Intuition Behind, or Canonical Examples of Finite Type Morphisms
This is a great question I hope there are many answers to it. I give here my own uneducated comments. For an example of a finite map consider the map $$z\mapsto z^n$$ from $\mathbb{C}\to \mathbb{C}$.
This is the classical example of a finite covering with ramification. It corresponds to the map of rings
$$\mathbb{C}[w]\to\mathbb{C}[z]$$ defined by $$w\mapsto z^n$$ (I use different variables to avoid confusion). If we identify $\mathbb{C}[w]$ with its image we have the extension $$\mathbb{C}[z^n]\subseteq \mathbb{C}[z].$$ Now this is a finite map. That is $\mathbb{C}[z]$ is finitely generated as a $\mathbb{C}[z^n]$ module. Its generators are $$1,z,z^2,\cdots ,z^{n-1}.$$
So finite map of schemes is really like a finite covering map, the fibres are zero dimensional and finite.
To take a second example consider the projection of $2$-dimensional affine space onto one dimensional line. $$\mathbb{A}^2\to\mathbb{A}^1$$
$$(x,y)\mapsto x$$
This morphism corresponds to the map $$k[x]\to k[x,y]$$ $$x\mapsto x$$ This is not a finite map as $k[x,y]$ is not finitely generated as a $k[x]$-module. The fibres in this case are one dimensional. It is however a map of finite type, as $k[x.y]$ is generated by $$1,y$$ as a $k[x]$-algebra.
All classical varieties and maps are of finite type, so non finite type morphism are a scheme phenomena.
It occurred to me as I was thinking about this question that localization is aptly named. Take an ring $A$, with its associated affine scheme $\rm{Spec} A$ now localize to $A_f$ this is the same as localizing geometrically to the open set $D(f)$, so as we localize the ring we zoom in like a microscope to smaller and smaller open sets, its a finite process since we are removing subvarieties defined by equations. Imagine now if we turned the microscope up to full and localized to $A_{\mathfrak{p}}$, that is like zooming in on a single point. Now $\rm{Spec}A_{\mathfrak{p}}$ is a strange, but important variety, it has a generic point and one closed point, and this closed point has a infinite sequence of neighbourhoods. A kind of microcosm of a single point. Note that this is not the same as an isolated single point defined by equations say as $x=0,y=0$ defines the origin of the plane. Or in scheme language as $\rm{Spec} \ k$. This is a point that belongs to a variety, its a specialization of a generic point. Scheme theory has given us higher order resolving powers to look at this point. This scheme is not of finite type, as for example $$\{\frac{p(x)}{q(x)}|q(0)\neq 0\}$$ is not a finite $k[x]$-algebra.
Another non finite type example is $A=k[x_0,x_1\ldots]$ the polynomial algebra in an infinite number or variables, sort of an infinite dimensional affine space, I have never seen it talked about, so its probably not that important, but hey its a ring.
So finite type is that intermediate region where we normally live with well defined dimensions. The local ring of a point cannot be said to have dimension zero because as before that would be an isolated point.
Hope this helps.
I suggest reflecting upon the following themes.
Consider, say, an affine scheme of finite type $X = \text{Spec}\,A$ over an algebraically closed field $k$. In the sense of classical algebraic geometry, its points are in bijection with maximal prime ideals.
But in the schematic environment, its points are all prime ideals, for example $(0)$, if $A$ has no zero divisors. This $(0)$ is called the generic point of $X$.
- Describe all points of the closure of the set, consisting of one point.
- Show that this closure is an affine scheme itself. What is its function ring?
- In particular, localization $\text{Spec}\,A_\mathfrak{p}$ generally has more points than just one closed and one generic point! There are generic points of other localizations: which ones?
This was about zooming.
- But you can also start, with say, one closed point, and then successively extend its "neighborhood", but keeping it infinitesimally small: consider spectra of $A/(\mathfrak{p}^n)$, $n = 1, 2, 3, \ldots$.