Prime spectrum of a ring, understanding geometry I am doing some exercises out of Atiyah & Macdonald these days. Doing the exercises isn't the problem, but I am having trouble understanding geometric about $\text{Spec}(A)$. 
Consider $A=\mathbb{C}[X,Y]/(Y^2 - X^3 + X + 1)$ or something. Are the elements of $\text{Spec}(A)$ (prime ideals of $\mathbb{C}[X,Y]$ containing $P(X,Y)=Y^2 - X^3 + X + 1$) supposed to be "points on the curve" $0 =  Y^2 - X^3 + X + 1$ or something? I can see why $\text{Spec}(\mathbb{C}[X])$ is an affine line over $\mathbb{C}$ (plus another "generic" point), but I haven't been able to conceptualise what's going on in general. Basically, the exercises keep coming back to commutative algebra, but I am not seeing the geometic ideas behind it. 
Here's my question: could someone point me to some elementary exercises that help interpret the geometry of $\text{Spec}(A)$ for the ring $A$ above?
 A: The basic idea might be described  as follows.
Consider, following Descartes,  the classical curve $C\subset \mathbb C^2$ consisting of pairs $(x,y)\in \mathbb C^2$ satisfying $y^2-x^3+x+1=0$.
Classically the regular functions functions $C\to \mathbb C$ are defined as the restrictions of polynomial functions $F\in \mathbb C[X,Y]$to $C$ and they form a ring $\mathcal O(C)$ isomorphic to $A$.
Now to each $P=(a,b)\in C$ you can associate the ideal $\mathfrak m_P\subset \mathcal O(C)$ of those functions $f=F\mid C\in \mathcal O(C)$ vanishing at $P : f(P)=F(P)=0$.
For good measure you can also consider the ideal of functions $f\in \mathcal O(C)$ vanishing on all of $C$ thus obtaining (rather tautologically)   the zero ideal $\eta=(0)$.
One of the profound idea of scheme theory is that the consideration of all those prime ideals, the $\mathfrak m_P$'s and $\eta$, constituting $Spec(A)$ is a superior version of the naïve curve $C$, once you endow $Spec(A)$ with its Zariski topology and its structure sheaf $\mathcal O_C$.
The older, classical, data are  easy to deduce from the scheme-theoretic ones: the classical points correspond to the closed points of $Spec(A)$ and the regular functions are given by $\mathcal O(C)=\Gamma(C, \mathcal O_C)$.
A: I urge you to look at Chapter 4 of Vakil's "Foundations of Algebraic Geometry" (http://math.stanford.edu/~vakil/216blog/). It has very well though out exercises that will help you develop an intuition for what $Spec(A)$ for a commutative ring $A$ is. One of the reasons for the introduction of $Spec(A)$ was to be able to deal with nilpotent elements of a ring (which give rise to functions that vanish on all of the space but are yet to zero). Ravi discusses this in chapter 4 as well as chapter 5 (visualizing nilpotents). You should read this topic if you get a chance.
But, to see how it relates to geometry in the sense of classical affine algebraic geometry (basically the closed points of the spectrum of finitely generated algebras over algebraically closed fields), it might be useful to read other sources too such as Fulton's "Algebraic Curves" or the first chapter of Hatshorne's "Algebraic Geometry".
But, try to work through chapter 4 of Vakil's FOA systematically. It is a gem. If I remember correctly, in particular it contains exercises for the specific case you mention.
A: Maximal ideals are always prime, so among your prime ideals will be the maximal ones.
What do maximal ideals in $A=\mathbb{C}[X,Y]/(Y^2-X^3+X+1)$ look like?  Much like the primes, maximal ideals in $A$ are maximal ideals in $\mathbb{C}[X,Y]$ which contain the ideal $P(X,Y)=Y^2-X^3+X+1$.  
The maximal ideals in $\mathbb{C}[X,Y]$ are all of the form $(X-a,Y-b)$ for some $a,b\in \mathbb{C}$. This is the kernel of the map $\mathbb{C}[X,Y]\rightarrow \mathbb{C}$ which sends $(X,Y)\mapsto (a,b)$. In this way, maximal ideals of $\mathbb{C}[X,Y]$ are identified with pairs $(a,b)\in \mathbb{C}^2$.
The maximal ideals which contain $P(X,Y)$ are the kernels of maps $\mathbb{C}[X,Y]\rightarrow \mathbb{C}$ which send $Y^2-X^3+X+1$ to $0$.  Equivalently, $b^2-a^3+a+1=0$.  In this way, maximal ideals of $A$ are identified with pairs $(a,b)$ which solve the equation $Y^2-X^3+X+1=0$; that is, with points on the solution curve.
What about non-maximal primes?  As it happens, $A$ has only one non-maximal prime ideal, the zero ideal.  Depending on how much commutative algebra you know, the easiest proof of this fact uses that the Krull dimension of $A$ is $1$. So in this case, the prime spectrum $Spec(A)$ consists of the complex solutions to the equation $Y^2-X^3+X+1=0$, together with a "generic point" (much like the affine line).
More generally, when a commutative ring $A$ is a finitely-generated $\mathbb{C}$-algebra, the set of maximal ideals $mSpec(A)$ will be in bijection with kernels of maps $A\rightarrow \mathbb{C}$, which in turn are in bijection with the solution set of some finite set of polynomial equations (this second bijection depends on a choice of finite presentation for $A$).  The non-maximal prime ideals will all behave like "generic points", in that they are smeared out over some irreducible subvariety of $mSpec(A)$.
