This question is from Introduction to Singularities and Deformations, Greuel et al.
Let $R=\mathbf{C}\langle x,y,z\rangle/\langle x^2-yz\rangle,$ where $\mathbf{C}\langle x,y,z\rangle$ is the ring of convergent power series over $\mathbf{C}$ with 3 variables $x,y,z$. Prove that the residue class $\bar{y}$ of $y$ is irreducible in $R$ but not prime.
My attempt: Suppose $\bar{y}=\bar{g}\bar{h}$ for some $g$ and $h$. Then $gh=(x^2-yz)f+y$ for some $f\in \mathbf{C}\langle x,y,z\rangle.$ Plugging in $x=z=0$ gives $g(0,y,0)h(0,y,0)=y,$ so $gh$ is $y$-general of order 1. Hence, by Weierstrass division theorem, there exists $q\in \mathbf{C}\langle x,y,z\rangle,r\in \mathbf{C}\langle x,z\rangle[y]$ such that $y=ghq+r$ with $r$ being constant with respect to $y.$ Observe that $r=y-ghq$ is divisible by $x^2-yz,$ so it must be the case that $r=0.$
$\therefore ghq=y,$ so exactly one of $g,h,q$ is $uy$, where $u$ is a unit, and the other two are also units. From the relation $\overline{ghq}=\overline{yq}=\overline{y},$ it follows that $\overline{q}=\overline{1}$ by cancellation law of an integral domain. Therefore, it must be the case that $q$ is a unit, so either $g$ or $h$ is a unit. $\therefore \bar{y}$ is irreducible in $R$.
Here I proved the irreducibility of $\bar{y}$, but how should I prove that it is not prime? Thanks in advance!
In your ring, $x^2 =yz$ so $x^2$ is in $\left<y\right>$. Now we need to check that $x$ is not in that ideal. This proves this ideal is not prime (and thus $y$ is not prime).
To prove that, we just need to look at the ideal generated by $y$ and $x^2-yz$ in $\mathbb C[x,y,z]$. This ideal is also generated by $y$ and $x^2$ and as a result $x$ is not in that ideal. This concludes the proof.