A question about the geometric representation of Spec $\Bbb{C}[x,y]/(x-y)$

Question

The representation of Spec $\Bbb{C}[x,y]/(x-y)$ is given geometrically as the line $x-y=0$.

I don't understand how this is. Can every prime ideal be represented as a point on the line $x-y=0$?

Answer 1

Points correspond to maximal ideals and since $\Bbb C$ is algebraically closed the maximal ideals are exactly these of the form $(x-a,y-b)$.

It is now enough to check that $(x-a,y-b)\supset(x-y)$ if and only if $a=b$.

Answer 2

Pretty close, yeah, but $Spec( \mathbb{C}[x,y]/(x-y) )$ contains even more geometric information than that. Every point on the line corresponds to a maximal ideal in the ring ( not just prime ), and conversely every maximal ideal corresponds to a point on the line.

There's an additional prime ideal, though... the ideal $(0)$. This ideal corresponds to... well, kind of the whole line all at the same time, or a "generic" point on the line.

EDIT: Since you mentioned drawing/graphing the scheme, Mumford had this cool picture of $Spec(\mathbb{Z}[x])$ in the Red Book. I don't want to risk any copywrite infringement, but I'd wager you can find it by a quick google search of Mumford Red Book Scheme. Honestly, the picture doesn't convey a lot of content, but as I learned more about schemes, the more sense that picture made to me, and now it's what I think of when I think of prime ideals geometrically.