find all prime ideals of $\mathbb{C}[x,y]/(x^3,y^3)$ and show uniqueness of maximal ideal

Question

I'm working on the following problem:

Let $R=\mathbb{C}[x,y]/(x^3,y^3)$, where $(x^3,y^3)$ is the ideal generated by $x^3$ and $y^3.$ Find all prime ideals of $R$. And show that $R$ has a unique maximal ideal.

I only know that $(x^3,y^3)$ should contain all polynomials in the form of $fx^3+gy^3$, where $f,g \in \mathbb{C}[x,y]$, so the elements in $R$ should look like $\sum_{i,j=0}^2 a_{i,j}x^i y^j+(x^3,y^3)$, $a_{i,j} \in\mathbb{C}.$ It seems to me that $R$ has divisor of zero, for example, $(xy+(x^3,y^3))*(x^2y^2+(x^3,y^3))=(x^3,y^3)$. I'm not sure how to deal with it if it's not a PID. Thanks for any help.

Answer 1

The prime ideals of $\mathbb C[x,y]/(x^3,y^3)$ are of the form $P/(x^3,y^3)$ with $P$ a prime ideal of $\mathbb C[x,y]$ containing $(x^3,y^3)$. From $(x^3,y^3)\subseteq P$ we get $(x,y)\subseteq P$, so $(x,y)=P$ since $(x,y)$ is maximal.
Conclusion: $\mathbb C[x,y]/(x^3,y^3)$ has only one prime ideal, namely $(x,y)/(x^3,y^3)$.

Answer 2

Let's try to formilize the comment and wait for experts' review! If the new quotient is an integral domain, then if $x \not = 0$, since $x^3 = 0$ we get $x \cdot x^2 = 0$ and $x \cdot x = 0$ that is absurd, thus $x=0$. In the same way $y=0$. If you add a new element to $(x,y)$ you get the whole ring (by subtraction the right element). So in order to get a prime ideal you need $x=y=0$, and this ideal is already maximal, so it is the only prime and the only maximal.