I kind like understand the case that in Spec $k[x]$, where we send every prime ideal $(x-a)$ to the point $a$ on the affine line, and the ideal $(0)$ goes to the generic point. If we have more than one parameter, say $k[x,y,z]$. In a 3-D space, we send the maximal ideal $(x-a,y-b,z-c)$ to the closed point $(a,b,c)$.
Now if I consider the picture of Spec $k[x,y,z]/f(x,y,z)$, why it is the picture of $f(x,y,z)=0$? Or equivalently, why those ideals $(x-a,y-b,z-c)$ such that $f(a,b,c) \neq 0$ can not be a prime ideal in $k[x,y,z]/f$?
Thanks a lot!
The reason is simply the correspondence theorem for quotient rings. That is, (prime) ideals $\bar{J}$ of $R/I$ correspond to (prime) ideals $J$ of $R$ such that $I\subset J \subset R$. So, if the maximal ideal $(x-a,y-b,z-c)$ does not contain $(f)$ (i.e. $f(a,b,c)\neq 0$), there is no corresponding maximal ideal of the quotient ring $k[x,y,z]/f$.