9
$\begingroup$

Let $X$ be a smooth projective curve over a field $k$. We let $\omega$ be the canonical line bundle of $X$ and we denote by $F$ the field of $k$-valued rational functions on $X$.

(1) When $k$ is algebraically closed then $\omega$ is a dualizing sheaf for $X$. From there it is easy to prove Riemann-Roch for regular (holomorphic) line bundles $L$ over $X$: By this I mean a precise formula which computes the Euler characteristic of $L$ in terms of the degree of $L$ and the genus of $X$ (I think of both as being topological invariants).

(2) When $k$ is a finite field then one may consider the topological ring $\mathbf{A}_F$, the ring of Adeles of $F$. Doing Fourrier analysis on this self-dual locally compact abelian group and doing a counting argument one may deduce Riemann-Roch.

Q1: Is it possible to generalize Riemann-Roch to other fields? What about real and $p$-adic numbers?

Q2: Is $\omega$ a dualizing sheaf when $k$ is finite? If not

(I guess that in general one has to replace the notion of dualizing sheaf by some kind of complex in a derived category)

Q3: Is there a way to prove simultaneously $(1)$ and $(2)$?

Q4: Is there some notion that would encompass simultaneuously $\mathbf{A}_F$ and $\omega$?

$\endgroup$
0

2 Answers 2

18
$\begingroup$

Yes. There is a Riemann-Roch for smooth projective curves over arbitrary fields. It was proved by the German school of function fields in the 30's. From (2) I deduce that you've been reading Weil's "Basic Number Theory". Anyway, the proof that Weil gives there is a shortened version of a proof he gave of the full theorem. It's in his collected works ("Algebraische Beweis der Riemann-Roch Satz", or some such title). The proof is reproduced in many books, particularly those with "function field" in the title (e.g. Stichtenoth) or Lang's "Introduction to algebraic and abelian functions" (watch for misprints, as usual). You can also prove it with the modern geometric machinery. A book that bridges the two approaches is Serre's "Groupes algebriques et corps de classes". There is absolutely no restriction on the ground field, except the proof is slightly more tortuous when the field is not perfect.

$\endgroup$
4
  • 3
    $\begingroup$ The proof in Serre's book is one of the best examples I know of of beauty as it can be embodied in an argument :) $\endgroup$ Feb 14, 2011 at 22:37
  • 1
    $\begingroup$ Van der Waerden's Algebra, vol. 2, 5th edition, presents Weil's proof, taken from J. Reine u. Angew. Math., vol.179, 1938. He mentions it was influenced by the metodo rapido of Severi, for which a later reference is in Acta Pont. Accad. Sci., 1952. The earliest proof cited is by F.K. Schmidt, Math. Z., vol.41, 1936. $\endgroup$
    – roy smith
    Feb 15, 2011 at 15:44
  • $\begingroup$ The unabridged name of the journal mentioned by roy is Acta Pontificia Accademia Scientiarum, Città del Vatican. Manin, Hawking and Witten are contemporary members of this Academy: it.wikipedia.org/wiki/Pontificia_Accademia_delle_Scienze $\endgroup$ Feb 15, 2011 at 17:47
  • $\begingroup$ Just to point out that Serre assumes the base field is algebraically closed on the first page of chapter 2 of his book (where RR is proved), so this is not the reference which the OP is seeking. $\endgroup$ Nov 22, 2023 at 12:54
28
$\begingroup$

Dear Hugo, the wonderful formalism of schemes allows us to have a Riemann-Roch theorem for a projective curve $X$ over an arbitrary field $k$, even without any assumption of smoothness. It says, like in the good old times of Riemann surfaces, that for a Cartier divisor $D$ on $X$ we have $$\chi (\mathcal O_X(D))= deg(D)+ \chi (\mathcal O_X)$$

There is a dualizing sheaf $\omega$ and Serre duality yields the formula $$ h^0(X,\mathcal O_X(D))-h^0(X,\omega \otimes\mathcal O_X(-D))=1-p_a(X)+deg D$$ where $p_a(X)=1-\chi(\mathcal O_x)$ is the so called arithmetic genus of the curve.

Everything is in our friend Qing Liu's fantastic book Algebraic Geometry and Arithmetic Curves but I bet he's too modest to give you this obvious answer!

Edit The first displayed formula is actually valid in even greater generality: it holds for any projective curve $X$ (smooth or not) over an arbitrary artinian ring $k$. The proof is on page 164 of

Altman, A.; Kleiman, S., Introduction to Grothendieck duality theory. Lecture Notes in Mathematics No. 146, Springer-Verlag, Berlin-New York, 1970

Complement As an answer to Hugo's question in his comment below, let me add that indeed, in the case of a smooth projective curve over a field $k$, the dualizing sheaf $\omega$ is nothing else than the canonical sheaf . More generally for a smooth projective variety of dimension $r$ over $k$, the dualizing sheaf is just the canonical sheaf $\omega=\Omega^r_{X/k}$. This is (a special case of) Theorem I.4.6, page 14 in Altman-Kleiman's monograph.

$\endgroup$
4
  • 1
    $\begingroup$ Of course Felipe is absolutely right about the German school's priority . However I find fascinating the ease and elegance with which schemes integrate all those classical results into their harmonious, coherent and powerful language. $\endgroup$ Feb 15, 2011 at 1:31
  • $\begingroup$ Thanks a lot Georges for Qing Liu's reference. So what is the dualizing sheaf in general for a smooth projective curve defined over an arbitrary field $k$? Is it still the canonical line bundle? $\endgroup$ Feb 15, 2011 at 18:08
  • 1
    $\begingroup$ @Hugo: yes. (I would have answered your question by saying: "Yes. It's called the Riemann-Roch Theorem." None of the standard proofs use the algebraic closure of the ground field at any point. But the answers you have been given are somewhat more generous than this...) $\endgroup$ Feb 16, 2011 at 7:29
  • 1
    $\begingroup$ You are probably right, but unfortunately, unless I'm mistaken, I thought that $k$ was assumed to algebraically closed in Hartshorne's book. Is there some tricky point to address in characteristic $p$ when $k$ is not perfect? $\endgroup$ Feb 17, 2011 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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