Find the associated primes of $\dfrac{\mathbb{C}[x,y,z]}{\langle xy,yz\rangle}$.
I have already found that $<x,z> $ and $<y>$ are associated primes. But i am not being able to find the others.
Find the associated primes of $\dfrac{\mathbb{C}[x,y,z]}{\langle xy,yz\rangle}$.
I have already found that $<x,z> $ and $<y>$ are associated primes. But i am not being able to find the others.
If you cannot find any others, look for ways to show there are none. Here's a sketch that should help you accomplish that.
First, show that $\langle x,z\rangle$ and $\langle y\rangle$ are both minimal prime ideals of $\Bbb C[x,y,z]$ that contain $\langle xy,yz\rangle$. So, it suffices to show that any zero divisor $p(x,y,z)\in\Bbb C[x,y,z]$ of the module $M=\Bbb C[x,y,z]/\langle xy,yz\rangle$ satisfies $p\in\langle x,z\rangle$ or $p\in\langle y\rangle$. (Why?)
Given $p(x,y,z)\in \Bbb C[x,y,z]$, to be a zero divisor of $M$ means that $pq\in\langle xy,yz\rangle$ for some $q(x,y,z)\in \Bbb C[x,y,z]$. Now, write $q$ in the form $$q = a + xq_x+zq_z+yq_y$$ where $a$ is the degree $0$ term, $xq_x$ is the polynomial consisting of all terms divisible by $x$ in $q$, $zq_z$ is the polynomial consisting of all remaining terms divisible by $z$, and $yq_y$ consists of all remaining terms divisible by $y$. There are two cases to consider: $a=0$ or $a\ne0$.
If $a=0$ reduce mod. the ideal $\langle x,z\rangle$ to show that $p\in\langle x,z\rangle$ (recall $pq\in\langle xy,yz\rangle$).
If $a\ne 0$ reduce mod. the ideal $\langle y\rangle$. What can you show?
Therefore, $p$ belongs to one of the named associated primes of $M$. It follows that $$\mathrm{Ass}(M)=\{\langle x,z\rangle,\langle y\rangle\}$$