## What elementary problems can you solve with schemes?

189

117

I'm a graduate student who's been learning about schemes this year from the usual sources (e.g. Hartshorne, Eisenbud-Harris, Ravi Vakil's notes). I'm looking for some examples of elementary self-contained problems that scheme theory answers - ideally something that I could explain to a fellow grad student in another field when they ask "What can you do with schemes?"

Let me give an example of what I'm looking for: In finite group theory, a well known theorem of Burnside's is that a group of order $p^a q^b$ is solvable. It turns out an easy way to prove this theorem is by using fairly basic character theory (a later proof using only 'elementary' group theory is now known, but is much more intricate). Then, if another graduate student asks me "What can you do with character theory?", I can give them this example, even if they don't know what a character is.

Moreover, the statement of Burnside's theorem doesn't depend on character theory, and so this is also an example of character theory proving something external (e.g. character theory isn't just proving theorems about character theory).

I'm very interested in learning about similar examples from scheme theory.

What are some elementary problems (ideally not depending on schemes) that have nice proofs using schemes?

Please note that I'm not asking for large-scale justification of scheme theoretic algebraic geometry (e.g. studying the Weil conjectures, etc). The goal is to be able to give some concrete notion of what you can do with schemes to, say, a beginning graduate student or someone not studying algebraic geometry.

My gut feeling is that this question has appeared on MO at least two times. But anyway, 1+, since I also wonder if there are elementary problems. – Martin Brandenburg – 2011-03-21T16:32:23.877

7Once you have enough scheme-theoretic machinery, there's an almost trivial proof that the nonsingular points of a variety form a dense open subset; see III.4 Prop. 3 of Mumford's Red Book, or Hartshorne Corollary II.8.16. The basic idea is to show that, in an appropriate sense, the variety is nonsingular at its generic point. – Charles Staats – 2011-03-21T18:18:59.680

41Fermat's Last Theorem is a completely elementary problem, whose very nice proof uses schemes in an essential way... – Kevin Buzzard – 2011-03-21T19:09:39.533

http://math.columbia.edu/~dejong/wordpress/?s=challenge

– Niels – 2011-03-21T20:35:41.493

4Community wiki? – Yemon Choi – 2011-03-22T05:25:41.460

9

I'm not sure how many of these actually need schemes, but you might like this:

http://math.stanford.edu/~vakil/725/funprobs.pdf

– David Zureick-Brown – 2011-03-23T16:39:52.220

– darij grinberg – 2013-04-04T00:20:16.610

93

A smooth projective variety over $\mathbb{Q}$ has only finitely many places of bad reduction. Shimura had a horribly complicated proof of this in the language of Weil's foundations in a paper from the 50's. With schemes, it's completely obvious, as smoothness is an open condition. Even stating this without schemes is painful. The whole field of arithmetic geometry is an example of what you want. Actually, this is my only serious complaint about Hartshorne. He doesn't do any Number Theory and $\operatorname{Spec}\mathbb{Z}$ is where schemes really shine.

Knowing as little about algebraic geometry as I do, I like the sound of this answer since it seems close in spirit to the example mentioned by the original poster (Burnside's proof of $p^aq^b$ via character theory). – Yemon Choi – 2011-03-22T03:22:51.663

9For us non-arithmetic geometers, can you explain what a "place of bad reduction" is? – Kevin H. Lin – 2011-03-22T04:31:17.743

33Kevin, it's a prime p where the reduced curve mod p is not smooth. So the reduction is "bad" in the sense that it is geometrically worse than what you started with. For example, consider the cubic curve y^2 = x^3 - 2, which is smooth in characteristic 0. When you reduce it mod p it is still smooth except when p = 2, in which case it turns into the singular curve y^2 = x^3 (cusp at origin). So we'd say 2 is a prime (place) of bad reduction. Actually, reducing curves mod p is a setting where schemes are a useful language, because it lets you discuss reduction without the crutch (continued...) – KConrad – 2011-03-22T04:51:12.550

24of depending on explicit equations. After all, maybe one equation reduces nicely mod p and another doesn't, so what should you do? With schemes you can make sense of what reduction mod p means in an equation-free way. – KConrad – 2011-03-22T04:52:35.900

24The word "place" is sort of a funny word for "prime". The point is that a prime for a number field can correspond to an embedding into a p-adic completion, but there are also embeddings into R or C which need to be tracked in the same way, so instead of speaking about "prime or real/complex embedding" they are all called by a neutral new word "place". – KConrad – 2011-03-22T04:55:49.060

1@KConrad, thank you for the nice explanation. – Kevin H. Lin – 2011-03-22T08:31:47.400

1Felipe Voloch's class of examples works equally well in algebraic geometry: replace $\mathrm{Spec}\mathbb{Z}$ by an (affine) algebraic curve. – Hagen – 2011-03-22T12:09:05.010

7+1 for the comment about Hartshorne – David Corwin – 2013-04-04T01:05:25.613

It is especially noteworthy that "Even stating this without schemes is painful". Indeed, without schemes it is not clear what a place of bad reduction is, so that IMO this example is not quite in the family "ideally not depending on schemes" asked by the OP. – Matthieu Romagny – 2013-04-04T07:23:52.070

7@KConrad nice example, but that curve has another "bad" prime p=3, because x^3-2 is (x-2)^3 mod 3 so there's a cusp at (x,y)=(2,0). – Noam D. Elkies – 2017-07-29T23:21:41.107

67

It seems to me that there are a lot of great answers here but I am not sure if they live up to the challenge of showing the usefulness of schemes through an elementary example.

Let me try my luck and risk the wrath of the MO crusaders.

Probably anyone having any kind of mathematical background have seen the classification of conics (a.k.a. plane quadrics) over $\mathbb R$. Perhaps still a large percentage of those who saw this early in their mathematical career wondered about the asymmetry involved in that you have the usual nice ones (ellipse, parabola, hyperbola), the reasonable degenerate ones (two different lines, either intersecting or parallel), and then there are the weird ones (double line, point, empty set).

This last bunch, while clear from the proof seems odd and one might feel that they should not be considered. Now if one looks at them scheme theoretically, then it becomes completely clear how these fit into the same mold and how those "weird" ones are really the same as the nice ones (some points, some empty sets) or the "reasonable" degenerate ones (double lines, other points, other empty sets). The double line especially is hard to explain without schemes, while the other two "weird ones" are really a consequence of $\mathbb R$ not being algebraically closed, although interestingly, the associated schemes actually kind of "see" the non-real points as well.

A more high brow version of the same idea is the fact that (say over $\mathbb C$) a family of elliptic curves (topologically tori) can degenerate to a nodal cubic curve (topologically a sphere with two points glued together) and a family of rational curves (topologically spheres) can also degenerate to the same object. This seems to lead to at least confusion if not to contradiction as it seems to indicate that a sphere can be continuously transformed into a torus. Looking at these families the proper way, that is, using schemes we see that the two degenerations are not producing the same scheme. The one coming from the rational curves will add an embedded point at the node while that embedded point does not appear in the family of elliptic curves.

60

This was my own motivation for learning schemes:

Theorem (Mazur): If $E$ is an elliptic curve over $\mathbb Q$, then the torsion subgroup of $E(\mathbb Q)$ (the set of rational points of $E$) is isomorphic to $\mathbb Z/n\mathbb Z$ for $n = 1, \dots, 10,$ or $12$, or $\mathbb Z/2n\mathbb Z \times \mathbb Z/2\mathbb Z$ for $n = 1,\ldots,4$.

A special case (due to Mazur and Tate) is

Theorem: If $E$ is an elliptic curve over $\mathbb Q$, then $E$ does not contain a rational point of order $13$.

This is certainly a simple statement, but their proof uses the theory of schemes in a crucial way.

1Do you have a reference? I would like to look at this paper... – Steven Gubkin – 2012-06-07T02:09:36.597

5Dear Steven, Mazur's paper is in Publications of the IHES 47. The Mazur--Tate paper is in Inventiones in the early 70s, and so will be easily found on MathSciNet. (And Mazur probably has a list of publications on his Harvard web-page, which would also be easy to find.) Regards, Matthew – Emerton – 2012-06-07T05:59:49.860

54

If $I,J \subset A$ are comaximal ideals in a commutative ring $A$, i.e. $I+J=A$, then for all $n,m \in \mathbb N$ the ideals $I^n$ and $J^m$ are also comaximal.
Proof: $\emptyset= V(A)=V(I+J)=V(I)\cap V(J)=V(I^n)\cap V(J^m)=V(I^n+J^m)$ hence $I^n+J^m=A$.

Warning: I wouldn't like to be drawn into a discussion on whether this is just terminology or trivial algebraically or a big cheat or what not. All I know is that when I had to prove this result a long time ago, I came up with this proof a few months after I had started learning affine schemes and I was exhilarated at the thought that I could literally see why the result held by drawing two disjoint little doodles representing $V(I)$ and $V(J)$ inside a potato representing $Spec(A)$.

Edit (April 8th, 2016)
Here is an example of how thinking scheme-theoretically led to a proof of a purely algebraic problem.

78Never underestimate the power of drawing potatoes... – Mariano Suárez-Álvarez – 2011-03-21T18:58:32.687

4For those of us who are not really into schemes, could you add something to explain what makes this an especially scheme-y idea? – Thierry Zell – 2011-03-21T19:43:29.243

13Related to this is the statement that every artinian ring A is a product of local artinian rings -- clear from the fact that Spec(A) is just a discrete finite set. – Dustin Clausen – 2011-03-21T19:52:09.967

8Dear Thierry, what I find schemey is:

a) That you can interpret that two ideals $I,J$ are comaximal as being exactly equivalent to their associated subschemes $V(I), V(J)$ being disjoint in $Spec (A)$.
b) That the underlying sets $|V(I)|$ and $|V(I^n)|$ of the different subschemes $V(I)$ and $V(I^n)$ are equal. c) That there are pleasant formulae like $V(I+J)=V(I) \cap V(J)$ relating the algebra of the ring $A$ to the geometry of the scheme $Spec(A)$. – Georges Elencwajg – 2011-03-21T20:16:49.077

11Dear Georges, while I am a big fun of doodles and a firm believer in the usefulness drawing potatoes representing all kinds of mathematical objects, and I generally enjoy reading your answers and agree with them, I feel that this is not really a scheme-y example. The same proof can be carried out by only knowing varieties. The formulae does not have anything to do with the scheme structure. In fact, the point of the proof is that the varieties (a.k.a. reduced schemes) defined by the ideals $I$ and $I^n$ are the same. – Sándor Kovács – 2011-03-22T04:37:50.370

16Dear Sándor, the idea of associating a geometric space to an arbitrary ring is due to Grothendieck and was developed in EGA I. Nobody in the world had considered this association in such generality before: I have heard this said explicitly by Serre, Deligne and Cartier. So I don't agree with you: Spec(A) is definitely not a variety for $A=\mathbb Z$ and my proof couldn't have been written the day before EGA was published. Amusingly you implicitly acknowledge the role of schemes yourself when you define a variety as a reduced SCHEME (which is not at all the standard definition of a variety). – Georges Elencwajg – 2011-03-22T07:15:38.943

6Georges, you are right, I buy that. I guess I was thinking of Zariski topology, but I can accept that Zariski did not consider arbitrary rings. Perhaps you should emphasize that more in your answer that this works for arbitrary rings. And of course, you are also right in your amusement: I should change my comment (if it were possible) to say "reduced scheme" instead of "variety". Finally, I would not dare challenge the fundamental importance of Grothendieck's inventions nor do I want to lessen the importance of schemes. The point I tried to make here is that this proof is using the topology – Sándor Kovács – 2011-03-22T07:55:18.093

5...of $\mathrm{Spec}(A)$ not its scheme structure. – Sándor Kovács – 2011-03-22T07:55:43.800

Dear Sándor, of course I understand your point and, given your remarkable command of Algebraic Geometry, I gave it the highest attention and thank you for it. Maybe you will agree with the following: it is improbable that a bright young undergraduate who had to solve this exercise on comaximal ideals would come up with such a geometric proof if he had never opened a book on elementary scheme theory. Even though you are quite right that logically the full force of nilpotents is definitely not used, it seems that you can only learn the geometric language from such books. – Georges Elencwajg – 2011-03-22T11:11:10.307

5Hi Georges, as much as I hate arguing with you, I cannot not answer this. :). Zariski topology for arbitrary rings is introduced on page 12 of Atiyah-MacDonald. I for one read that book as a 3rd year undergrad, but did not know what schemes were until grad school. – Sándor Kovács – 2011-03-22T15:59:41.337

5Dear Sándor, you are right. Indeed a bright young undergraduate could learn this geometric language from Atiyah-MacDonald. And apparently, one of them did... – Georges Elencwajg – 2011-03-22T16:35:31.633

9While I enjoyed an argument, note that it's actually much similar to Furstenberg "topological" proof of infinitude of primes, namely, if you unravel it you'll get this(schemeless proof): if $I^n$ and $J^m$ are not comaximal, we hav $I^n+J^m\not{=} A$ hence there exist a maximal $\mathbf{m}$ such that $I^n+J^m\subset\mathbf{m}$ hence $I^m \subset \mathbf{m}$, by primeness of $\mathbf{m}$ we get $I\subset \mathbf{m}$, and $J^m\subset \mathbf{m}$ so again $J\subset \mathbf{m}$ so $I+J\subset \mathbf{m}$. Contradiction.

Here we see that under all algebraic manipulation there is hidden geometry. – Ostap Chervak – 2012-04-12T19:13:31.320

4This is a nice example, but it doesn't really answer the question, because an easy and entirely elementary proof is available: $I+J=A$ means $a+b=1$ for some $a \in I$ and $b \in J$, and then $1 = (a+b)^{m+n-1}$ is a sum of $m$ terms from $I^n$ (monomial multiples of $a^n$) and $n$ terms from $J^m$ (monomial multiples of $b^m$), so it's in $I^n + J^m$, QED. – Noam D. Elkies – 2017-07-29T23:29:34.840

@Noam: I knew your elementary proof, but the point is that once you know the language of elementary scheme theory the result doesn't need any proof: you just see it. It is then easy, if challenged, to translate that vision into a formal proof. (To be continued) – Georges Elencwajg – 2017-07-30T09:35:52.917

(Continuation) With a pinch of salt one could say that Jean-Jacques Rousseau wonderfully represented that geometric point of view in a paragraph reproduced at the beginning of Fulton's Algebraic Curves. And I stand in awe at Fulton's erudition: to my knowledge no francophone algebraic geometer had ever noticed the relevance of that extract from Rousseau's Les Confessions

– Georges Elencwajg – 2017-07-30T09:43:11.093

38

The fact that a first-order statement about algebraically closed fields holds in characteristic zero if and only if it holds in all large enough finite characteristic can be proved using Chevalley's theorem on images of constructible sets, together with a routine cataloguing of the constructible subsets of $\mathrm{Spec}(\mathbb{Z})$. This gives a bunch of examples, including the Ax-Grothendieck theorem on injective polynomial mappings.

(The idea of the proof is you assign to a free formula in n variables $x_1,\ldots,x_n$ the subset of $\mathrm{Spec}(\mathbb{Z}[x_1,\ldots,x_n])$ consisting of points which evaluate to "true" in an algebraically closed field over that point, and show by induction on the length of the formula that this subset is constructible. The only tricky point is quantifiers, and these are handled by Chevalley's theorem.)

I find this application especially nice. – Thierry Zell – 2011-03-21T23:12:52.303

20But this is just an application of the compactness theorem in mathematical logic! – Martin Brandenburg – 2011-03-22T09:31:09.440

4While I also agree that the logic proofs are simpler in this case (I'd probably use ultraproducts myself), the technique of specializing into characteristic $p$ can pushed much further, and scheme theory provides a very powerful language for this. – Donu Arapura – 2011-03-22T17:11:50.000

@MartinBrandenburg: logic proves the if part, but does it also prove the only if part? – Zsbán Ambrus – 2013-07-16T08:01:51.920

3@Zsban: Yes. Suppose $\phi$ is a first-order sentence true in all algebraically closed fields of characteristic zero. The set of algebraically closed fields of characteristic zero is the set of models of $T=\lbrace \psi\rbrace\cup\lbrace \theta_i: i\in\omega\rbrace\cup\lbrace \chi_j: j\in\omega\rbrace$, where $\psi$ is the field axiom(s), $\theta_i$ asserts that all degree-$i$ polynomials have roots, and $\chi_j$ asserts that the characteristic is at least $j$. (cont'd) – Noah Schweber – 2013-07-16T11:51:04.603

3By the compactness theorem, since $\phi$ is a consequence of $T$ we know that $\phi$ is a consequence of some finite $T'\subset T$; this $T'$ can only contain finitely many $\chi_j$, so in fact $\phi$ is a consequence of the characteristic of the algebraically closed field being at least $n$ for $n=\max\lbrace j: \chi_j\in T'\rbrace$. (Incidentally, this also proves that we only need roots of polynomials of sufficiently large degree!) – Noah Schweber – 2013-07-16T11:52:30.580

@NoahS: True, thanks for the explanation. – Zsbán Ambrus – 2013-07-16T12:49:14.347

27

Another answer to your question is provided by Mumford's book "Lectures on curves on an algebraic surface". This book explains Grothendieck's proof (via schemes) of the fact that for a sufficiently positive curve $C$ on a smooth projective surface $S$, the algebraic equivalence class of $C$ modulo the linear equivalence class of $C$ has maximal possible dimension, namely the dimension of the Picard variety of $S$.

According to Mumford's introduction, the only known earlier proofs were analytic in nature.

Mumford's book is really excellent by the way; it is not a substitute for Hartshorne or any other introductory textbook, but is a wonderful "second course", which introduces ideas such as the Hilbert and Picard schemes, some deformation theoretic ideas, and related techniques, and (perhaps more than Hartshorne) really shows how you can use all of Grothendieck's new methods to actually do something!

Of course, this is not an elementary problem in the sense of your question, but in thinking about algebraic geometry and its foundations, its worth remembering that algebraic geometry already had an extremely rich history by the time Grothendieck introduced schemes, and so the big outstanding problems (of which there were definitely many!) that demanded the introduction of this new technology were not simple ones (hence it's not so easy to justify the introduction of schemes with one or two very simple examples). (If one wants a truly simple example, one can just discuss how the size of a fibre under a map of projective curves is constant, provided that one counts the size of the fibre using its scheme structure, i.e. taking into account the nilpotents that appear at ramified points. But I don't know how compelling this example would be to non-algebraic geometers; it still may seem more like convenient book-keeping than a genuinely important new technique if you can't demonstrate it's utility through some specific application.)

24

Let me give two examples, however not so elementary,

1) The proof of the degeneration of the Hodge-De Rham spectral sequence by Deligne and Illusie in positive characteristic.

The basic assumption in Deligne-Illusie's theorem is the existence of a lift over the Witt vectors $W_2$. Hard to state within varieties. Moreover, Raynaud gives applications to instances of the Kodaira vanishing in positive characteristic.

2) Grothendieck's study of the fundamental group.

The notion of an étale morphism can probably be discussed within varieties, but one needs formal schemes, and the invariance of the fundamental group by a nilpotent immersion. But at the end, the result can be stated within varieties!

21

The "classical" example is surely duality of abelian varieties. If you want this duality to work over finite fields (or in characteristic p generally), it becomes apparent that you can't work with varieties alone. Technically taking the quotient by a group scheme that is not reduced is just too hard to express in the older geometric language.

16

I am far from an expert in algebraic geometry, but I think that there is something that really should be said in answer to this question.

Namely, there are at least two big reasons one might introduce a new theory--one is to answer old questions, and the other is to ask new questions (often, of course, motivated by the old classics). Of course, scheme theory was introduced to a large extent for the first reason. But I, at least, think the greatest virtue of the theory is in the second area.

Almost every question we could have asked about complex varieties, or at best about varieties over perfect fields (or maybe dvrs, in the lead-up to the development of schemes), can now be asked about schemes over arbitrary rings (or non-affine bases). We can see the geometry in diophantine equations; we can use cohomological tools to answer arithmetic questions. We can analyze deep algebraic structures of commutative rings in a geometric way. See e.g. Minhyong Kim's answer here.

I think this is what Kevin Buzzard's comment is getting at, in an extremely pithy way, in his comment about Fermat's last theorem; one might make a similar comment about, say, Falting's theorem. In order to even ask the right questions to approach these classics, we must have access to geometry over all sorts of supposedly pathological bases: imperfect fields, or even non-Noetherian rings. 100 years ago, who could have even imagined the statements of theorems about modular curves, for example, which seem totally natural today?

So back to your question: what elementary questions can be addressed using scheme theory? I guess I would say: any question about families, all of arithmetic geometry, any question about varieties over $\mathbb{C}$ you might be interested in over another base, any application of cohomological methods from the analytic theory (e.g. Riemann-Roch) you want to generalize, almost any problem where moduli spaces come up, etc.

15

If $C$ is a sufficiently general rational curve in $\mathbb P^3$ of degree $d$, then the vector space of degree $k$ homogeneous polynomials vanishing on $C$ has dimension precisely equal to $$\max\left\{0,\binom{k+3}{3} - dk - 1\right\}\;.$$ More geometrically, there exists a degree $k$ surface containing $C$ if and only if $\binom{k+3}{3} - dk - 1 > 0$.

This is a statement that could have been understood a hundred years ago. However, the proof involves degenerating rational curves to non-reduced schemes whose Hilbert functions can be computed more easily; see Hirschowitz's 1980 paper "Sur la postulation generique des courbes rationales."

12

This is not a great answer, but it was getting a bit long to be a comment, so I just made it community wiki instead. If anyone (with rep >= 100) wants to elaborate, feel free to do so in the answer itself rather than in the comments.

There is a technique for showing a closed subset $Z$ of an (irreducible) variety $X$ is all of $X$ that does not seem to have any analogue without using schemes. Let's suppose that $Z \subset X$ has a natural structure as a closed subscheme that comes from its definition (the induced reduced structure won't work here, as will become obvious). To show $Z = X$, it suffices to show that $Z$ contains a nonempty open set. If $z \in Z$, then to show $Z$ contains a neighborhood of $z$, it suffices to show that $Z$ contains every subscheme of $X$ supported on $z$--i.e., in a sense, $Z$ contains every infinitesimal neighborhood of $z$. A version of this is used in Mumford's book Abelian varieties to prove the Theorem of the Cube (II) in chapter III.

7

Ellenberg-Venkatesh-Westerland have a very nice result about realization of abelian $p$-groups as the $p$-part of class groups of certain function fields. All this fits in the general program of studying Cohen-Lenstra heuristics over function fields. One key scheme theoretic player in E-V-W result is a certain type of scheme called Hurwitz scheme.

To read great stuff related to all this check:

Ellenberg-Venkatesh, Statistics of Number Fields and Function Fields (Proceedings of ICM 2010)

Ellenberg-Venkatesh-Westerland http://arxiv.org/pdf/0912.0325

I'm almost sure you've already read this, but just in case https://quomodocumque.wordpress.com/2009/12/12/homological-stability-for-hurwitz-spaces-and-the-cohen-lenstra-conjecture-over-function-fields/

1@David: I think I should add to my answer: go to the third floor -just by the computer lab- and ask the question :) – Guillermo Mantilla – 2011-03-22T08:45:03.370

4

This is merely too long to be a comment.

I'm not sure that I completely agree with or understand the basis of the question. Is this one of those questions where the OP is taking for granted that varieties are interesting and then wondering what schemes are useful for in this context? If so, there have certainly been other questions of this sort on mathoverflow. However, I get the feeling this is not what's being asked for, in which case I'm not sure that the example of character theory being used to solve a problem of group theory is completely analogous. In that example, groups are already inherently part of character theory, so it seems reasonable that problems of group theory might be attacked using character theory. When asking a similar question about schemes, I'm not sure what the starting point is. To me, asking what you can do with schemes is more like asking 'what can you do with groups?' than 'what can you do with characters?'. At this point, I would say that both groups and schemes are naturally occurring basic mathematical objects that are of great interest to many mathematicians.

13That groups are part of character theory does not make it reasonable that problems about groups might be attacked using character theory. Right at the beginning of the subject (ca. 1900) Burnside didn't include characters in his group theory book precisely because, as he put it in the preface, he knew no examples of any problems involving groups that could be solved more efficiently with characters rather than without them. (See quotations from Burnside on pp. 216--217 of "Finite Group Theory" by Isaacs.) Later Burnside proved the $p^aq^b$-theorem and included characters in the 2nd edition. – KConrad – 2011-03-22T02:56:09.973

4I think the OP's analogy with character theory is this: here is a new technical concept (characters, schemes), and what does it buy me in terms of progress or understanding on problems which I could already have appreciated without having these concepts available? – KConrad – 2011-03-22T02:59:10.930

Of course depending on a student's background, even the question "why groups" fits into this line of inquiry. – KConrad – 2011-03-22T02:59:58.073

8Personally I never cared much about the $p^aq^b$ theorem on its own (I hate questions asking for a proof that all groups of such-and-such a size are solvable) so I never found that to be an inspiring reason to learn representation theory. But when I learned about Artin $L$-functions, to me that was a great reason to care about representations as well as to care about various induction theorems for representations (such as Braeuer's), and in fact those induction theorems were originally motivated by conjectures about Artin L-functions in the first place. – KConrad – 2011-03-22T03:05:21.393

Yikes, I misspelled Brauer in the previous comment. – KConrad – 2011-03-22T08:02:15.277

3

In line with Felipe Voloch's remark "Spec $\mathbb{Z}$ are where schemes really shine", I thought I'd also add this beautiful (expository and very readable!) paper of Serre:

"How to use finite fields for problems concerning infinite fields"

which makes crucial use of being able to do algebraic geometry over (finitely generated algebras over) $\mathbb{Z}$ in order to prove geometric statements (many of which are easily understable without any background in algebraic geometry!) over fields like $\mathbb{C}$ and $\mathbb{Q}$.

2

I should mention the Hamilton-Caylay theorem for matrices: proof: base change to an algebraic closure of your underlying field, and use the fact that diagonalizable matrices are Zariski dense. (However, this doesn't used scheme that are glued from affines.)

5You can prove this result without the usage of schemes. One just need to know what's an irreducible variety is. – Asaf – 2011-05-04T10:16:21.610

20You don't even need that: You prove density in $\mathbb{R}^{n\times n}$ w.r.t. the usual topology and derive the Cayley-Hamilton-theorem from that and then use the fact that there are infinitely many algebraicly independent real numbers. This allows you to view $\mathbb{Z}[X{11},...,X{nn}]$ as a subring of $\mathbb{R}$ and by specializing $X{ij}$ to the value $x{ij}\in R$ in any fixed commutative ring $R$ you can transport the Cayley-Hamilton-equation from $\mathbb{Z}$ to $R$. – Johannes Hahn – 2011-05-04T14:30:29.657

1That's a good elementary approach, but you need to use ${\mathbb C}^{n\times n}$, not ${\mathbb R}^{n\times n}$, because the matrices diagonalizable over the reals are not dense in ${\mathbb R}^{n\times n}$ once $n \geq 2$. Deducing the result for arbitrary fields doesn't require the infinite transcendence degree of $\mathbb R$: for each $n$, Cayley-Hamilton is a polynomial identity with integral coefficients, so once it's proved over $\mathbb Q$ it works for all fields (induction on the number of variables). – Noam D. Elkies – 2017-07-29T23:36:56.157

0

Purity theorem: A map between smooth complex algebraic manifolds of the same dimension has ramification locus of pure codimension 1.

One can prove this by a clever induction on dimension using punctured spectra of local rings and exact sequences in local cohomology both of which are difficult to deal with without schemes.

"The goal is to be able to give some concrete notion of what you can do with schemes to, say, a beginning graduate student or someone not studying algebraic geometry." Will such a student know what a smooth complex algebraic variety is? or what a ramification locus is? or what punctured spectra of local rings are? – Gerry Myerson – 2017-07-30T00:28:09.550

I think believe anyone interested in the question "what elementary problems can you solve with schemes?" can, in the least, be expected to know something about complex curves. All the notions in my example already appear there and with the precise same meaning. Except "punctured spectra of local rings" which I mentioned as an example of why the use of schemes in this particular case can't be easily avoided. – Saal Hardali – 2017-07-30T01:36:12.237

I think this is a nice result because it tells us that the picture one has for a map between complex curves (i.e. local isomorphism except for finitely many points) has a naive generalization to higher dimensions which is true for much deeper reasons. – Saal Hardali – 2017-07-30T01:40:12.127