Ideals in $\mathbb{C}[x,y]/I$ where $V(I)$ is finite set. $R = \mathbb{C}[x,y]$, $I \subset R$ ideal and $V(I)$ - is finite set.
I want to prove that all prime ideals in $R/I$ are maximal.
I know that it's true for $R/radI$. Let $\pi$ be natural map $R/I \to R/radI$. And suppose $\mathfrak{p} \subset \mathfrak{m} \subset R/I$ - prime ideals and $\mathfrak{m}$ - maximal. $\pi(\mathfrak{p}) = \pi(\mathfrak{m})$ - maximal ideal in $R/radI$, but i have no ideas what to do next.
 A: Recall since $R=\mathbb{C}[x,y]$ is a finitely generated $\mathbb{C}$-algebra, every quotient of $\mathbb{C}[x,y]$ is a finitely generated $\mathbb{C}$-algebra. For any prime ideal $\mathfrak{p}\subset R/I$ we have $$\text{height } \mathfrak{p} + \dim (R/I)/\mathfrak{p} = \dim R/I$$
Since $\dim R=2$ we have $\dim R/I \in\{0,1,2\}$. However, it cannot be $2$ as $V(I)$ is assumed finite. If $\text{height }\mathfrak{p}=1$ then $\dim(R/I)/\mathfrak{p}=0$ and is a field and therefore $\mathfrak{p}$ is maximal. If $\text{height } \mathfrak{p}=0$ then $\mathfrak{p}$ is maximal.
A: $\sqrt{0}=\ker\pi$ is the intersection of all the prime ideals of $R/I$.
Therefore, let $\mathfrak{p}\in \text{Spec}(R/I)$.
Consider the projection $\pi:R/I\to R/\sqrt{I}$. You know that $\pi^{-1}(\pi(\mathfrak{p}))=\mathfrak{p}$, and therefore that $\pi(\mathfrak{p})\in\text{Spec}(R/\sqrt{I})$.
You said you already knew that this implies $\pi(\mathfrak{p})\in\text{SpecMax}(R/\sqrt{I})$.
This implies that $\text{SpecMax}(R/I)\ni\pi^{-1}(\pi(\mathfrak{p}))=\mathfrak{p}$. $\square$
