what is algebraic geometry

What is algebraic geometry? What motivates it? Most importantly what are its foundations?

 Have you tried Wikipedia? http://en.wikipedia.org/wiki/Algebraic_geometry
 I'd like a more complete explanation.

what is algebraic geometry

Mathwonk should really write a set of introductory notes on Algebraic Geometry...

 Recognitions: Homework Help Science Advisor i have done so. maybe i will put them online soon. i will post here the introductory chapter.
 Recognitions: Homework Help Science Advisor example: given the equation X^2 + Y^2 + Z^2, one can asociate the set of points in complex projective 2 space which make this equation = 0. this is the old fashioned approach. I.e. an algebraic variety is a point set defined by polynomial equations. Or one can view this equation as a functor from fields to sets, that associastes to each field K, the subset of K^3 that satisfies this equation. this is a more modern approach. i.e. an algebraic variety is a certain functor of rings.
 Recognitions: Homework Help Science Advisor 8300 2003 Day 2: Roots of polynomials in several variables Algebraic geometry arises from the problem of solving systems of polynomial equations. In modern times, this study has become the theory of “schemes”. Many of us have seen the definition of an “affine scheme” as the set of prime ideals of a commutative ring with identity. To see what these have to do with one another, we want to retrace in the classical case of polynomials over a field, how prime ideals arise in looking for common solutions of systems of equations. Let k be any field, and {fi} in k[T1,...,Tn] = k[T] a collection of polynomials in n variables over k. A “solution” or “common zero” of the system {fi=0} will be a vector, or point, p = (p1,...,pn) in E^n, where E is a field extension of k, such that for all i, fi(p) = 0, in E. The key to connecting a solutions vector p with a prime ideal of k[T] is to look at the associated evaluation map π:k[T]—>E, taking f(T) to f(p), and its kernel ker(π) = Ip. Then p is a common zero of the polynomials {fi} if and only if {fi} is in Ip. Moreover since E is a field, the ideal Ip is prime in k[T]. Thus a “solution” of the system {fi} in k[T], which might be thought of as a “point” of the algebraic variety defined by this system, corresponds to a prime ideal of k[T] containing {fi}, or equivalently, to a prime ideal of the quotient ring k[T]/({fi}), where ({fi}) is the ideal generated by the set {fi}. Conversely, if {fi} is in I is in k[T] where I is a prime ideal, then there is an associated k algebra map π:k[T]—>k[T]/I is in E where E is the field of fractions of the domain k[T]/I. If we define the point p in E^n by setting pi = π(Ti) in E, then this map π is evaluation at p, in the sense above. Thus prime ideals of the quotient ring k[T]/({fi}) correspond to common zeroes of the system {fi}, with components in extensions of k. This correspondence is not however one-one, since e.g. both i and -i in C define the same prime ideal in R[T], since R[T]/(T^2+1) has two isomorphisms with C, taking T to either i or -i. Thus considering solutions in a field extension of k involves some Galois theory over k. Indeed if E is an algebraic closure of k, then maximal prime ideals m Ă k[T] correspond to orbits of Gal(E/k) in E^n, parametrizing different embeddings of k[T]/m in E. (See Mumford, II.4., pp. 96 - 99.) Natural questions: 1) Which prime ideals correspond to solutions in k^n, i.e. when do the solutions have coordinates in the original field k itself? 2) Which prime ideals correspond to solution vectors p = (p1,...,pn) with coordinates pi lying in some algebraic extension E of k? 3) What is the geometric meaning of solution vectors whose coordinates lie in transcendental extensions of k? Question 1). Let p in k^n be a solution of the system {fi}. Then the evaluation map k[T1,...,Tn]—>k, is surjective since it equals the identity on k, hence its kernel is a maximal ideal m … {fi}. Thus p in kn corresponds to a maximal ideal m … {fi} such that the composition k—>k[T]/m—>k is an isomorphism, and since k[T]/m—>k is injective, equivalently such that k—>k[T]/m is an isomorphism. Conversely let {fi} in m in k[T] and assume the map k—>k[T]/m is an isomorphism. Thus for each i, there is a unique pi in k that maps to Ti, mod m. Thus the composition with the inverse isomorphism k[T]—>k[T]/m—>k takes Ti to pi, and this is a k algebra map. Thus it must be evaluation at p. Moreover it takes every element of m to 0. Thus p is a common zero of the ideal m, and hence of the system {fi} having coefficients in k. Thus common solutions of the system {fi} in k[T] with coefficients in k, correspond one - one to maximal ideals m in k[T], with {fi} in m, and such that k—>k[T]/m is an isomorphism. Equivalently they correspond to maximal ideals m in k[T]/({fi}) such that k—>k[T]/m is an isomorphism. Of course there might not be any such ideals, as in the case of {T^2+1} in R[T], since ({T^2+1}) is already maximal and has quotient larger than R. This is because as we know, many systems of equations have no solutions in the coefficient field. Note that a maximal ideal of k[T] with p = (p1,...,pn) in kn as common zero, contains the functions Ti-pi. But the ideal (T1-p1,...,Tn-pn) is already maximal since the composition k—>k[T]/(T-p) takes pi to Ti, thus surjects, so is isomorphic. So arguments like those above give one - one correspondences between the following sets: {common solutions (a1,...,an) in kn of {fi}} ≈ {k algebra maps π:k[X1,...,Xn]—>k with π(fi) = 0, for all i} ≈ {maximal ideals {fi} in MĂk[X1,...,Xn] such that k—>k[X1,...,Xn]/M is an isomorphism} ≈ {ideals of form {fi} in (X1-a1,...,Xn-an) Ă k[X1,...,Xn]} These correspondences are as follows: a point a in kn at which all fi = 0 yields the k algebra map π = evaluation at a, which takes all fi to 0; a k algebra map π:k[X1,...,Xn]—>k is always surjective since it already is on k, so π has a maximal ideal kernel M such that the composition k—>k[X1,...,Xn]—>k[X1,...,Xn]/M —> k is an isomorphism, and since π(fi) = 0 for all i, thus {fi} in M; a maximal ideal M … {fi} such that the composition k—>k[X1,...,Xn]/M is an isomorphism is always of form (X1-a1,...,Xn-an) where ai = the unique element of k such that Xi = ai mod M; finally a maximal ideal of form M = (X1-a1,...,Xn-an) … {fi} determines the point a by setting aj equal again to the unique element of k congruent mod M to Xj, and since the point a belongs to M, all fi vanish at a. Question 2). The question is more complicated but the answer is simpler: solutions p of the system {fi}, with coordinates pi in some algebraic extension of k, correspond to all maximal ideals of k[T]/({fi}), i.e. to those maximal ideals of k[T] containing the system {fi}. This time however the correspondence is not one to one, since several solutions can correspond to the same maximal ideal. To prove it will take a little work, but one direction is elementary. Let p be a common zero of all fi, where pj is in E, and E is algebraic over k. Consider the evaluation map π:k[T]—>E taking f(T) to f(p). The image is a domain k[p1,...,pn], since contained in the field E, and its fraction field k(p1,...,pn) is a finitely generated algebraic extension field of k, hence a finite dimensional k vector subspace of E. Multiplication by any non zero element u in the domain k[p1,...,pn] is an injective k linear map of the finite dimensional k vector space k[p1,...,pn] to itself, hence also surjective. Thus there is some v in k[p1,...,pn] such that uv = 1. Thus the image k[p1,...,pn] of the evaluation map is actually a field, i.e. k[p1,...,pn] = k(p1,...,pn). Hence the kernel, ker(π) is a maximal ideal of k[T] containing {fi}. The converse is less elementary, but we claim it is true: i.e. if {fi} in m in k[T] is any maximal ideal, then the quotient k[T]/m is always a finite dimensional algebraic extension field of k, hence embeds in E = algebraic closure of k, via some embedding π:k[T]/m—>E. If pi = π(Ti), then the map π with kernel m, is evaluation at p in En, which is a common solution of the system {fi} with coordinates in E. Here we see that different embeddings of k[T]/m into E give rise to different solutions p corresponding to the same ideal m. The simplest statement arises if k = E is itself algebraically closed. Then we can combine the statements in 1) and 2) and see that there is a one one correspondence between solutions of the system {fi} with coordinates in k, and the set of all maximal ideals of k[T] containing {fi}. In particular, if the set {fi} does not generate the unit ideal, then there must be some maximal ideals containing the system, hence there are some common solutions. This is a several variables analog of the fundamental theorem of algebra. In its simplest form it says if k is algebraically closed, there is a one one correspondence between points of kn and maximal ideals in k[T1,...,Tn], where p = (p1,...,pn) in kn corresponds to (T1-p1,...,Tn-pn) in k[T]. This is Hilbert’s famous (“weak”) nullstellensatz, the foundation result of the whole subject of algebraic geometry, the precise dictionary between geometry and polynomial algebra. If {fi} in k[T] where k is any field and E its algebraic closure, maximal ideals of k[T]/({fi}) correspond to (finite) Gal(E/k) orbits of the common solution set of {fi} in En. Solutions of the system {fi} lying in k^n are the fixed points of the action. Question 3), the interpretation of points on the algebraic variety V({fi}) with coordinates in non algebraic (“transcendental”) extensions of k. For simplicity we assume k is algebraically closed, as we will henceforth always assume in this course. We know from the discussion above, such points p correspond to prime ideals Ip in k[T]/({fi}), hence the maximal ideals containing this prime ideal constitute a certain subcollection of the k valued points of the variety V({fi}) Ă kn. I.e. a point p with values in a transcendental extension E of k corresponds to a subcollection V(Ip) of k valued points. Thus if {fi} in Ip in k[T], then the prime ideal Ip corresponds to a subvariety V(Ip) in V({fi}) in k^n, and this subvariety V(Ip) is our geometric interpretation, in kn, of the “point” p with values in the transcendental extension E of k. Moreover, if we think of a k valued point, i.e. a maximal ideal, as having dimension zero, then a “point” p with coordinates pi in E … k which generate a transcendence degree r extension k(p1,...,pn) of k in E, corresponds to a prime ideal I in ?k[T]/({fi}) of “coheight r”, so the variety V(Ip) has dimension r. I.e. the prime ideal Ip can be joined to a maximal ideal by a chain of prime ideals of length r, but no longer, and since k(p1,...,pn) = fraction field of k[p1,...,pn] ≈ k[T]/Ip, thus as commutative algebra students may know, the transcendence degree of k(p1,...,pn) equals the Krull dimension of k[p1,...,pn], equals the coheight of Ip in [T]. (We will prove this later.) The subvariety V(I) in V({fi}) thus has dimension r. Note that “dimension” is a relative term, and here it is taken relative to the base field k. Since not all fields are algebraic over k, not all “points” are zero dimensional over k. To sum up, if k is algebraically closed, and {fi} in I in k[T], where I is a prime ideal of coheight r, then I corresponds equivalently to an (“irreducible”) r dimensional subvariety of the variety V({fi}), and to a point with values in the extension field E = fraction field(k[T]/I) of k of transcendence degree r. (Since a prime ideal is not the proper intersection of two other ideals, here “irreducible” means the variety it defines is not the proper union of two other varieties.)
 Recognitions: Homework Help Science Advisor hows that? i realize it stinks. but maybe someone can make something of some of it.,
 Hey, that's cool, I'll try to make into good lookin' PDF document.
 Recognitions: Homework Help Science Advisor containment symbol seems to have become Atilda.
 Maybe you could give me a list of what needs to be corrected? I can't promise you anything, but I'll try to finish it by the end of the week.
 Recognitions: Homework Help Science Advisor i have already tried to correct it.
 Recognitions: Homework Help Science Advisor Here is a revised version of the fresh one i wrote just for you guys: Naive introduction to algebraic geometry: the geometry of rings I used to say algebraic geometry is the study of the geometry of polynomials. Now I sometimes call it the "geometry of rings". I also feel that algebraic geometry is defined more by the objects it studies than the tools it uses. The naivete in the title is my own. I. BASIC TOOL: RATIONAL PARAMETRIZATION Algebraic geometry is a generalization of analytic geometry - the familiar study of lines, planes, circles, parabolas, ellipses, hyperbolas, and their 3 dimensional versions: spheres, cones, hyperboloids, ellipsoids, and hyperbolic surfaces. The essential common property these all have is that they are defined by polynomials. This is the defining characteristic of classical algebraic sets, or varieties - they are loci of polynomial equations. A further inessential condition in the examples above is that the defining polynomials have degree at most 2 and involve at most 3 variables. This limitation arose historically for psychological and technical reasons. Before the advent of coordinates, higher dimensions could not be envisioned or manipulated, and even afterwards it was commonly felt that space of more than 3 dimensions did not "exist" hence was irrelevant. The dimension barrier was lifted by Riemann and Italian geometers in the 19th century such as C. Segre, who realized that higher dimensions could be useful for the study of curves and surfaces. Riemann's use of complex coordinates for plane curves simplified their study, and Segre understood that some surfaces in 3 space were projections of simpler ones embedded in 4 space. One reason for restricting attention to equations in (X,Y) of degree at most 2 is a limitation of the basic method of "parametrization", expressing a locus by an auxiliary parameter. E.g. the curve X^2 + Y^2 = 1 can be parametrized by the variable t by setting X(t) = 2t/[1+t^2], Y = [1-t^2]/[1+t^2]. This substitutiion, along with dX = 2[1-t^2]dt/[1+t^2]^2, allows one to simplify the integral of dX/sqrt(1-X^2), to that of 2dt/[1+t^2] = 2d[arctan(t)]. The cubic Y^2=X^3 can also be parametrized, say by X = t^2, Y = t^3. But to simplify in this way the integral of dX/sqrt(1-X^3), requires us to parametrize the cubic Y^2 = 1-X^3, a problem which is actually impossible. These questions were considered first by the Bernoullis, and resolved by new ideas of Abel, Galois, and especially Riemann as follows. (Interestingly, in three variables the difficulty arises in degree 4, and 19th century geometers knew how to parametrize most cubic surfaces.) II. NEW METHODS FOR PLANE CURVES: TOPOLOGY and COMPLEX ANALYSIS Riemann associated to a plane curve f(X,Y)=0 its set of complex solutions, compactified and desingularized. This is its "Riemann surface", a real topological 2 manifold with a complex structure obtained by a branched projection onto the complex line. For instance the curve y^2 = 1-X^3 becomes its own Riemann surface after adding one point at infinity, making it a topological torus. Projection on the X coordinate is a 2:1 cover of the extended X line, branched over infinity and the solutions of 1-X^3 = 0. This association is a functor, i.e. a non constant rational map of plane curves yields an associated holomorphic map of their Riemann surfaces, in particular a topological branched cover. Riemann assigns to a real 2 manifold its "genus" (the number of handles), and calculates that branched covers cannot raise genus, and the only surface of genus zero is the sphere = the Riemann surface of the complex t line. Hence if the Riemann surface of a plane curve has positive genus, it cannot be the branched image of the sphere, hence the curve cannot be parametrized by the coordinate t. Riemann also proved a smooth plane curve of degree d has genus g = (d-1)(d-2)/2, so smooth cubics and higher degree curves all have positive genus and hence cannot be parametrized. He proved conversely that any curve whose Riemann surface has genus zero can be parametrized, e.g. hyperbolas, circles, lines, parabolas, ellipses, or any curve of degree < 3. Moreover a singularity, i.e. a point where the curve has no tangent line, like (0,0) on Y^2 = X^3, lowers the genus during the desingularization process, and this is why such a "singular" cubic can be parametrized. One also obtains a criterion for any two irreducible plane curves to be rationally isomorphic, namely their Riemann surfaces should be not just topologically, but holomorphically isomorphic. By representing a smooth plane cubic as a quotient of the complex line C by a lattice, using the Weierstrass P function, one can prove that many complex tori are not holomorphically equivalent, by studying the induced map of lattices. It follows that there is a one parameter family of smooth plane cubics which are rationally distinct from each other. This shows briefly the power and flexibility of topological and holomorphic methods, which Riemann largely invented for this purpose, an amazing illustration of thinking outside traditional confines. III. RINGS and IDEALS To go further in the direction of arithmetic questions, one would like more algebraic techniques, applicable to fields of characteristic p, algebraic numbers fields, rings of integers, power series rings,.... One can pose the question of isomorphism of plane curves algebraically, using ring theory, as follows. Since all roots of multiples of the polynomial f vanish on the zero locus of f, it is natural to associate to the curve V:{f=0} in k^2, the ideal rad(f) = {g in k[X,Y]: some power of g is in (f)}. Then the quotient ring R = k[X,Y]/rad(f) is the ring of polynomial functions on V. Moreover if p is a point of V, evaluation at p is a k algebra homomorphism R-->k with kernel a maximal ideal of R. In case k is an algebraically closed field, like C or the algebraic numbers, this is a bijection between points of V and maximal ideals of R. In fact everything about the plane curve V is mirrored in the ring R in this case, and two irreducible polynomials f,g, in k[X,Y], define isomorphic plane curves if and only if their associated rings R and S are isomorphic k algebras. Indeed the assignment of R to V is a "fully faithful functor", with algebraic morphisms of curves corresponding precisely to k algebra maps of their rings. To recover the points from the ring one takes the maximal ideals, and to recover a map on these points from a k algebra map, one pulls back maximal ideals. (Since these rings are finitely generated k algebras and k is algebraically closed, a maximal ideal pulls back to a maximal ideal.) Any pair of generators of the k algebra R defines an embedding of V in the plane. Similarly, if f (irreducible) in k[X,Y,Z] defines a surface V:{f=0} in k^3, (k still an algebraically closed field), then not only do points of V correspond to maximal ideals of R = k[X,Y,Z]/(f), but irreducible algebraic curves lying on V correspond to non zero non maximal prime ideals in R. Again this is a fully faithful functor, with polynomial maps corresponding to k algebra maps. In particular the pullback of maximal ideals is maximal, but now the pullback of some non maximal ideals can also be maximal, i.e. some curves can collapse to points under a polynomial map. To give the algebraic notion full flexibility, in particular to embrace non Jacobson rings with too few maximal ideals to carry all the desired structure, Grothendieck understood one should discard the restriction to rings without radical and expand the concept of a "point", to include irreducible subvarieties, i.e. consider all prime ideals as points, as follows. IV. AFFINE SCHEMES If R is any commutative ring with 1, let X (= "specR") be the set of all prime ideals of R, with a topological closure operator where the closure of a set of prime ideals is the set of all prime ideals containing the intersection of the given set of primes. (Intuitively, each prime ideal contains the functions vanishing at the corresponding point, so their intersection is all functions vanishing at all the points of the set, and the prime ideals containing this intersection hence are all points on which that same set of functions vanishes. So the closure of a set is the smallest algebraically defined locus containing the set.) This closure operator defines the "Zariski topology" on X. Now any ring map defines a morphism of their spectra by puling back prime ideals, and in particular a morphism is continuous, although this alone says little since the Zariski topology is so coarse. Notice now maximal ideals may pull back to non maximal ones, e.g. under the inclusion map Z-->Q of integers to the rationals, taking the unique point of specQ to a dense point of specZ. Maximal ideals now correspond to closed points, and in particular there are usually plenty of non closed points. Intuitively, every irreducible subvariety has a dense point, and together these "points", one for each irreducible subvariety, give all the points of specX. If K is a ring, a "K valued point" of X is given by a ring homomorphism R-->K, not necessarily surjective. E.g. if K is a field, the pullback of the unique maximal ideal of K is a not necessarily maximal, prime ideal P of R, the K valued point. Even if the point is closed, i.e. if P is maximal, we get information on which maximal ideals correspond to points with coefficients in different fields. If say k = the real field, and f is a polynomial over k, then a k algebra map g:k[X,Y]/(f)-->k has as kernel a maximal ideal corresponding to a point of {f=0} in k^2, i.e. a point of {f=0} in the usual sense, with real coefficients. The coordinates of this point are given by the pair of images (g(X),g(Y)) in k^2 of the variables X,Y, under the algebra map g, which after all is evaluation of functions at our point. But if say f = Y-X^2, the map from k[X,Y]/(f) -->C taking X to i, and Y to -1, corresponds to the C (complex) - valued point (i,-1), in C^2 rather than k^2. More generally, if I is any ideal in Z[X1,...,Xn] generated by integral polynomials f1,...fr, and A is a ring, a ring homomorphism Z[X1,...,Xn]/I -->A takes the variables Xj to elements aj of A such that all the polynomials fi vanish at the point of A^n with cordinates (a1,...an). I.e. the map defines an "A valued point " of the locus defined by I. E.g. if M is a maximal ideal of R,we can always view the coordinates of the corresponding point in the residue field R/M, i.e. the point M of specR is "R/M valued". This approach lets us recover tangent vectors too, in case say of a variety V with ring R = k[X1,...,Xn]/I, where radI = I, and k is an algebraically closed field. Consider the ring S = k[T]/(T^2), with unique maximal ideal (t) generated by the nilpotent element t. Then we claim tangent vectors to V correspond to S valued points (over spec(k)), i.e. to k algebra maps R-->S. E.g. if R = k[X], and we map R-->S by sending X to a+bt, then the inverse image of the maximal ideal (t) is the maximal ideal (X-a), and two elements of (X-a) have the same image in S if and only if they have the same derivative at X=a. Thus S valued points of V are points of the "tangent bundle" of V.