8
$\begingroup$

I'm new to the world of schemes, so I'm still trying to grasp some of the basics. I understand most of the simple topological properties of schemes, as well as some of the sheaf-theoretic properties like reducedness, integrality, normality, etc. But I'm having a hard time digesting what finte and finite type morphisms really entail concretely. Also, Hartshorne doesn't seem to include many nice examples from what I can tell.

Pretty much the only example I have down at this point is what it means for a scheme $X$ to be of finite type over a field $k$. In such a case, we'd have a morphism of schemes $f:X \to \rm{Spec}(k)$, and since the spectrum of a field is simply a point, the inverse image is all of $X$. Hence, just applying the definitions, we see that $X$ has a finite open cover by affine schemes $\rm{Spec}A_{i}$, where each $A_{i}$ is a finitely generated $k$-algebra. If we add integrality of $X$, we observe that a variety over $k$ is simply a integral scheme of finite type over a field $k$.

So this is nice, but beyond this I'm failing to grasp any intuition. Does anyone have any canonical examples that I should work through? Or maybe any pointers on how to think about finite and finite type. Eisenbud and Harris point out that pretty much all geometrically interesting morphisms are usually in these two regimes. For example, I guess spectra of local rings are "non-geometric" as they aren't of finite type over a field. This is a great example of something I don't understand well enough yet.

$\endgroup$
  • $\begingroup$ Finite type over a field really just means covered by some finite collection of open sets, each of which is a closed subscheme of some affine space over the field. $\endgroup$ – Nefertiti Jan 27 '17 at 16:14
  • $\begingroup$ Almost all reasonable maps of varieties are of finite type. On the other hand finite morphisms are very special. They correspond to finite covering spaces, such as any dominant map of a projective curve to another projective curve. $\endgroup$ – KeD Jan 30 '17 at 19:10
4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $X$.

  1. Describe all points of the closure of the set, consisting of one point.
  2. Show that this closure is an affine scheme itself. What is its function ring?
  3. 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.

  1. But you can also start, with say, one closed point, and then successively extend its "neighborhood", but keeping it infinitesimally small: consider spectra of $A/(\mathfrak{p}^n)$, $n = 1, 2, 3, \ldots$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.