Here is the question I am trying to understand the solution of the second part in it:
Let $R = \mathbb Q[x,y,z]$ and let bars denote passage to $ \mathbb Q[x,y,z] / (xy - z^2).$ Prove that $\overline{P} = (\overline{x}, \overline{z})$ is a prime ideal. Show that $\overline{xy} \in \overline{P}^2$ but that no power of $\overline{y}$ lies in $\overline{P}^2.$ (This shows $\overline{P}$ is a prime ideal whose square is $\textit{not}$ a primary ideal.)
The solution of the second part is written here.
I just do not understand why in the comments the coset representing $\bar{y}^n$ has $a_1$ in it? Why this $a_1$ is not equal to 1 just like in the coset of $\bar{y}$ or why it is not equal to $n$?
Could someone help me in answering this question please?
I don't understand what you mean by "just like in the coset of $\bar{y}$". Let me try to fill out the proof sketched in the comments of the linked post.
By definition, $\bar{y}^n\in (\bar{x},\bar{z})$ means that $$\bar{y^n}=\bar{y}^n = \bar{f} \cdot \bar{x}+\bar{g}\cdot \bar{z}, \quad \text{ for some }\quad \bar{f},\bar{g}\in \Bbb{Q}[x,y,z]/(xy-z^2).$$ Now, we want to lift everything to $\Bbb{Q}[x,y,z]$, and remembering the definition of the bar, we get $$y^n+h_1 = (f+h_2)(x+h_3)+(g+h_4)(z+h_5)$$ for $h_i\in (xy-z^2)$. Note that these $h_i$ could be written $a_i(xy-z^2)$ for some $a_i\in \Bbb{Q}[x,y,z]$ as well. In particular, we may expand and collect all the terms in this ideal into one, $H\in (xy-z^2)$, and get $$y^n = xf+zg+H.$$ Now you wonder, is such a decomposition possible? To which the answer is no: if this was true, it would be an identity of polynomials, in particular as functions on $\Bbb{Z}^3$, but the left hand side yields $1$, and the right $0$, when evaluated at $(0,1,0)$.