Spec of $\mathbb{C}[x,y]/(xy)$ 
Find an open sets and points of $\mathrm{Spec}(\mathbb{C}[x,y]/(xy))$.

I know that prime ideals in $\mathbb{C}[x,y]$ are $(0),(x-a,y-b),(f)$ where $a,b \in \mathbb{C}$ and $f$ is irreducible. Prime ideals in $\mathbb{C}[x,y]/(xy)$  are of form $\mathfrak{p}+(xy)$, where $\mathfrak{p}$ is prime ideal in $\mathbb{C}[x,y]$ that $(xy) \subset \mathfrak{p}$.
Why $(xy)$ is not contained in $(x-a,y-b)$ for arbitrary complex $a,b$, but $a=0$ or $b=0$? I have a problems to see this.
How do open sets look? I know the definition of open sets in Zariski topology, but I can't see it. I would be very grateful for hints and help.
 A: One of your questions is:

Why $(xy)$ is not contained in $(x-a,y-b)$ for arbitrary complex $a$, $b$, but $a=0$ or $b=0$?

I would like to offer a very elementary answer to this. (Other comments and answers have talked about the open sets, I will not talk about it here.) Specifically I would like to answer the following question:

Why is it true that $(xy)$ is contained in $(x-a,y-b)$ if and only if $a=0$ or $b=0$ (or both)?

First of all, I will say something very basic, just in case that you or someone else reading this, is still learning about ideals. If $a=0$, then $(x-a,y-b) = (x,y-b)$. This contains $x$. So it contains the multiple, $xy$. Specifically, $xy = x(y) + (y-b)(0)$. Similarly if $b=0$ then $(x-a,y-b)=(x-a,y)$ contains $y$, so it contains the multiple $xy$.
But really, the point is to prove the converse: if $xy \in (x-a,y-b)$ then $a=0$ or $b=0$ (or both). So suppose that $xy \in (x-a,y-b)$. We can write $xy = (x-a)f(x,y) + (y-b)g(x,y)$ for some polynomials $f,g$. Let us substitute the values $x=a$ and $y=b$ in this equation. We obtain
$$
  ab = (a-a)f(a,b) + (b-b)g(a,b).
$$
On the right hand side we have $0+0$. Therefore $ab=0$. Therefore $a=0$ or $b=0$.
A: The prime ideals of $\mathbb C[x,y]/(xy)$ are in bijection with the prime ideals of $\mathbb C[x,y]$ containing $(xy)$. So if $xy\in \mathfrak p$ for some prime $\mathfrak p\subset \mathbb C$, then either $x\in \mathfrak p$ or $y\in \mathfrak p$. These are exactly the (non-maximal) prime ideals $(x),(y)$ and the maximal ideals $(x,y-a),(y,x-b)$ for all $a,b\in\mathbb C$. 
To write out all the details carefully, it's useful to re-examine the statement "$xy\in \mathfrak p$ for some prime $\mathfrak p\subset \mathbb C$, then either $x\in \mathfrak p$ or $y\in \mathfrak p$." Suppose the former holds, then quotient out by $x$ and now you are reduced to finding the prime ideals of $\mathbb C[y]$ which are just $(0)$ and $(y-b)$ which in the larger ring are the ideals $(x)$ and $(x,y-b)$.
The open sets of a scheme are just the complements of the closed sets which, in an affine variety $\mathrm{Spec}(R)$ are just sets of the form $$V(I)=\{\mathfrak p\in \mathrm{Spec}(R)\mid\mathfrak p\supseteq I  \}$$ for ideals $I$. Can you finish now?
