Delta^2=\Delta\Delta=(1-L)^2=1-2L+L^2$.