I was just wondering what the real prerequisites are for reading Qing Liu's 'Algebraic Geometry and Arithmetic Curves', and if it is a good first book on the subject. In his preface he states that the prerequisites are few and any graduate student possesses the background necessary to read it, but this being algebraic geometry I am reticent to believe him. For example, does he assume knowledge of Differential geometry? Algebraic Topology? I expect that he assumes commutative algebra, but at what level? Anyways, for people that have read him, please share your thoughts/comments on this.
[Math] Question about Qing Liu’s Algebraic Geometry book
algebraic-geometryarithmetic-geometrygeometryreference-request
Related Solutions
A compact Riemann surface $X$ is in particular a compact real orientable surface. These surfaces are classified by their genus.
That genus is indeed the number of handles cited in popular literature; more technically it is
$$g(X)=\frac {1}{2}\operatorname {rank} H_1(X,\mathbb Z) = \frac {1}{2}\operatorname {dim} _\mathbb C H^1_{DR}(X,\mathbb C) $$ in terms of singular homology or De Rham cohomology.
Under the pressure of arithmetic, geometers have been spurred to consider the analogue of compact Riemann surfaces over fields $k$ different from $\mathbb C$: complete smooth algebraic curves.
These have a genus that must be calculated without topology.
The modern definition is (for algebraically closed fields) $$ g(X)=\operatorname {dim} _k H^1(X, \mathcal O_X)= \operatorname {dim} _kH^0(X, \Omega _X)$$
in terms of the sheaf cohomology of the structural sheaf or of the sheaf of differential forms of the curve $X$.
Of course this geometric genus is always $\geq 0$.
There is a more general notion of genus applicable to higher dimensional and/or non-irreducible varieties over non algebraically closed fields: the arithmetic genus defined by $$p_a(X)=(-1)^{dim X}(\chi(X,\mathcal O_X)-1)\quad {(ARITH)}$$ (where $\chi(X,\mathcal O_X)$ is the Euler-Poincaré characteristic of the structure sheaf).
[ Hirzebruch and Serre have, for very good reasons, advocated the modified definition $p'_a(X)=(-1)^{dim X}\chi(X,\mathcal O_X)$, which Hirzebruch used in his ground-breaking book and Serre in his foundational FAC]
For smooth projective curves over an algebraically closed field $g(X)=p_a(X)\geq 0$ : no problem.
It is only in more general situations that the arithmetic genus $p_a(X)$ may indeed be $\lt 0$
Edit
The simplest example of a reducible variety with negative arithmetic genus is the disjoint union $X=X_1\bigsqcup X_2$ of two copies $X_i$ of $\mathbb P^1$.
The formula $(ARITH)$ displayed above yields: $p_a(X)=1-\chi(X,\mathcal O_X)=1-(dim_\mathbb C H^0(X,\mathcal O_X)-dim_\mathbb C H^1(X,\mathcal O_X))=1-(2-0)$
so that $$p_a(X)=p_a(\mathbb P^1\bigsqcup \mathbb P^1)=-1\lt0$$
I recommend you take a look at the course notes on Algebraic Geometry by Edixhoven and Taelman; see
http://www.math.leidenuniv.nl/~edix/teaching/2010-2011/AG-mastermath/ag.pdf
In these notes all the necessary tools from algebraic geometry are explained and a complete proof is given of the Riemann hypothesis for curves over finite fields.
Best Answer
No differential geometry or algebraic topology is necessary, though the former might help with motivation for some things, depending upon your mathematical inclinations. You need to know basic graduate abstract algebra (he develops most of the commutative algebra he needs so you don't need to know that in advance) and you need to know some basic point set topology (definition of a topological space, continuous map, Hausdorff, not much more). That's it. For serious.
Would it help to already know some commutative algebra? Yeah, probably, but it's not essential by any means. Liu's is a remarkably self-contained book. I consider it to be an excellent first book on modern algebraic geometry. Whether or not you should first learn some classical algebraic geometry depends on you and your tastes. Liu does do some of that (algebraic sets) in the second chapter. Being familiar with the classical theory might help with motivation, but I don't think it's necessary.