Finding maximal ideals and Krull dimension

596 Views Asked by At

I have difficulties in finding the maximal ideals and compute the dimension of a quotient ring $\mathbb{C}[x,y,z]/(x^2-y^2,z^2x-z^2y)$.

Here $(x^2-y^2,z^2x-z^2y)$ is a product of $(x+y,z^2)$ and $(x-y)$. But they are not coprime, so the product is not the intersection. And this ideal is not prime. Then I don't know how to deal with it. Can anyone help me with this? Thanks!

1

There are 1 best solutions below

3
On BEST ANSWER

$(x-y)$ is a prime of height $1$, which contains $(x^2-y^2,z^2x-z^2y)$, so your ring has dimension at least two, since we have a chain $(x-y) \subset (x-y,z) \subset (x,y,z)$.

On the other hand it has dimension at most two, since it is a non-trivial quotient of the $3$-dimensional polynomial ring.

For the maximal ideals you should use $V(IJ) = V(I) \cup V(J)$. You can also make use of the Nullstellensatz: The maximal ideals, that contain a given ideal, correspond to the solutions of the system of equations given by the generators of the ideal.