How to compute $\dim_{\mathbb C}\mathbb{C}[x,y,z]/(z^4,x^2+y^2+z^2-1,xy)$?
I tried to decompose $$(z^4,x^2+y^2+z^2-1,xy)=(z^4,x^2+y^2+z^2-1,x)\cap(z^4,x^2+y^2+z^2-1,y)=(z^4,x^2+z^2-1,y)\cap(z^4,y^2+z^2-1,x)$$ so I obtained a direct sum of two vector spaces. But I do not know how to continue to understand better the structure of that quotient. Thank you!
By the CRT, the quotient is isomorphic to
$$\mathbb C[x,y,z]/(z^4,x^2+z^2-1,y) \times \mathbb C[x,y,z]/(z^4,y^2+z^2-1,x).$$
The two factors are isomorphic, hence it suffices to compute the dimension of one factor, i.e. compute the dimension of $\mathbb C[x,z]/(z^4,x^2+z^2-1).$
It is easy to choose some monomial order (i.e. such that $Z^4 > X^2 > Z^2$ holds), such that those two generators form a Groebner basis, hence the initial ideal is $(z^4,x^2)$ and $\dim \mathbb C[x,z]/(z^4,x^2)=8.$
Hence the dimension of your algebra is $16$.