I hope I'm not overbearing in this site. Yes, I'm still struggling.
If you can, I have a question about primary decomposition that still needs help, you can find it in my page.
Now I wanted to find the height of a certain ideal. In the ring $ A = K[x,y,z]/(xy,xz,z-y^2)$ I need to find the height of $I = (x,y,z)$. Here is my reasoning up to now:
the Krull theorem states that the height of a prime ideal, such as $I$, is not greater than the number of generators of $I$, so it's not greater than the least amount of polynomials that generate $I$. Since in $A$, $z=y^2$, we can consider $I$ having 2 generators, so the height of $I$ is at best two. So I'd need to find two proper prime ideals that are contained in $I = (x,y)$. I really don't know how to use the conditions $xy=0$ and $xz=0$, so I'm having second thoughts about my solution. I initially said that the height is indeed two because $(0) \subseteq (x) \subseteq I$ and I don't need to find if there are any more ideals there, but I just realized $(0)$ is not prime because the ring is not a domain since $x*y=0$, but $x,y \neq 0$. Can somebody help me? Thank you!
Let $P$ be a minimal prime over $J=(xy,xz,z-y^2)$. Clearly $z-y^2\in P$. Since $xy\in P$, either $x\in P$ or $y\in P$. Suppose $x\in P$. Then we see that $J\subseteq (x,z-y^2)\subseteq P$, and since $(x,z-y^2)$ is a prime ideal, we conclude $P=(x,z-y^2)$.
Next, suppose $x\notin P$. Then $y,z\in P$ and a similar argument as above yields $P=(y,z)$. So the minimal primes of $J$ are $(x,z-y^2)$ and $(y,z)$.
Now a maximal chain of prime ideals in $A$ descending from $(x,y,z)$ must end at one of the minimal prime ideals, but both minimal primes have height two, and $(x,y,z)$ has height three in $k[x,y,z]$, so the height of $(x,y,z)$ in $A$ must be one.