SOLUTION TO MATH 797N, PROBLEM SET #1.

SOLUTION TO MATH 797N, PROBLEM SET #1. SIMAN WONG Comments: As Silverman points on p. 461,... it is hoped that this book will lead the student on into the realm of active mathematics, and the benefits of working without aid clearly outweigh any advantage that might be gained by having solutions readily available. So it is with great trepidation that I present the solution key here. I do so in part because I m lazy; I don t feel like repeating the same comments on multiple papers. Another reason is that periodically I would assign more challenging problems like 4(b) in this problem set for which it would make sense to include a complete solution. And sometimes I will try to provide motivation as to how one would go about discovering an argument that seems to jump out from nowhere. There is yet another reason why I make solutions available, namely to show you how to write mathematics. By this point you know how to write rigorous proofs yourself and are capable of checking other people s arguments. But as the material becomes more technical and proofs get longer, it is not practical or even desirable to spell out every single detail. The trick is to be able to highlight the key ideas without getting lost in a sea of epsilons and deltas. If you have not seen it before, do check out Professor Goss excellent tips on mathematical writing styles www.math.ohio-state.edu/ goss/hint.ps I make no claim about being a good mathematical writer myself; I do hope that my solutions would help give you some how idea how to do this yourself. One last thing: I ll try, but I don t promise that I will provide solution to every question on every problem set. It takes a lot of work to type these up... And let me know if you find any typos and/or mistakes. 1(a) Note that V is defined by f = y 2 x 3. Now f = x 3x2 and f = 2y; set both to zero y and we get (x, y) = (0, 0), which is a point on V. So the origin is a singular point of V. (b) Take f = x 2 + y 2 (z 1) 2 ; then f f f = 2x, = 2y, and = 2(z 1). Set these x y z to zero and we get (x, y, z) = (0, 0, 1), which is on V. So this is a singular point. This is compatible with the fact that V (R) is a (double) cone with vertex (0, 0, 1). 2(a) As is written, the coordinate functions of φ are all polynomials, so it is defined everywhere except when s 2 t = s 3 = t 3 = 0. That happens precisely when s = t = 0, which is not possible since [s, t] P 1. And since (s 3 ) 2 (t 3 ) = (s 2 t) 3, the image of φ does lie in V, so φ is a morphism from P 1 to V. (b) Dehomogenize φ and you get a map from A 1 to V that sends s K to the point (s 2, s 3 ) on the affine equation y 2 = x 3. Note that s 3 /s 2 = s; this suggests that we consider the map ψ : V P 1 : [x : y : z] [y : x]. This is defined everywhere except at [0 : 0 : 1] V, and we check that (1) φψ V {[0:0:1]} = id and ψφ P 1 {[0:1]} = id, as desired. Remarks: In plain language, ψ is simply projection onto the y-axis draw a picture! For future reference, note that φ is both injective and surjective: The former is clear; as for the latter, from the first equation in (1) we see that the image ψφ V {[0:1]} = id contains V {[0 : 0 : 1]}. And since φ([0 : 1]) = [0 : 0 : 1], we are done.

2 SIMAN WONG (c) In order for φ to be an isomorphism, we need to find a rational map ψ : V P 1 which is in fact regular at all points of V and such that the compositum with φ in both ways are identity. If we have such a ψ, then after dehomogenizing we get an affine map ψ a : (y 2 = x 3 ) A 1. Upon identifying A 1 with K, we see that ψ a is given simply as a two-variable rational function over K: ψ a(x, y) = f(x, y). The requirement that φψ = id V now reads (2) f(x, y) 2 = x, f(x, y) 3 = y whenever ψ a is defined. But the set of points on which ψ a is not defined is a proper closed subset of V (K), and hence at most a finite set even over K. From (2) we then see that f(x, y) = y/x for all but finitely many pairs of (x, y) over K. But f(x, y) is a rational function over a field; since K[x, y] is a UFD, that means f(x, y) = y/x (see?!). To recapitulate, if φ is an isomorphism, its inverse restricted to the affine set y 2 = x 3 must be given by y/x, which is not defined at the origin, so φ is not an isomorphism. 3. Preliminary remark: Given a variety V defined a field K, remember that unless otherwise stated, when we say that P is a point of V, that means the (projective) coordinates of P are a priori defined over K. If we want to talk about a K-rational point we write P V (K). (a) The coordinate functions of φ are all polynomials, and x q 0 = = x q n = 0 if and only if x 0 = = x n = 0, so φ is a morphism taking P n (F q ) to P n (F q ). And for any homogeneous polynomial f defined over F q, note that φ(f(p )) = φ(f)(φ(p )) = f(φ(p )), so f(p ) = 0 if and only if f(φ(p )) = 0. Since V is defined over F q, that means φ takes V to V, and hence the image of the restriction of the morphism φ to V is contained in V, and we are done. (b) We begin with surjectivity. Fix a point Q := [y 0 : : y n ] V (F q ). Since F q is algebraically closed, for each i the equation X q y i has a solution x i F q. Then the point P := [x 0 : x n ] clearly satisfies φ(p ) = Q. We do need to check that P V ; that follows from the fact that ( f(p ) ) q = (f q )(P q ) = f(q) for any F q -rational polynomial f, so (since V is defined over F q ) P is a solution to any F q -rational defining polynomial for V if and only if Q is a solution. We now turn to injectivity. Fix two points P = [x 0 : : x n ], Q = [y 0 : : y n ] in V (F q ), and suppose that φ(p ) = φ(q). That means we can find non-zero λ F q such that x q i = λyq i for every i; equivalently, (x i /y i ) q = λ whenever one (and hence both) of x i, y i is non-zero. But the equation Z q λ is purely inseparable, i.e. it has just one root this follows easily by induction, using the fact that q is the power of a prime p and that the equation Z p λ is purely inseparable in characteristic p. In any case, denote by τ F q this unique solution; then x i /y i = τ for every i for which one (and hence both) of x i, y i is non-zero. That means P = Q as projective point, and we are done. (c) First, this exercise is false as is stated! Here is a concrete counter-example: Take for V the single point [1 : 0 : : 0]. This is the solution set of x 1 = = x n = 0 and is clearly irreducible, and φ acts as the identity on V, so φ restricted to V is its own inverse! But this is basically the only counter-example; more precisely, we will prove a slightly stronger result: (a) Let V/F q be a variety (necessarily irreducible) of dimension > 0. Then there exists a finite extension F q /F q such that the q -Frobenius is not an automorphism of V//F q.

SOLUTION TO MATH 797N, PROBLEM SET #1. 3 By theory of finite fields we have q = q m for some integer m > 0, so the q -Frobenius is just the q-frobenius composed with itself m times. In particular, the statement quoted above is indeed stronger than the original one (for dim V > 0). And yes, from the definition of dimension (via transcendent degree) it is clear that it is invariant under constant field extension. Remarks. I m pretty sure that there is a more elementary solution than the one above. But in light of the counter-example above, it must make use of the positivity of dim V and thus necessarily involve non-trivial information. To prove (a) we begin with a sequence of reduction. First, a simple but important observation. Since the q-power map is a field automorphism for F q /F q, that means φ takes any F q -rational hyperplane L P n to itself, and hence (b) φ must take L V to itself. We now make use of a non-trivial fact we have not discussed but should at least be intuitively obvious: (c) there exists a hyperplane L P n over some finite extension F q /F q such that L V is irreducible and that dim(l V ) = dim V 1. Results of this type, namely if a variety has a certain property then so do the general hyperplane sections, are commonly known as Bertini s theorem. Of course it is not true without hypotheses; the version we cited above can be derived from, for example, Shafarevich s Basic Algebraic Geometry I, p. 139 (but note that while the statement holds in general, the proof there requires the field to have characteristic 0). Combine (b) and (c) and we see that to prove (a), we are reduced to the case of varieties V of dim V = 1 over some F q. We need one more piece of standard mechinary from algebraic geometry: (d) there exists a finite extension F q /F q over which there exists a point P P n /F q and a 2-dimensional linear subspace T P n /F q such that the projection map P n T from the vertex P, when restricted to V, is birational. From this we see that we are further reduced to the case where V F q. To recap, we have a projective curve V over F q for which the q -Frobenius is an isomorphism for V. Write the projective coordinates of P 2 as [x 0 : x 1 : x 2 ]. Consider the projection from V to each of the coordinate axes x i = 0; each of these is a rational map from V to P 1, corresponding to the inclusion of function fields F q (x i ) F q (V ). These projection maps all have finite degree so F q (V )/F q (x i ) is a finite field extension. If any one of these extensions, say x 1, is also separable, then by Prop. II.4.2(b) we see that dx i is a basis of Ω V where V /F q is a smooth model of V/F q. By hypothesis the Frobenius φ when restricted to V has an inverse ψ, in which case ψ φ = (φψ) = id ΩV. But φ (dx i ) = d(x q i ) = 0, a contradiction. Thus F q (V )/F q (x i ) is not separable for every x i. That means in the projective ideal J that defines V, in every element f J each variable x i must occur to a p-power. By Hilbert s basis theorem, J is finitely generated; pass to finite extension of F q if necessary we can assume that every coefficient of every such generator is also a p-power, in which case every generator is the p-power of a polynomial. But varieties are defined by prime ideals, which are necessarily radical, so we have just reduced the degree of each generator of J by a factor of p. Repeat this process until every generator of J has degree 1, in which case V is just a linear subspace of P 2 of dimension > 0 in other words, a line! Thanks to (b), we can take V = P 1.

4 SIMAN WONG To handle the case of P 1 we play with polynomials. If the q -power map is an isomorphism for P 1, that would mean, upon passing to the affine line A 1, we can find an one-variable rational function f(x)/g(x) such that f(x q )/g(x q ) = x for all but finitely 1 many x F q. ( ) In other words, the polynomial f(x q ) xg(x q ) has infinitely many roots, which is impossible. Thus the Frobenius morphism is not an isomorphism when V = P 1. Remarks. As is written ( ) is an equality of values of rational functions, not (yet) an equality of rational functions. To get from equality of values to equality of functions, in this case we need to invoke facts about finiteness of roots of one-variable polynomials. This as we will see is the source of difficulty of the proof in higher dimension. 4(a) If V p is Q-isomorphic to P 1, then necessarily V p (Q). Clear the denominators and we see that there are integers x, y, z, not all zero, such that (3) x 2 + y 2 = pz 2. Since each variable occurs to the same degree > 1, we can assume that x, y, z are pairwise coprime (see?). We can assume that at least one of x, y is, and hence both are, 0 (mod p); otherwise the other one of x, y is also 0 (mod p), in which case the left side is 0 (mod p 2 ), whence z 0 (mod p), contradicting the pairwise coprimeness. So both x, y 0 (mod p); reduce (3) modulo p and we see that 1 (y/x) 2 (mod p), whence p 1 (mod 4). Conversely, suppose p 1 (mod 4). A classical theorem of Gauss states that p is the sum of two integer squares, in which case V p (Q) has a non-trivial point w p. Using this point w p as the base-point and mimic the sweeping-line construction I presented on the very first lecture (for the unit circle) and we get a rational map φ p from P 1 to V p which is bijective; with a tiny bit more calculation we find that it is in fact an isomorphism. (b) Preliminary comments: Every so often I will assign hard problems problems that require nontrivial work clever ideas, delicate calculations, computer algebra, heavy machines, etc. beyond unwinding the definitions. #4(b) is definitely one such a problem. I assign these to help stimulate your thoughts and ideas, and I have had assigned hard problems which turned out to be really easy. And when you starting doing research, it s not always easy to judge how hard a problem is until you try it. What I m trying to say is that, don t feel bad if you don t get a hard problem; do try to learn from the solution and pickup ideas and tools for further use. Let p, q be distinct, 3 (mod 4) primes. By the law of quadratic reciprocity, we have ( p )( q = 1, q p) so exactly one of these two quadratic symbols is 1. Without loss of generality, say ( q p ) = 1. Suppose V q and V p are Q-isomorphic. Then they are also K-isomorphic for any field K/Q. Take in particular K = Q( p); note that V p is K-isomorphic to P 1, so V q is also K-isomorphic to P 1, whence V q (K) is infinite. In particular, we can find x, y, z K, not all zero, such that (4) x 2 + y 2 = qz 2. 1 The finitely many possible exceptional values are due to the fact that when we write a morphism in terms of rational functions, a priori the denominators of these functions could vanish at some point; to say that we have a morphism is to say that at each of these bad points we can scale the rational functions one scaling for each bad point to remove these vanishing denominators. For any variety these bad points form a proper closed subset; for P 1 that means a finite set.

SOLUTION TO MATH 797N, PROBLEM SET #1. 5 We can further stipulate that the non-trivial solution (x, y, z) to (4) in fact lies in the ring of integers O K of K. Our solution for Part (a) suggests that we reduce this O K -equation modulo q := any 2 prime ideal of O K lying above q. But just like the solution for Part (a), for the rest of the congruence argument to work we need that the algebraic integers x, y, z be pairwise coprime. However, this GCD condition does not make sense in general since O K need not be a PID. To get around this problem, and keeping in mind our ultimate goal of taking reduction modulo q i, we pass to the completion O qi of O K by q i : by basic theory of completion, O K injects into O qi, so the O K -solution to (4) still gives a non-trivial O qi -solution; and O qi is a PID, so starting from the non-trivial O K -solution, we can throw away powers of (a generator of) q i until we arrive at a pairwise coprime O qi -solution for (4). And just like Part (a), this pairwise coprime O qi -solution must satisfy both x, y 0 (mod q i ). Reduce modulo q i and we see that the congruence 1 (x/y) 2 (mod q i ) is solvable over O qi. But by basic facts about completions, O qi /q i O K /q i, so this congruence is in fact solvable in O K. That means q i splits in K( 1)/K, and hence the rational prime q must split completely in K( 1)/Q. But q 3 (mod 4) means that q is inert in the subfield Q( 1)/Q. This is a contradiction, so V p and V q cannot be isomorphic over Q, as desired. Remarks. This is certainly a correct solution, but not necessary a good solution. It seems ad hoc and depends on a lucky fact about sums of two squares. What if we change the V p to (say) 31x 2 + 47y 2 = pz 2? (but we did prove a stronger result, namely V p is not isomorphic to V q to Q( p)). There turns out to be a well-oiled machine for studying varieties which are are isomorphic over Q but not over Q (like out V p ), called the theory of descent in Galois cohomology. It plays a very important role in the theory of elliptic curves (if we have time we will talk a bit about it, in Chapter 10) and in other problems in algebraic geometry. While my solution above is a bit ad hoc, it does make good use of a very versatile technique, namely passing to completion. It s a very useful tool when you want to (and only need to) focus on a single prime ideal; it also gets around the problem of not having a PID and still let you do all the congruence argument you want, two features we took full advantage of above. Keep this in your tool kit. One final thought, for those of you well-versed in the history of mathematics: cohomology descent (as far as I know) has nothing to do with Fermat s infinite descent; the latter turns out to be an important ingredient in the proof of the Mordell-Weil theorem and for which we definitely will discuss, in Chapter 8. Department of Mathematics & Statistics, University of Massachusetts. 01003-9305 USA E-mail address: s i m a n AT m a t h DOT u m a s s DOT e d u Amherst, MA 2 there are two such q since ( q p ) = 1, but we don t need this extra quadratic residue condition until the end.