Let $k$ be any field. I want to determine the (Krull) dimension of the following rings:
(i) $k[x,y,z]/(x^3-1)$,
(ii) $k[x,y,z]/(x^2-y^4,x-y^3)$ and
(iii) $k[x,y,z]/(x^3-1,x^2y^4-1,x^2y^3-z^5)$.
I realise in the all the cases $(x-1,y-1,z-1)$ is a maximal ideal containing the quotiented ideal. But how do I determine the dimension from this.
I am actually trying to determine the coheight ($\dim A/I$) of binomial ideals in $k[x_1,\ldots ,x_n]$. So is there a general method to determine the dimensions above so that I can use it in the general case.
Thank you in advance.
Tool 1
A polynomial ring over a noetherian ring $A$ has dimension : $$\operatorname {dim}A[T_1,\cdots ,T_n]=\operatorname {dim}A+n$$ Tool 2
Let $k$ be a field with algebraically closure $\bar k$ and let $I$ be an ideal $I\subset k[T_1,\cdots,T_n]$. Then $$\operatorname {dim} \frac {k[T_1,\cdots,T_n]}{I}=0 \iff V(I)\subset \mathbb A^n(\bar k)=\bar k^n \quad \text {is finite}$$ Answer to (i) $$ \operatorname {dim}k[x,y,z]/(x^3-1)=\operatorname {dim}_k( \frac {k[x]}{(x^3-1)})[y,z]=0+2=2$$ where the second equality follows from Tools 1 and 2.
Answer to (ii)
We have $k[x,y,z]/(x^2-y^4,x-y^3)=k[y,z]/(y^6-y^4)=\frac {k[y]}{(y^6-y^4)}[z]$ which, by the same reasoning as in (i), has dimension $0+1=1$
Answer to (iii)
Tool 2 immediately implies that the ring $k[x,y,z]/(x^3-1,x^2y^4-1,x^2y^3-z^5)$ has dimension $0$, since $V(x^3-1,x^2y^4-1,x^2y^3-z^5)\subset \bar k^3$ contains $3\cdot 4\cdot 5=60$ points.