What is algebraic geometry?

I am a second year physics undergrad, loooking to explore some areas of pure mathematics. A word that often pops up on the internet is algebraic geometry.

What is this algebraic geometry exactly? Please could you give a less technical answer to describe the what this field does and how?

I have done linear algebra, some group and representation theory, and some basic point set topology all from mathematical physics textbooks. Also, a brief overview of the prerequisites to study and do research in the field. I know that commutative algebra, and topology is used, but in exactly what way and how are they inter-connected? How exactly do you mix algebra with topology? Thanks!

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"Less technical" than what? –  Chris Eagle Oct 19 '12 at 15:53
What's wrong with the Wikipedia article? –  Chris Eagle Oct 19 '12 at 15:54
It was relevant because your question desperately needed linebreaks. Reading single chunk stream-of-conscious questions turns off potential readers of your questions. I've inserted some linebreaks for you this time, but I hope you try them out in the future. –  rschwieb Oct 19 '12 at 16:37
@anon: That's one way of doing things, but that is precisely not what algebraic geometry does... –  Zhen Lin Oct 19 '12 at 17:49
The question answers itsself once one opens any book on algebraic geometry. I don't understand why we have to tell the whole story here again which is covered in so many books, articles, including wikipedia and other easy digestible encyclopedias. –  Martin Brandenburg Oct 19 '12 at 18:41

It's a massive subject, and there are many different perspectives; here are a few that don't require too much background.

Perspective one: It's a generalization of linear algebra.

Linear algebra is about dealing with systems of linear equations. This is easy: the set of solutions to a (homogeneous) system is just some subspace of $F^n$ (where $F$ is the field of scalars), and you can compute its dimension by row-reducing your system into echelon form.

Algebraic geometry is about dealing with systems of polynomial equations. As you may imagine, this is much harder. In linear algebra, much of the theory is entirely independent of the field $F$, at least until you want to diagonalize operators; in algebraic geometry, non-algebraically-closed $F$ are a massive headache, and there are phenomena in characteristic $p$ that don't show up in characteristic $0$.

Perspective two: It's a computational tool in classical geometry.

In geometry and topology we may wish to study invariants of manifolds. We define lots of invariants, e.g., homology groups, but how can we get our hands on them? For most examples, we can't do it easily at all, but if the example happens to be a complex manifold given by polynomial equations, there's a lot more that we can say. This is especially important if you want to do things with computers.

Perspective three: It's a conceptual way to think about commutative algebra.

If I give you some ring, OK, great, it has prime ideals, maximal ideals, zero divisors, etc. What does all this mean, and how do you ever remember the barrage of technical theorems about integrality, Artin rings, regular local rings, etc?

If the ring is the ring of functions on some space, then the geometry of the space may reflect properties of the ring, and we can remember the commutative algebra by picturing the geometry. What Grothendieck realized is that if we define "space" correctly (which is not so easy), every ring is the ring of functions on some space! For an example of how you might relate geometry to intrinsic properties of the ring: the space attached to a ring is connected if and only if all of the zero divisors in the ring are nilpotent.

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Nice answer! Please could you also tell me the prerequisites needed to study algebraic geometry? –  ramanujan_dirac Oct 19 '12 at 17:03
Perspective three is particularly interesting. –  ramanujan_dirac Oct 19 '12 at 17:04
you will have to know the languages of point-set topology and commutative algebra; it will also be slow-going if you don't have some prior exposure to other areas of geometry –  user29743 Oct 19 '12 at 17:09
"If we define 'space' correctly (which is not so easy), every ring is the ring of functions on some space!" sounds marvelous. Can you give some more sugar? After all: what is the - not so easy - way to define "space" correctly? And which kind of functions? Is this a representation theorem? –  Hans Stricker May 13 '13 at 22:37
Given a ring $A$, there's a topological space Spec$(A)$ whose underlying set is the set of prime ideals of $A$; for an arbitrary subset $S \subset A$, let $V(S)$ be the set of prime ideals containing $S$. You get a topology on Spec$(A)$ by declaring sets of the form $V(S)$ closed. This is the space. An element $f \in A$ gives a "function" on it by the rule $\mathfrak{p} \mapsto \text{Image(f)}\in A_\mathfrak{p}/\mathfrak{p}$, where $A_\mathfrak{p}$ denotes the localization. This is a strange definition of "function" since it lands in different fields when you evaluate it at different points! –  user29743 May 13 '13 at 23:07

Suppose you look at all polynomials in two variables, $\mathbb C[x,y]$. Any element $p$ in that ring can be evaluated at any element of $\mathbb C^2$ - if we have two complex values, $a,b$ we can compute $p(a,b)$. And if $p,q$ evaluate to the same value at every point of $\mathbb C^2$ then they are actually the same polynomial.

Now, look at how $\mathbb C[x,y]$ acts when it is evaluated only on some subset of $\mathbb C^2$, say the set of points $X=\{(a,b): b = a^2\}$.

Then $\mathbb C[x,y]$ evaluation is no longer "faithful," but rather, two different polynomials, such as $p_1(x,y)=x^3y$ and $p_2(x,y)=xy^2$, can evaluate as the same on $X$. On X, then, the ring of evaluation functions is actually $R=\mathbb C[x,y]/\left<y-x^2\right>\cong \mathbb C[x]$.

That's a VERY simple case, but Algebraic Geometry is founded on the idea that we can learn something about the geometry of the solutions to a set of equations (in this case, $b=a^2$) by looking at the ring of evaluation functions on that set.

Now, given any point $(a,b)\in X$, there is a natural ring homomorphism $\phi:R\rightarrow \mathbb C$ which corresponds to evaluation. Since this map is onto, this means that the point $(a,b)$ corresponds to some maximal ideal of $R$.

That correspondence between points of the graph of your equations and maximal ideals of the "evaluation ring" associated with your equations, is the big starting point of algebraic geometry.

I will skip the topology part of the question, because it is probably too confusing to try to mix in, except to say that the types of topology used in Algebraic Geometry are very different from the typical starting topologies used in Algebraic Topology.

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