I am working with a finite field $\mathbb{Z}_p$ where $p$ is a prime number and I am trying to understand how to extend this field to $\mathbb{Z}_{p^2}[X]/(X^2-\beta)$ and beyond.
So far, I understood that I can represent elements in $\mathbb{Z}_{p^2}[X]/(X^2-\beta)$ as polyonimal: $$ a = a_0+a_1X $$ where $a_0,a_1$ are elements in $\mathbb{Z}_p$. I can then compute addition and multiplication as usual polynomial add/mul, always considering that $X^2=\beta$.
If that's all correct, now I would like to build $\mathbb{Z}_{p^4}[X]/(X^4-\beta)$. Of course I could represent an element as $$ a = a_0+a_1X+a_2X^2+a_3X^3 $$ but I would like to understand how to construct this field as a quadratic over a quadratic.
A paper that I'm reading state that $\mathbb{Z}_{p^4}=\mathbb{Z}_{p^2}[Y]/(Y^2-\psi)$, where $\psi=\sqrt{\beta}$. However, I can not understand why this is correct. In fact, I can write:
$$ a = \alpha_0 + \alpha_1 Y $$ where $\alpha_0$ and $\alpha_1$ are in $\mathbb{Z}_{p^2}[X]/(X^2-\beta)$ but I can't understand why this is equal to the above expression (even though I guess that it is related with the fact that $\psi=\sqrt{\beta}$).
Any help is really appreciated!