I am an independent reader of some Algebraic geometry and commutative algebra.
I want to use the procedure of in this example ($R:= \mathbb{C}[[x,y,z]]/(xz,zy)$) to help me compute for more dimensions of other rings.
So, given a local ring $$R:= \mathbb{C}[[x,y,z]]/(xz,zy),$$ how can I compute its dimension?
I want to use the result above and learn how to compute the dimensions of the following independently:
$R:= \mathbb{C}[[x,y,z]]/(x^2, xy)$
$R:= \mathbb{C}[[x,y]]/(x^2)$
$R:= \mathbb{C}[[x,y]]/(x^2,y^2)$
The prime ideals in $R=\mathbb{C}[\![ x,y,z]\!] /(xz,yz)$ may be identified with the prime ideals in $\mathbb{C}[\![ x,y,z]\!]$ containing the ideal $(xz,yz)$. These are precisely those primes which contain $z$ or $x$ and $y$. In particular, the minimal prime ideals in $R$ correspond to the ideals $(z)$ and $(x,y)$. I claim that a maximal chain of prime ideals in $R$ is provided by the chain $(z)\subsetneq (x,z)\subsetneq (x,y,z)$ (modulo $(xz,yz)$). In particular, the ring $R$ has dimension $2$.
We know that a maximal chain has to either start with $(z)$ or with $(x,y)$. If it starts with $(z)$, then we may mod out $z$ and so we look at a maximal chain of prime ideals in $\mathbb{C}[\![ x,y]\!]$ and it is well-known that such a chain is of the form $0\subsetneq (x) \subsetneq (x,y)$.
If the chain starts with $(x,y)$, then we may mod out $(x,y)$ and so the ideal $(x,y)$ corresponds to the zero ideal in $\mathbb{C}[\![z]\!]$. But the latter ring is one-dimensional and so the only prime ideal in $\mathbb{C}[\![ x,y,z]\!]$ properly containing $(x,y)$ is the ideal $(x,y,z)$ yielding a chain of length one in $R$, $(x,y)\subsetneq (x,y,z)$.