# Best way to learn Algebraic Geometry?

I've been reading the book Commutative Algebra with a view towards Algebraic Geometry.

I was wondering is the best way to learn algebraic geometry through commutative algebra? As the book I'm currently trying to read is Reid and he assumes Commutative Algebra.

Was wondering what you guys think is the best way to learn Algebraic Geometry. I'm patient so could probably read David Eisenbud and do all the exercise in it before learning Algebraic Geometry.

To make it more concrete. What would you suggest to a fourth year undergrad student who wanted to learn Algebraic Geometry. Like add I know noncommutative algebra up to Artin Weddingburn Theorem. Also, know Group theory up to sylow theorem. Topology up to classification of 2-surfaces.

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Take a look at, mathoverflow.net/questions/35288/… –  Ehsan M. Kermani Dec 26 '11 at 10:12

Reid's Undergraduate Algebraic Geometry requires very very little commutative algebra; if I remember correctly, what it assumes is so basic that it is more or less what Eisenbud assumes in his Commutative Algebra!

The trouble with algebraic geometry is that it is, in its modern form, essentially just generalised commutative algebra. Indeed, in a very precise sense, a scheme can be thought of as a generalised local ring. (The structure sheaf $\mathscr{O}_X$ of a scheme $X$ is a local ring object in the sheaf topos $\textrm{Sh}(X)$, and a $\mathscr{O}_X$-module is literally a module over $\mathscr{O}_X$ in the topos.) If you are willing to restrict yourself to smooth complex varieties then it is possible to use mainly complex-analytic methods, but otherwise there has to be some input from commutative algebra.

That said, it is not necessary to learn all of Eisenbud's Commutative Algebra before starting algebraic geometry. Classical algebraic geometry, in the sense of the study of quasi-projective (irreducible) varieties over an algebraically closed field, can be studied without too much background in commutative algebra (especially if you are willing to ignore dimension theory). Reid's Undergraduate Algebraic Geometry, Chapter I of Hartshorne's Algebraic Geometry and Volume I of Shafarevich's Basic Algebraic Geometry all cover material of this kind.

Modern algebraic geometry begins with the study of schemes, and there it is important to have a thorough understanding of localisation, local rings, and modules over them. A scheme is a space which is locally isomorphic to an affine scheme, and an affine scheme is essentially the same thing as a commutative ring. The theory of affine schemes is already very rich – hence the 800 pages in Eisenbud's Commutative Algebra! For general scheme theory, the standard reference is Chapter II of Hartshorne's Algebraic Geometry, but Vakil's online notes are probably much more readable. Volume II of Shafarevich's Basic Algebraic Geometry also discusses some scheme theory. Also worth mentioning is Eisenbud and Harris's Geometry of Schemes, which is a very readable text about the geometric intuition behind the definitions of scheme theory.

Perhaps the most important piece of technology in modern algebraic geometry is sheaf cohomology. For this, some background in homological algebra is required; unfortunately, homological algebra is not quite within the scope of commutative algebra so even Eisenbud treats it very briefly. The first few chapters of Cartan and Eilenberg's Homological Algebra give a good introduction to the general theory but is strictly more than what is needed for the purposes of algebraic geometry. (For example, they allow their rings to be non-commutative.) The last chapters of Lang's Algebra also cover some homological algebra. Chapter III of Hartshorne's Algebraic Geometry is dedicated to the cohomology of coherent sheaves on (noetherian) schemes.

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I always find your answers to be very thorough, composed, and intuitive. Are you a student? –  Alex Youcis Dec 26 '11 at 8:24
@Alex: Yes, and one who has recently been learning algebraic geometry! –  Zhen Lin Dec 26 '11 at 8:43

One should be warned, though,that Algebraic Geometry is not merely "very structured Commutative Algebra", but there's more to it.

For instance, the theory of abelian varieties requires a more intrinsic geometric approach to be appreciated. Maybe abelian varieties are not objects one needs to deal with at the beginning, but they pop up as soon as one tries to get a bit more sophisticated (e.g. in the case of the Jacobian variety attached to an algebraic curve).

Also, a well-motivated student with inclinations toward Algebraic Geometry should have some basic knowledge of Algebraic Number Theory. After all, one of the motivations to introduce the language of schemes was to develop a good environment where the Weil Conjectures could be attacked, and many interesting examples come from situations with a number-theoretic content.

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I'm currently working my way through Ravi Vakil's course notes for Foundations of Algebraic Geometry at Stanford at the recommendation of an Algebraic Geometry professor, which can be found here. The notes are very long (over 600 pages) and thorough, and include plenty of exercises of varying difficulty. They also indicate what material is important and what can be skipped on first reading. Not much Commutative Algebra is assumed, so I was able to skip some sections as I've taken Commutative Algebra, but these sections might well give you sufficient background for the rest of the notes. If you have the time, I really recommend working through Vakil's notes. I love his style and think I;ve really gained something in the first few chapters.

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Thanks for that. The notes look good. However, he seems to assume Hilbert’s Weak Nullstellensat. Which, is what really scares me about commutative algebra. –  danielwond Dec 26 '11 at 8:01
The idea of the W.N. is not that complicated. The right perspective is to go from the S.N. to the W.N. as follows: for a field $k$ and maximal ideal $M\subset k[x_1,\ldots,x_n]$, we must have that $k[x_1,\ldots,x_n]/M = k(a_1,\ldots,a_m)$ is a field which is a finite extension of $k$. Since these extensions are finite, we get $a_i^{n_i}\in k$ for some $n_i$ hence if $k$ is algebraically closed then $a_i\in k$ so $k[x_1,\ldots,x_n]/M = k$. This is only true when $M = (x_1-a_1,\ldots,x_n-a_n)$ as having multiple/repeated points on which all polynomials vanish gives us an extension of $k$ –  Alex Becker Dec 26 '11 at 8:25
Alex's proof is slightly flawed. What you know from S.N. is that $k(a_1,\ldots,a_m)$ is a finite hence algebraic extension of $k$. So each $a_i$ is algebraic over $k$. This means that $P(a_i) \in k$ for some non constant polynomial $P \in k[X]$ but you cannot assume that this polynomial is of the form $P(X) = X^{n_i}$. (Ex: $a = 1+\sqrt{2}$ is algebraic over $\mathbb{Q}$ but $\forall n\in N^*$, $a^n \notin \mathbb{Q}$). What you can say instead is that since $k$ is algebraically closed and $a_i$ is algebraic over $k$ then $a_i \in k$ (this is the definition of being algebraically closed). –  YBL Dec 27 '11 at 1:51
I agree with you, this is the best way to learn Algebraic geometry from scratch! –  Abramo Dec 1 '13 at 11:31