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David Speyer commented the following here.

I saw Brian Conrad give an excellent one hour talk to undergraduates where he proved that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$ and $c(t) \in \mathbb{C}[t]$ such that $$a(t)^3 + b(t)^3 = c(t)^3.$$ He gave an elementary proof, then outlined the better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve.

Here is my elementary proof, which I gave here.

Suppose there are some solutions of$$a(t)^3 + b(t)^3 = c(t)^3$$in $\mathbb{C}[t]$. Choose a solution $(a(t), b(t), c(t))$ such that the maximum $m > 0$ of the degrees of $a$, $b$, $c$ is minimal possible among all solutions. Clearly, this choice ensures that $a(t)$, $b(t)$, $c(t)$ are coprime. Then we have$$a(t)^3 = c(t)^3 - b(t)^3 = (c(t) - b(t))(c(t) - \omega b(t)) (c(t) - \omega^2 b(t)),$$where $\omega$ is a third primitive root of unity. Now, we look at the factors $c(t) - b(t)$, $c(t) - \omega b(t)$. Suppose that they have a common factor $q(t)$. Considering their sum and difference, we conclude that $c(t)$ and $b(t)$ have a common factor too. Moreover, $q(t)$ is a factor of $a(t)$. Thus, $a$, $b$, $c$ are not relatively prime, which is a contradiction. Repeating the same game with other pairs of factors, we see that all three factors $c(t) - b(t)$, $c(t) - \omega b(t)$, $c(t) - \omega^2 b(t)$ are pairwise coprime. Therefore,$$c(t) - b(t) = d_1(t)^3,\text{ }c(t) - \omega b(t) = d_2(t)^3,\text{ }c(t) - \omega^2b(t) = d_3(t)^3,\text{ where }d_1,\,d_2,\,d_3 \in \mathbb{C}[t].$$Note that$$\omega^2 + \omega + 1 = 0.$$Multiplying the second equation by $\omega$ and the third equation by $\omega^2$ and adding all three, we arrive at $$d_1(t)^3 + \omega d_2(t)^3 + \omega^2d_3(t)^3 = 0.$$Choosing $\eta_1$ and $\eta_2$ as any third roots of $-\omega$ and $-\omega^2$, respectively, and letting$$a_1 = d_1^3,\text{ }b_1 = \eta_1d_2^3,\text{ }c_1 = \eta_2d_2^3,$$we get$$a_1^3 = b_1^3 + c_1^3.$$By construction, at least one of $a_1$, $b_1$, $c_1$ is a nonconstant polynomial, and the maximum of their degrees is smaller than that of $a$, $b$, $c$. This is a contradiction to the choice of $a$, $b$, $c$.

Here is my better motivated "algebraic geometric" proof, which I gave here.

Let $a(t)$, $b(t)$, $c(t)$ have degree $n_1$, $n_2$, $n_3$, respectively, and $n = \text{max}(n_1, n_2, n_3)$. Then we can define $A(u, v)$, $B(u, v)$, $C(u, v)$ as homogeneous polynomials of degree $n$ such that$$A(u, 1) = a(u),\text{ }B(u, 1) = b(u),\text{ }C(u, 1) = c(u).$$Then, by construction,$$A(u, v)^3 + B(u, v)^3 = C(u, v)^3.$$Let$$E = \{(x, y, z) \in \mathbb{P}^2 : x^3 + y^3 = z^3\}.$$$E$ is a smooth curve of genus $1$, i.e. an elliptic curve. Now, define a map$$\varphi: \mathbb{P}^1 \to E,\text{ }(u, v) \mapsto (A(u, v), B(u, v), C(u, v)).$$This map is well-defined since $A(u, v)$, $B(u, v)$, $C(u, v)$ are homogeneous polynomials of the same degree which do not vanish simultaneously. Moreover, this map is nonconstant and proper since its source is projective. As the image of a proper map is closed and it is not a point, and $E$ is an irreducible $1$-dimensional variety follows that $\varphi$ is a surjective morphism. $E$ is a topologically a torus, i..e it has genus $1$ and there is a one up to scaling differential form $\tau$ on it. This will imply that its pullback $\varphi^*\tau$ is a differential form on $\mathbb{P}^1$, which is impossible since it has genus $0$. In more formal terms, a surjective map $\mathbb{P}^1 \to E$ gives rise to the injection $H^0(E, \Omega^1) \to H^0(\mathbb{P}^1, \Omega^1)$. But this is absurd, since the former is a vector space of dimension $1$ and the latter is a vector space of dimension $0$.

Perhaps you might want to say something about why the pullback of a non-zero holomorphic differential is non-zero.

$X$ and $Y$ are Riemann surfaces and $f: X \to Y$ holomorphic differential form looks like $g(z)\,dz$, where $g$ is a holomorphic function. Then its pullback is locally given by $g(f(z))\,df$. It is obviously nonzero as long as $f$ is nonconstant.

My question is, is there anyone who knows more algebraic geometry than me who can comment on the precise/possibly deep relationship between these two different proofs, i.e. why they are fundamentally the same or different?

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    $\begingroup$ You have seriously misunderstood the EGA reference, and the deduction you claim (that a faithfully flat map cannot induced the zero map via pullback of relative 1-forms) is seriously wrong (faithful flatness is far from sufficient in scheme generality). There is also the matter of aesthetics: invoking EGA to prove a concrete fact about algebraic curves is poor style, as one should understand such things in a more tangible way. This is all undergraduate-level math, not research-level, so it belongs on Math StackExchange, not MO. I recommend to put EGA aside until you know basic things better. $\endgroup$ – nfdc23 Nov 25 '15 at 2:13
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    $\begingroup$ I'm sure you didn't intend to, but your tone comes across rather condescending, nfdc23. While of course corrections are never unwarranted, the question didn't even invoke EGA for the fact (a tangential fact, at that) about curves; it seems to just be an offhand (incorrect) mention of a generalization. And it is conceivable that this question, philosophical as it is, can be interesting at graduate or higher levels despite invoking relatively simple concepts. A friendly stance never hurts! $\endgroup$ – peterx Nov 25 '15 at 8:06
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    $\begingroup$ @peterx: Fair enough. I was recently approached by a high school student who had learned about finite groups and modules over PID's and decided to start "reading" EGA as preparation for Lurie's HTT book, and a freshman with no proof experience who told me his main ambition for math in college is to learn category theory, so it put me in a foul mood to see EGA invoked for non-vanishing of pullback on 1-forms on curves in char. 0. But I shouldn't have allowed that to cloud the tone of my response. Since the OP has excellent faculty at his university to consult with in person, I hope he does so. $\endgroup$ – nfdc23 Nov 25 '15 at 19:43
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    $\begingroup$ It can be helpful to see the difference in the arithmetic side. The first proof reminds me Kummer's proof of FLT when the exponent is a regular prime (or even better, when unique factorization holds). The second proof corresponds to deducing FLT from ABC in the arithmetic side (R-H formula for function fields and the truncated second main theorem for meromorphic functions are usually seen as analogues of ABC). One approach works in many cases over Z, the other remains incomplete as far as I know. $\endgroup$ – Pasten Nov 25 '15 at 20:33
  • $\begingroup$ @OP: The fact quoted in EGA cannot even by applied to the case of $\Bbb{P}^1 \to E$ above. You are taking $\mathcal{G} = \Omega^1_E = \mathcal{O}_E$. Then the pullback is $\mathcal{O}_{\Bbb{P}^1}$ which is not the sheaf of Kahler differentials on $\Bbb{P}^1$ ($\Omega^1_{\Bbb{P}^1} = \mathcal{O}_{\Bbb{P}^1}(-2)$). $\endgroup$ – Ben Lim Nov 27 '15 at 5:06
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This answer is basically a longer version of Felipe Voloch's, but maybe it will be useful. Both proofs take a class in $H^1(E)$, pull it back to $H^1(\mathbb{P}^1)$ and note that $H^1(\mathbb{P}^1)$ is zero to conclude that the map $\mathbb{P}^1 \to E$ was constant. The difference is what form of $H^1$ we work with.

The topological version is that a continuous map from a sphere to a torus must have degree zero; then holomorphic maps of degree zero are constant.

Proof with differentials We take the class $dy/x$ in $H^0(E, \Omega^1)$ and pull it back to $H^0(\mathbb{P}^1, \Omega^1)=0$. This gives some equations on derivatives which imply that we are pulling back by the zero map. By Hodge theory, for any curve $X$, the sheaf cohomology $H^0(X, \Omega^1)$ is half of $H^1(X, \mathbb{C})$.

The proof by descent The map $$ \phi(x:y:z) = (x+\omega y + \omega^2 z : x+\omega^2 y + \omega z : - 3 x y z)$$ is a degree $3$ unbranched cover of $E$, giving a class in $H^1(E, \mathbb{Z}/3)$. If we pull this cover back to $\mathbb{P}^1$, then it must become trivial. Your computations are explicitly showing that, if $a(t)^3+b(t)^3+c(t)^3=0$, then we can find some $x(t)$, $y(t)$, $z(t)$ such that $(a:b:c)=\phi(x:y:z)$. So you have shown that this $\mathbb{Z}/3$ cover has a section, and is thus trivial. Abstractly, the reason there must be a section is that $H^1(\mathbb{P}^1, \mathbb{Z}/3)=0$.

Here is an abstract proof that $H^1(\mathbb{P}^1, \mathbb{Z}/3)=0$. Look at the short exact sequence $1 \to \mu_3 \to \mathcal{O}^{\ast} \to \mathcal{O}^{\ast} \to 1$, where $\mu_3$ is the group of cube roots of unity, $\mathcal{O}^{\ast}$ is the sheaf of nonvanishing holomorphic functions and the second map is cubing. (The topology is either analytic or etale.) Looking at the long exact sequence in cohomology, we see that $H^1(\mathbb{P}^1, \mu_3)$ is the $3$-torsion in $Pic(\mathbb{P}^1)$. But $\mathbb{C}[t]$ is a PID, so $Pic(\mathbb{A}^1)=0$ and $Pic(\mathbb{P}^1) \cong \mathbb{Z}$. You wind up using the exact same fact -- that $\mathbb{C}[t]$ is a PID -- when you write down the proof in an elementary way.

I remark that the first proof using a "de Rham" description of $H^1$ (by differential forms) and the second proof uses a "Betti" description (by covering maps). Relating de Rham and Betti descriptions is always complicated.

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I more or less gave your geometric proof more generally for solutions of $$aX^n+bY^m=c$$ over 1-dim'l function fields (and subsequently adapted it to the function field ABC theorem). The idea is that one constructs a map between algebraic curves (or, if you prefer, Riemann surfaces) and uses the Riemann-Hurwitz genus formula, where of course, Riemann-Hurwitz is really a statement about the degrees of differential forms after pull-back. The two articles are:

Joseph H. Silverman, The Catalan equation over function fields, Trans. Amer. Math. Soc. 273 (1982), no. 1, 201--205.

Joseph H. Silverman, The $S$-unit equation over function fields, Math. Proc. Cambridge Philos. Soc. 95 (1984), no. 1, 3--4.

(Wow,that new "Insert Citation" button is great!)

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    $\begingroup$ @KevinDong RIght, wasn't meant as an answer to your specific question, but thought you might be interested in seeing how your second version generalizes. So maybe it should have been a comment. $\endgroup$ – Joe Silverman Nov 24 '15 at 21:49
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I find it highly unlikely that the two proofs are really thematically linked in any way. The first, being an infinite descent proof, is in some sense doing geometry "over $\mathbb{C}[t]$" - in that one could conceivably rephrase it into language involving intersection theory and heights of points in projective space over $\text{Spec }\mathbb{C}[t]$. In the second, you are clearly doing geometry over $\mathbb{C}$, taking $\mathbb{C}[t]$ as a coordinate ring object instead of as a base.

In particular, note that the second method has the natural geometric generalizations as noted in Joe Silverman's answer by extending the "$a,b,c$ parametrize a map between curves" idea, while the former is much more ad hoc because the "central idea" is nothing more than particular intersection/congruence type properties of the sum of cubes which feel very fragile.

Mason-Stothers also gives an effective proof here which has a different flavor than both of these; perhaps it is thematically a little closer to the ad hoc proof because it implicitly takes the idea of "working over a function field" seriously.

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As mentioned by others, the first proof uses a 3-descent, namely the map $E(K)/3E(K) \to H^1(K,E[3])$ (where $E$ is the elliptic curve and $K = \mathbb{C}(t)$) obtained by taking (Galois) cohomology of the exact sequence $0 \to E[3] \to E \to E \to 0$. This is standard and covered, e.g. in Joe Silverman's book. Here $H^1(K,E[3]) = K^*/(K^*)^3 \times K^*/(K^*)^3$ and the map is given by $(a:b:c) \mapsto (b-c,b-\omega c)$ or something similar, which connects with the first proof (all points are divisible by three, hence infinitely divisible by three, hence torsion).

The second proof uses a map $E(K) \to K$ which goes by identifying $K$ with the function field of a curve $X$ and a point in $E(K)$ as a map $f: X \to E$ and mapping such a point to $f^*(dx/y)$. I think one can describe this map cohomologically by considering the exact sequence $0 \to \pi_1(E) \to \mathbb{C} \to E \to 0$ view that as coherent sheaves over $X$ and getting a map $E(K) \to H^1(X,\pi_1(E))$. Now use $\pi_1(E) = H_1(E,\mathbb{Z}) \subset H^1(E,\mathbb{C})$ and the fact it's a constant sheaf to identify the two maps. I am missing many details. I've thought about this a long time ago and I may be misremembering something.

Finally $\pi_1(E)/3\pi_1(E) = E[3]$ which connects the two approaches.

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