Intuition for fibers over a point in $\operatorname{Spec}(A)$ This is a rather soft question, but I would like to see what I get. I'm currently reading/working through Atiyah Macdonald, and I just did exercise 21 of Chapter 3. I won't repeat the entire exercise here as it is rather long, but the point is to get to the identification of $(f^*)^{-1}(\mathfrak{p})$ with $\operatorname{Spec}(k(\mathfrak{p}) \otimes_A B)$, where $f : A \to B$ is a homomorphism of commutative rings, $\mathfrak{p}$ is a prime ideal of $A$, and $k(\mathfrak{p})$ is the residue field of $A_\mathfrak{p}$.
My question is: Can someone give geometric intuition about what the fiber over $\mathfrak{p}$ "is", or why we would expect it to be homeomorphic to $\operatorname{Spec}(k(\mathfrak{p}) \otimes_A B)$?
I don't know much about schemes, but I have a solid understanding of the basics in classical algebraic geometry (projective varieties, nullstellensatz, etc.), so I would love to get some intuition that makes sense from that perspective.
Thanks!
 A: An alternative approach for why this is the "right object" goes through the formalism of category theory.
If you recall, the category of Affine Schemes is exactly opposite the category of commutative rings. Depending on your book this is either a definition or a theorem. Next, following our geometric nose, let's see if we can figure out what the fibre should be in this category.
Well, in the category of topological spaces, we can compute fibres "by pullback".

Here $x$ is the map sending the single point $\star$ to $x \in X$, and the pullback $f^{-1}(x)$ admits the explicit description $\{y \in Y \mid f(y) = x \}$. So we see that this pullback really is the fibre of $f$ at $x$, and experience shows that this pullback gives the fibre in all sensible geometric categories.
So then, to build the fibre in the category of affine schemes, we want to imitate this picture. That is, we want to build the inclusion $\{ \mathfrak{p} \} \to \text{Spec}(A)$, and look at the pullback of $f^* : \text{Spec}(B) \to \text{Spec}(A)$ along it.
Of course, we know how to find an affine scheme whose only point is $\mathfrak{p}$! We take $A$, and we localize at $\mathfrak{p}$. This gives us a ring $A_\mathfrak{p}$ whose only maximal ideal is $\mathfrak{p}$. Then, we quotient in order to make this maximal ideal the only ideal, leaving us with $k(\mathfrak{p})$, the residue field of $A_\mathfrak{p}$, whose only ideal is $\mathfrak{p}$ (though this ideal now happens to be equal to $(0)$).
Since we have a map $A \to A_\mathfrak{p} \to k(\mathfrak{p})$, we can apply $\text{Spec}(-)$ to get a map from
$\text{Spec}(k(\mathfrak{p})) \to \text{Spec}(A)$, where $\text{Spec}(k(\mathfrak{p})) = \{ \mathfrak{p} \}$.
Then, by analogy with the topological case, we see the fibre of $f^* : \text{Spec}(B) \to \text{Spec}(A)$ at $\mathfrak{p}$ should be a pullback

Now since the category of affine schemes is opposite the category of commutative rings, we know a pullback in $\mathsf{Aff}$ is really the same thing as a pushout in $\mathsf{CRing}$! But a pushout in $\mathsf{CRing}$ is nothing more than a tensor product!

So if we turn the arrows back around (by applying $\text{Spec}$ to everything in sight) we see that
$$(f^*)^{-1}(\mathfrak{p}) \cong \text{Spec}(B \otimes_A k(\mathfrak{p}))$$
as desired.

I hope this helps ^_^
A: A prime ideal is the "generic point" of an irreducible closed subset $V \subset \operatorname{Spec}A$. Don't worry too much about what this means; the important thing is that the closure of this generic point is the whole set $V$. The fiber over $\mathfrak p$ is close to being the generic point of the inverse image of $V$ under the induced map on spectra (it is called the generic fiber; in the example below it is "uncertain" in the $x$ and $y$ directions but not the $z$ direction), and one might say that the classical intuition is just that you are considering the equations that generate $\mathfrak p$ (an ideal of $A$) as equations in the ring $B$, and then modding out by those equations in $B$.
For instance, the inverse image of the parabola $\operatorname{Spec}k[x,y]/(y-x^2) = V(y-x^2) \subset \mathbb A^2 = \operatorname{Spec}k[x,y]$ under the projection $\mathbb A^3 \to \mathbb A^2$ corresponding to $k[x,y] \hookrightarrow k[x,y,z]$ is just $\operatorname{Spec}k[x,y,z]/(y-x^2) = V(y-x^2) \subset \mathbb A^3$, the "cylinder over the parabola" in $\mathbb A^3$.
