8
$\begingroup$

I am studying the congruent number problem and I heard that there is a paper by Kazuma Morita which claims to solve this problem from my colleague. I saw the paper on his homepage but it is very short and I cannot belive it is true because it is too short. While I try to find his mistakes, I don't have the knowledge about the Hodge-Tate weight (p,q). Although there is an explanation on the Hodge-Tate weight (p,q), he writes that it is based on the Japanese article by T.Yoshida. I know the notion about Hodge-Tate weight p but not Hodge-Tate weight(p,q) well. So, let me know the litaratures in English on Hodge-Tate weight (p,q).

$\endgroup$

2 Answers 2

-1
$\begingroup$

Objection to nfdc23. Sorry if it is wrong!

His $p$-adic avatar carries the uniformizer to Frob and one gets the equality of L-functions but this is related just by $\chi=\iota\circ \sigma\circ Art$ which is not a continuous character in general because of a field isomorphism $\iota:\bar{Q}_{p}\simeq C$. He looked at only the finite components but not infinite or $p$-adic components. So, I think that his proof is incomplete. The effect of the complex conjugation does not necessarily swap two Galois characters (as a continuous character). On the other hand, as I wrote in the comment, Morita takes care of this issue.

$\endgroup$
2
  • 6
    $\begingroup$ The main content of my answer has nothing at all to do with $\chi$ (which I brought up just to provide my guess as to the context for the motivation of the main idea of the paper). The author defines the $\sigma^{(i)}$'s exactly as I do in my answer, and this does not logically relate to the $\chi$'s for the purpose of the error that I have pointed out. I am unable to understand what you say, but the author's crucial claim that each $\sigma^{(i)}$ has the same HT-weight (0 or 1) are both $p$-adic places of $K$ is wrong, for the reason I have explained. I have nothing more to say on this. $\endgroup$
    – nfdc23
    Commented Jul 18, 2016 at 18:55
  • 1
    $\begingroup$ I think the Tate module $V_{p}(E)$ splits at any prime $p$ and it is an easy exercise that the Hodge-Tate weights are (0,0) and (1,1) at primes $p$ which inert in $K$. $\endgroup$
    – J.S.R.
    Commented Jul 20, 2016 at 13:47
22
$\begingroup$

The fatal error in the reasoning in the paper "L-functions and rational points on a CM elliptic curve via the classical number theory" at the top of the list of papers at

http://kazuma-morita.jimdo.com

occurs near the top of page 3 (when "proving" a statement near the bottom of page 2 that we'll see is false); it comes down to an elementary but unfixable blunder in $p$-adic Hodge theory.

The phrase "weight $(a, b)$" refers to the fact that for a quadratic field $K$ and rational prime $p$ split in $K$, a continuous Galois character $G_K \rightarrow \mathbf{Q}_p^{\times}$ arising from algebraic geometry over $K$ (e.g., inside the $p$-adic Tate module of an abelian variety over $K$, or in the geometric $p$-adic etale cohomology of a smooth projective $K$-scheme) has two Hodge--Tate weights: weight $a$ at one place and weight $b$ at the other (when restricting such a global character to the decomposition groups at the two places of $K$ over $p$). So this indexing is a notion which comes up in the global theory, not the purely local theory.

To explain the context for the error (including what I guess is the author's motivating idea), say $E$ is an elliptic curve over $\mathbf{Q}$ such that for an imaginary quadratic field $K$ we have ${\rm{End}}^0(E_K) = K$ (e.g., the key cases of interest $E_n: y^2 = x^3 - n^2 x$ with $K = \mathbf{Q}(\sqrt{-n})$ for a non-square positive integers $n$).

By the CM theory of elliptic curves, upon choosing an inclusion $\iota:K \rightarrow \mathbf{C}$ ("CM type") we have $L(E/K,s) = L(s, \chi)L(s,\overline{\chi})$ for the algebraic Hecke character $\chi: \mathbf{A}_K^{\times} \rightarrow K^{\times}$ attached to $(E, \iota)$; this $\chi$ is unramified at all good places for $E$ over $K$.

Suppose $p$ is an ordinary good prime for $E$, so $p$ splits in $K$; i.e., $K$ embeds in $\mathbf{Q}_p$ and precomposing with complex conjugation on $K$ swaps the two such embeddings. Let $j:K \rightarrow \mathbf{Q}_p$ be one such embedding (so $x \mapsto j(\overline{x})$ is the other). For a good place $v$ away from $p$, the (unramified) action of ${\rm{Frob}}_v$ on $V_p(E)$ has eigenvalues in $\mathbf{Q}_p$ given by $j(\chi(\pi_v))$ and $j(\overline{\chi(\pi_v)})$ for any local uniformizer $\pi_v$ at $v$ (using suitable class field theory conventions).

The author's idea seems to be to recover the $L$-functions attached to $\chi$ and $\overline{\chi}$ with a well-known (and quite simple) $p$-adic construction in the case of ordinary $p$, and then (the big "idea") prove a strong $p$-adic Hodge theory property of this $p$-adic construction from which striking consequences follow for those $L$-functions (and hence for $L(E/K,s)$). But the proof of the strong $p$-adic Hodge theory property is flawed, and more specifically the asserted property is false.


Here is the well-known $p$-adic construction. Consider the $p$-adic representation $V_p(E)$ of $G_K$ for any prime $p$. This $V_p(E)$ is a free module of rank 1 over $K_p := K \otimes_{\mathbf{Q}} \mathbf{Q}_p$ with $G_K$ acting $K_p$-linearly, so if $p$ is ordinary then the $\mathbf{Q}_p$-algebra decomposition $K_p = \mathbf{Q}_p \times \mathbf{Q}_p$ correspondingly decomposes the $K_p[G_K]$-module $V_p(E)$ as a direct sum of two $G_K$-stable $\mathbf{Q}_p$-lines $V^{(1)}$ and $V^{(2)}$ on which the $G_K$-action is through respective continuous characters $\sigma^{(1)}, \sigma^{(2)}: G_K^{\rm{ab}} \rightrightarrows \mathbf{Q}_p^{\times}$. Up to here there is nothing to intrinsically distinguish these two characters from each other (apart from that each is attached to one of the two factor fields of the $\mathbf{Q}_p$-algebra $K_p$, so each character is attached to one of the places of $K$ over $p$ based on the factor field of $K_p$ "corresponding" to each character).

The author claims near the start of 2.1 that there is an intrinsic way (in terms of $p$-adic Hodge theory) to distinguish $\sigma^{(1)}$ from $\sigma^{(2)}$ in a manner that very different from the elementary bijection defined above between this set of two characters and the set of two places of $K$ over $p$. This proposed alternative way to break the symmetry will turn out to not be compatible with the effect on this pair of characters by the action of complex conjugation on $G_K^{\rm{ab}}$ (or equivalently on $\mathbf{A}_K^{\times}/K^{\times}$). So let's first see why that is really bad news for the author (and then see what the author's incorrect proposed way is), by showing that as characters on $\mathbf{A}_K^{\times}/K^{\times}$ (via global class field theory) these $\sigma^{(i)}$'s are swapped by the effect of complex conjugation (how could it be otherwise?).

Since the effect of complex conjugation on the $\mathbf{Q}_p$-algebra $K_p$ swaps its two factor fields, and the $\sigma^{(i)}$ are defined intrinsically in terms of the $K_p[G_K]$-module $V_p(E)$, this swapping claim amounts to showing that the abelian character $\psi_p:\mathbf{A}_K^{\times}/K^{\times} \rightarrow K_p^{\times}$ encoding the $K_p$-linear $G_K$-action on $V_p(E)$ is equivariant for the effect of complex conjugation. Since $\psi_p$ is the "$p$-adic avatar" of $\chi$ (concretely, at good places $v$ of $K$ the local restriction of $\psi_p$ is unramified and carries a local uniformizer to the element of $K^{\times}$ giving the effect of the Frobenius endomorphism of the reduction of $E$ at $v$), by regarding the complex conjugation isomorphism $K \simeq K$ as an inclusion of number fields (i.e., overlook that the source and target fields happen to be the same) this is just expressing the "general nonsense" fact that the formation of the algebraic Hecke character attached to a CM abelian variety equipped with CM type on the CM field is compatible with ground field extension.


OK, now finally we come to the error. Consider $p$ that is good ordinary for $E$ (hence split in $K$), with places $\{v, v'\}$ over it in $K$. It is an elementary fact via the connected-etale sequence for $p$-divisible groups (or even just finite flat group schemes) over $O_{K_v}$ that among the two local characters $\sigma^{(1)}_v, \sigma^{(2)}_v: G_{K_v} \rightrightarrows \mathbf{Q}_p^{\times}$ exactly one is unramified, so the other is $p$-adic cyclotimic times unramified; the same goes for $v'$. Then one is led to the question: is the unique character among $\{\sigma^{(1)}, \sigma^{(2)}\}$ that is unramified at the place $v$ over $p$ also the unique one that is unramified at the other place $v'$ over $p$? This is exactly what the author is asserting at the start of 2.1 (i.e., that one of these has "Hodge-Tate weights" $(0,0)$ and the other has weights $(1,1)$). But it is wrong!

Indeed, we saw above the combined effect of complex conjugations on $G_K^{\rm{ab}}$ and $K_p$ swap $\sigma^{(1)}$ and $\sigma^{(2)}$, and clearly also swap the decomposition groups at $v$ and $v'$, so for whichever one is unramified at $v$ we see that the other must be unramified at $v'$. In other words, indexing Hodge-Tate weights by the two embeddings of $K$ into $\mathbf{Q}_p$, one of these $\sigma^{(i)}$'s has weights $(0,1)$ and the other has weights $(1,0)$. (Not that it matters, but this labeling of the HT-weights as $(0,1)$ and $(1,0)$ to break the symmetry can be made more explicit: for a given $\sigma^{(i)}$ the place over $p$ at which it is unramified -- i.e., has HT-weight 0 rather than 1 -- is the one corresponding to the factor field of $K_p$ that underlies the initial definition of the character.)

One cannot point to the error in the "proof" of the false statement because in the proof the author is initially analyzing each local place on its own and just boldly asserts without any reason that the weight which occurs for a given $\sigma^{(i)}$ at one $p$-adic place must be the same at the other place. Maybe the author is misled by the equalities $K_v = \mathbf{Q}_p = K_{v'}$ and didn't notice how exchanging these two incarnations of $\mathbf{Q}_p$ requires interacting with complex conjugation and that complex conjugation swaps the two $\sigma^{(i)}$'s in the way that is explained above.

$\endgroup$
9
  • 7
    $\begingroup$ Excellent answer! I hope the author of the paper concerned will read this and take it on board. $\endgroup$ Commented Jul 17, 2016 at 8:51
  • 4
    $\begingroup$ @J.S.R.: It is unclear what your comment is saying or asking. If you are asking whether there is something incompatible between what I have written and the Hodge-Tate decomposition at a given place of the ground field then the answer is "no" (and I gave a proof that the claim near the start of 2.1 of the paper in question is wrong). $\endgroup$
    – nfdc23
    Commented Jul 17, 2016 at 13:03
  • 1
    $\begingroup$ in terms of differential forms, what J.S.R is saying also seems to be correct. $\endgroup$
    – s.jonathan
    Commented Jul 17, 2016 at 14:28
  • 4
    $\begingroup$ @s.jonathan: I don't understand what J.S.R. says. Since you understand it, can you convey what is being claimed by J.S.R.? I'm also unable to understand what "in terms of differential forms" refers to, since the HT decomposition (which admittedly has a relation to differential forms) is not relevant: "(1,0)" and "(0,1)" in my answer are each just notation for indexing the HT-weights of a single global character at two different $p$-adic places where $p$ is split in $K$ (inert $p$ do not fit with this, even thought the HT-decomposition works for all $p$). $\endgroup$
    – nfdc23
    Commented Jul 17, 2016 at 15:21
  • 3
    $\begingroup$ @nfdc23. This is an excellent answer. Thanks. $\endgroup$
    – user87684
    Commented Jul 18, 2016 at 22:46

Not the answer you're looking for? Browse other questions tagged .