1
$\begingroup$

I try to make my question clear:

When reading a paper or listening a seminar talk, people showed me some set, and claim it to be a scheme; or some map, and claim it to be a morphism. I query why this is the case, and the answer sometimes goes as following : because every construction is algebraic, the result should be algebraic.

I don't know if this can be deemed as a proof or something easy to be checked (I am certainly a novice in AG) or a principle like Lefschetz principle. I do find a list of constructions called algebraic construction from Wikipedia. However I do not have any idea of how it is defined, and how to use them in the proof.

If you have heard similar explanation as I did, I hope you could understand my puzzle; and if you have use the similar explanation, I hope you could elaborate a little bit what does this exactly mean.

Improve this question
$\endgroup$
5
  • 5
    $\begingroup$ The people speaking to you are engaging in an abuse of notation/terminology that makes communication among experts efficient but can puzzle non-experts. One is using Yoneda's Lemma in certain technically powerful ways that experts express by describing a functor only on "complex points" rather than points valued in arbitrary (or finitely generated) $\mathbf{C}$-algebras -- the latter is what is actually needed in rigorous arguments but the notation can get clunky in that generality, so experts omit it; Mumford explains it extremely well in his book on abelian varieties. $\endgroup$
    – user36938
    Commented Jul 18, 2013 at 14:35
  • $\begingroup$ @user36938 Thank you very much! Though I do not quite follow what your wrote, I feel your functor perspective is closed to how it is used. Could you elaborate on it? In particular, I did not see how was Yoneda lemma involved. There is one example in my mind: suppose for any $\mathbb{C}$-point $x$ in a variety $X$, one can associate a variety $V_x$, then why the set $\{(x,V_x)|x \in X\}$ is a variety. I think this construction is usually used in the context of semicontinuity theorem. Besides, if you could provide some example to illustrate your functor perspective, it will be very useful! $\endgroup$
    – Li Yutong
    Commented Jul 19, 2013 at 6:18
  • $\begingroup$ Certainly without more information about the nature of $V_x$ it is impossible to make sense out of the assertion that the collection of pairs $(x,V_x)$ (whatever on earth it could mean) "is a variety". It would be better for you to give an example that was told you. (What you said about $(x,V_x)$'s is too vague to be an example; more specificity about the $V_x$'s is needed.) $\endgroup$
    – user36938
    Commented Jul 19, 2013 at 6:29
  • $\begingroup$ @user36938 It might be too complicate to explain the construction I mentioned. But general speaking, do you mean even if one views a construction is algebraic, to show it is really a variety or morphism, one still has to look into details and examine it. There is no fancy philosophy behind "algebraic construction". $\endgroup$
    – Li Yutong
    Commented Jul 19, 2013 at 11:35
  • $\begingroup$ Certainly there is some work to do, but it could be routine with experience. I don't know what you mean by "fancy philosophy", but there's nothing deep lurking in the shadows here, just a matter of familiarity or not with a certain technique. If you cannot provide any specific example then your request for clarification may be too vague to get a useful answer. $\endgroup$
    – user36938
    Commented Jul 19, 2013 at 12:21

4 Answers 4

Reset to default
1
$\begingroup$

My naive point of view is: One avoids the use of ODEs or integrals (e.g., disguised as using the exponential mapping for Lie groups, ahem, algebraic groups) which quickly leads outside of the realm of polynomials and rational functions.

Improve this answer
$\endgroup$
1
$\begingroup$

I think that there is no real perfect answer to your question.

Here is a simple example, that I often say in talks, I hope it can give an idea of what people mean when saying that a construction is algebraic.

Given any smooth cubic $C$ in the projective plane, and a point $p\in C$, I can define a birational map of the projective plane. A general line passing through $p$ is invariant and the restriction of the map to the line is the unique involution which fixes the two points of $C$ lying on the line which are distinct from the point $p$.

For the beginner, it is not clear that the map is really birational, but as soon as you deal with such constructions, you understand that if you give explicitly the point and the curve, then you can write the map with polynomials.

To see this, you can say that given a point in the plane, the equation of the line, of the points of intersections, and then of the image depend "algebraically" from the point. Hence the map is rational and does not involve for instance square roots or anything else.

Improve this answer
$\endgroup$
3
  • $\begingroup$ As you know, square roots would be OK - we would just call it ``$y$'' and write $y^2 = f(x)$, etc. Exp and log would be another matter. :) $\endgroup$
    – user36938
    Commented Jul 19, 2013 at 3:40
  • $\begingroup$ No, square roots are not allowed for a rational map. If you want a rational map from a variety $X$ to a variety $Y$, the image of a point $x\in X$ should depend algebraically from the coordinates in $X$ (i.e. be quotient of polynomials). $\endgroup$
    – Jérémy Blanc
    Commented Jul 19, 2013 at 19:25
  • $\begingroup$ @Jeremey: Sure, all I meant is that what looks like a "square root" from one point of view may be a "coordinate" from another; for the plane curve $C$ defined by $y^2 = f(x)$ one could say that the map $y:C \rightarrow \mathbf{A}^1$ is $P \mapsto \sqrt{f(x(P))}$. $\endgroup$
    – user36938
    Commented Jul 20, 2013 at 0:05
0
$\begingroup$

I think I understand what you are asking. When one first learns about schemes, it is often very hard to link the language of schemes and the geometry. Suppose you are sitting in a seminar, and the speaker comments:

We are going to consider the Segre embedding $ \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3 $ which is given by the formula $ ([x_0,x_1],[y_0,y_1]) \mapsto [x_0y_0,x_0y_1,x_1y_0,x_1y_1] $

this paints a very vivid picture. Indeed, if you are thinking about just sets and functions, everything is golden and this makes perfect sense. Even if you are thinking about manifolds, everything is fine. However, it is not immediately obvious how to translate such a statement into the language of schemes and that is one of the biggest hurdles when learning algebraic geometry, so lets walk through how the translation works in this case.

The formula above already gives a hint as to what is going on in the scheme world. Write $\mathbb{P}^3 = {\rm Proj} \mathbb{C}[z_{00},z_{01},z_{10},z_{11}]$. Being slightly imprecise (you can fix this by talking about line bundles) the homogeneous polynomial $z_{ij}$ pulls back to $x_i y_j$. This means that regular function $ z_{pq} / z_{ij}$ pulls back to $ x_p / x_i \cdot y_q / y_j $ which is a regular function on $ D_+(x_i) \times D_+(y_j)$. We have described a morphism $D_+(x_i) \times D_+(y_j) \to D_+(z_{ij})$. Both of these schemes are affine so we only need to explain how regular functions pull back to describe the morphism. Now we want to glue all of these morphisms together. It is a little messy to write down but you can algebraically verify that both of the morphisms $D_+(x_i) \times D_+(y_j) \to D_+(z_{ij})$ and $D_+(x_i') \times D_+(y_j') \to D_+(z_{i'j'})$ restrict to the same morphism $$ D(x_i' / x_i) \times D(y_j' / y_j) \to D(z_{i'j'} / z_{ij}) $$ modulo changing coordinates (when you change coordinates on a product, you change coordinates on each component independently). This means they glue to give us an affine morphism $ \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3 $ under which $z_{ij}$ pulls back to $ x_i y_j$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ i just looked at your previously asked questions. Now I don't think this is what you were looking for... $\endgroup$
    – Daniel Barter
    Commented Jul 18, 2013 at 14:00
  • 3
    $\begingroup$ The "right" way is to use the functorial meaning of projective space as classifying line bundles equipped with an ordered tuple of generating global sections. Rather than busting out affine opens, make a functorial upgrading of the classical formula: for any scheme $S$ and line bundles $L,L'$ on $S$ equipped with generating global sections $(s_0,s_1)$ of $L$ and $(s'_0,s'_1)$ of $L'$, we get an ordered 4-tuple of global sections $(s_0\otimes s'_0,s_0\otimes s'_1,s_1\otimes s'_0, s_1\otimes s'_1)$ of $L\otimes L'$ that generate it. This works equally well as a definition for manifolds, etc! $\endgroup$
    – user36938
    Commented Jul 18, 2013 at 14:40
  • $\begingroup$ @user36938: Thanks for the comment! I am still trying to absorb the sheafy terminology $\endgroup$
    – Daniel Barter
    Commented Jul 18, 2013 at 15:26
  • $\begingroup$ Daniel, as an exercise, you should check that the $i$th "standard affine open" in $\mathbf{P}^n$ has the functorial characterization as classifying those pairs $(L, (s_0,\dots,s_n))$ such that $s_i$ generates $L$ globally (i.e., $s_i$ is "everywhere non-vanishing"). $\endgroup$
    – user36938
    Commented Jul 19, 2013 at 3:30
0
$\begingroup$

Sorry, if my answer is to ambiguous.

An object $X$ e.g a curve, a morphism, is a subset of a bigger set e.g $\mathbb{C}^n$ or the set of pairs $(x,y)$. To describe $X$ is to give some restrictions that defined it. Whenever someone says "the contruction is algebraic" I imagine those restrictions are "locally" given by algebraic relations e.g polynomials. Similarly, whenever someone says "an object or relation is algebraic". I imagine that it can be constructed by "a finite combination of steps or operations"

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .