Show that the closed points of $\operatorname{Spec} \mathbb{C}[x,y]/(f)$ are in bijection with the complex solutions of the equation $f(x,y)=0$

Question

I know that for an algebraically closed field $k$, the closed points of $\operatorname{Spec} k[x,y]$ are in bijection with the topological space $\mathbb{A}^2$.

I am trying to show that the closed points of $\operatorname{Spec} \mathbb{C}[x,y]/(f)$ are in bijection to the complex solutions of the equation $f(x,y)=0$.

Does, the above fact somehow helps to show this? If yes, then how do I proceed? If not, then can you suggest a different way.

Answer 1

The closed points of $\operatorname{Spec} k[x,y]$ is of the form $(x-a, y-b)$ with $(a,b) \in k^2$, and is thus bijective with $\mathbb{A}^2$.

Now look at closed points for $\operatorname{Spec} k[x,y]/(f)$.

• They correspond to maximal ideals $(x-a, y-b) \in \operatorname{Spec} k[x,y]$ that contains $f$.
• Write down Taylor expansion of $f$ at $(a,b)$: $$f(x,y) = f(a,b) + c_1 (x-a) + c_2 (y-b) + c_3(x-a)^2 + \cdots$$ It should be clear that $f(x,y) \in f(a,b) + (x-a, y-b)$, and in particular $f(a,b) = 0$ iff $f \in (x-a, y-b)$.

Therefore maximal ideals $(x-a, y-b)$ that contains $f$ corresponds bijectively to $(a,b) \in \mathbb{A}^2$ where $f(a,b) = 0$.