# Why are AR(p) processes always invertible?

My question is the following:

If we have an AR(p) process, then we have the following $$\Phi(B)X_{t}=Z_{t}$$. I understand that to check for causal/non-causal stationarity, we consider the roots of $$\Phi(B)$$ to see if their modulus/absolute value lie within, or outside the unit circle. However, to check for invertibility, we need to check the roots of the MA polynomial, $$\Theta(B)$$. However, in the case of an AR(p) process, we can see that $$\Theta(B)=1$$ So it has no roots to check. I have seen from various sources that this automatically means that the process is invertible (and any AR process in invertible). But why is that the case. I want to understand why is the process invertible when the MA polynomial has no roots to check. Similarly, if we are given an MA process, would this mean all MA processes are stationary since $$\Phi(B)=1$$? Once again, I want to understand the result, as opposed to just remembering that these statements are true. Thank you in advance.

Note: The $$B$$ refers to the backshift operator. That is given a process $$X_{t}$$, then $$BX_{t}=X_{t-1}$$

In invertibility, all we want is to be able to write $$Z_t$$ (the noise) as a causal function of $$X_t$$ (the output). In other words, we want $$\pi_j$$ such that $$Z_t=\sum_{j=0}^\infty \pi_j X_{t-j}\ \ \ \ \text{s.t.}\ \ \ \ \sum_{j=0}^\infty |\pi_j|<\infty$$ AR(p) is already in this form. So, it's invertible.
From the perspective of roots (consider polynomials with backshift operator $$B$$, not $$z$$ transform), it says all roots must be outside the unit circle. This statement doesn't explicitly say what to do in the case of no roots. However, a better interpretation is no roots should be inside the unit circle, because they're the ones making the sum $$\sum_{j=0}^\infty |\pi_j|$$ diverge. Since $$\Theta(B)=1=0$$ has no roots, it doesn't violate the second statement in italics.