Persiflage

BSD and the p-converse (Guest Post: Ashay Burungale)

Posted on October 15, 2026 by Persiflage

(Here is the first of a series of guest blog posts by number theorists on the recent OpenAI results, this one by Ashay Burungale.)

Let $E/\mathbf{Q}$ be an elliptic curve. Its rational points form a finitely generated abelian group. The Birch–Swinnerton-Dyer (BSD) conjecture, formulated in the 1960s, predicts that its rank equals the order of vanishing of the Hasse–Weil $L$-function $L(E,s)$ at $s=1$ (analytic rank). It also predicts finiteness of the Tate–Shafarevich group $\text{Ш}(E/\mathbf{Q})$ and expresses the leading term of $L(E,s)$ at $s=1$ in terms of $\#\text{Ш}(E/\mathbf{Q})$.

The work of Gross–Zagier and Kolyvagin in the 1980s, together with modularity, gives a fundamental result towards BSD: if the analytic rank is zero or one, then the aforementioned rank equality and finiteness hold.

The announcements

For a prime $p$, recall the Kummer exact sequence

0\longrightarrow E(\mathbf{Q})\otimes\mathbf{Q}_p/\mathbf{Z}_p \longrightarrow\mathop{\mathrm{Sel}}\nolimits_{p^\infty}(E/\mathbf{Q}) \longrightarrow\text{Ш}(E/\mathbf{Q})[p^\infty]\longrightarrow0.

Put $s_p(E)=\mathop{\mathrm{corank}}\nolimits_{\mathbf{Z}_p}\mathop{\mathrm{Sel}}\nolimits_{p^\infty}(E/\mathbf{Q})$.

A $p$-converse to the Gross–Zagier and Kolyvagin theorem seeks to recover analytic rank from this Selmer corank.

The first converse paper [OA1] announces the following for every $E/\mathbf{Q}$ and every prime $p$, treating odd $p$ and importing the case $p=2$ from a companion paper:

s_p(E)=r\in\{0,1\}\quad\Longrightarrow\quad \mathop{\mathrm{ord}}\nolimits_{s=1}L(E,s)=\mathop{\mathrm{rank}}\nolimits E(\mathbf{Q})=r, \qquad \#\text{Ш}(E/\mathbf{Q})<\infty. \tag{1}

The following is my account of the proposed odd-prime strategy. The correctness of the announced proofs is still being analysed; I intend to report on this separately.

The companion [2-converse] also announces Goldfeld’s density conjecture. The leading-term paper [BSD1] treats the odd-primary components of the BSD formula and imports the $2$-primary component from [BSD2].

The Goldfeld paper states that $50\%$ of the quadratic twists $E^{(d)}$ of each $E/\mathbf{Q}$ have analytic rank zero and $50\%$ have analytic rank one, with non-zero squarefree $d$ ordered by $|d|$. Smith’s theorem [Selmer-distribution] gives these proportions for $s_2(E^{(d)})$. The announced $2$-converse and leading-term results would yield full BSD for a density-one set of twists of every $E/\mathbf{Q}$ (see also [rank-zero-converse]).

Another application concerns Sylvester’s 1879 conjecture: every prime $\ell\equiv4,7,8\pmod9$ is a sum of two rational cubes. The recent proofs by Yin [Sylvester1] for $4,7\pmod9$ and Burungale–Tian [Sylvester2] for $8\pmod9$ explicitly construct such cube-sum expressions. For $E_\ell:X^3+Y^3=\ell Z^3$, we have $s_3(E_\ell)=1$, and so the announced $3$-converse would also give a proof of Sylvester’s conjecture, albeit inexplicitly.

Prior work

The above question has been studied since the mid-1980s. The $p$-converses are now largely known for odd primes $p$ of good reduction. The survey [BSD-survey], Sections 3–4, describes some of these developments.

In rank zero, Rubin’s main-conjecture methods give CM converses (see also [rank-zero-converse]). For general elliptic curves, Kato’s Euler system [Euler-system] gives one divisibility in the cyclotomic main conjecture, while Skinner–Urban’s Eisenstein congruences [main-conjecture] give the reverse divisibility for ordinary primes under their hypotheses, leading to rank-zero converses and cases of the BSD formula (see also [supersingular]).

The first general rank-one converses grew out of the work of Skinner [converse1] and Zhang [converse2] around 2012. Their ideas rely on Iwasawa theory, albeit more prominently in Skinner’s approach. In view of the Gross–Zagier formula, the goal is to show that a suitable Heegner point $y\in E(K)$ has infinite order, where $K$ is a well-chosen imaginary quadratic field.

In Skinner’s approach, suppose $p=v\bar v$ splits in $K$ and put $V=T_pE\otimes_{\mathbf{Z}_p}\mathbf{Q}_p$. Write $H_f^1(K,V)$ for the associated Bloch–Kato Selmer group, defined here by the local Kummer images. Put

S_v=\{x\in H^1(K,V):\mathop{\mathrm{loc}}\nolimits_{\bar v}x=0,\quad \mathop{\mathrm{loc}}\nolimits_wx\in H_f^1(K_w,V)\ (w\not\mid p)\}.

The Bertolini–Darmon–Prasanna (BDP) formula [BDP] relates the trivial-character value of an anticyclotomic $p$-adic $L$-function to the Heegner logarithm:

L_p(E/K,\mathbf{1})=C\bigl(\log_v y\bigr)^2,\qquad C\ne0.

Let $X_v$ be the Pontryagin dual of the corresponding discrete Selmer group over the anticyclotomic $\mathbf{Z}_p$-extension $K_\infty/K$. Under suitable hypotheses at an odd prime $p$ of good reduction, Wan’s divisibility [divisibility] gives

(L_p(E/K))\mid\mathop{\mathrm{char}}\nolimits_\Lambda(X_v), \qquad\Lambda=\mathbf{Z}_p[[\mathop{\mathrm{Gal}}\nolimits(K_\infty/K)]].

Indeed, Wan constructs congruences between Klingen Eisenstein series and cusp forms on $\mathrm{GU}(3,1)$, then extracts Galois extensions. If $H_f^1(K,V)=\mathbf{Q}_p a$ and $\mathop{\mathrm{loc}}\nolimits_v(a)\ne0$, Poitou–Tate duality gives $S_v=0$. Specialization control then makes $\mathop{\mathrm{char}}\nolimits_\Lambda(X_v)$ non-zero at the trivial character, and the above divisibility and the BDP formula imply that $\log_v y\ne0$. Hence, $y$ is non-torsion.

To get around the possible vanishing of $\mathop{\mathrm{loc}}\nolimits_v(a)$, a key idea is to employ anticyclotomic Iwasawa theory of norm-compatible Heegner points, together with Wan’s divisibility. See [Heegner-main-conjecture1], [CM-converse], and [Heegner-main-conjecture2], as well as the supersingular [supersingular-converse] and Eisenstein [Eisenstein-converse] cases. This approach requires control of the Bloch–Kato local conditions at $p$ along $K_\infty$, which is not yet available in general at primes of additive reduction. This has been the principal obstacle to establishing the rank-one $p$-converse at primes of additive reduction.

The Sylvester strategy

The Sylvester proof [Sylvester2] for primes $\ell\equiv8\pmod9$ is not Iwasawa-theoretic. An auxiliary prime $q$ split in $\mathbf{Q}(\omega)$ is chosen, with $\omega$ a primitive cube root of unity. It provides a point $Z$ in a CM orbit on an elliptic curve $A$, with vanishing trace under a cyclic group $C$. The idea is to divide before taking the trace: writing $N_C=\sum_{g\in C}g$ and $\lambda=1-\omega$, one chooses $[\lambda]R=Z$ and proves

N_C Z=0,\qquad N_C R\in A[\lambda]\setminus\{0\}.

Frobenius at $\ell$ shows that the division boundary $N_C(R)$ is nonzero. This non-vanishing implies that a suitable cubic component of the CM orbit is non-torsion, giving the required cube-sum expression.

The rank-one strategy in OA1

We illustrate the rank-one case for odd $p$. Rather than varying along an anticyclotomic extension at $p$, a new idea of OA1 is to employ ring-class characters ramified at auxiliary primes away from $p$. These primes are chosen to obstruct lifts of Selmer classes. Theta lifts connect weighted logarithms of Heegner points to congruences with cusp forms, whose Galois representations give a lower bound for the Selmer group.

Assume $s_p(E)=1$, and let $N$ be the conductor of $E$. Choose $K$ so that every prime dividing $2Np$ splits in $K$ and $E^{(K)}$ has analytic rank zero. Then $H_f^1(K,V)=\mathbf{Q}_p a$. The aim is to prove that the conductor-one Heegner trace $y\in E(K)$ has infinite order. Poitou–Tate duality gives

\dim_{\mathbf{Q}_p}S_v= \begin{cases} 0,&\mathop{\mathrm{loc}}\nolimits_v(a)\ne0,\\ 2,&\mathop{\mathrm{loc}}\nolimits_v(a)=0. \end{cases} \tag{2}

Tame deformation. Choose primes $q_i\not\mid Np$ split in $K$, whose ring-class groups (of conductor $q_i$) admit cyclic quotients $\Delta_i=\langle X_i\rangle$ of order $p^{n_i}$, with $n_i\to\infty$; see OA1, Proposition 3.2. Write $H_{q_i}$ for the corresponding ring-class fields. Let $\tau_i:G_K\to\mathbf{Z}/p^{n_i}\mathbf{Z}$ be the quotient character, where $G_K=\mathop{\mathrm{Gal}}\nolimits(\overline K/K)$. It is chosen to vanish on decomposition groups above $p$ and at bad primes. Put $\nu_i(g)=X_i^{\tau_i(g)}$ and $t=X_i-1$, the augmentation parameter at the trivial character.

For a finite extension $k/\mathbf{Q}_p$ with ring of integers $\mathcal O$, the finite-stage substitution

\mathcal O[\Delta_i]\longrightarrow (\mathcal O/p^{n_i})[t]/(t^3),\qquad X_i\longmapsto1+t \tag{3}

is valid for odd $p$, since $p^{n_i}\mid\binom{p^{n_i}}2$. On a Galois-stable lattice the twisted action is $g\mapsto\rho_E(g)(1+t)^{\tau_i(g)}$.

For $b=1,2,3$, keep only the first $b$ coefficients in $t$. OA1 takes coefficientwise limits of the finite-stage cocycles on sequences of Galois elements, using a fixed ultrafilter and uniformly bounded denominators. The resulting coefficient ring is $R_b=k[t]/(t^b)$.

Let $S_b$ be the modified Selmer module, relaxed at $v$, strict at $\bar v$, and allowing auxiliary ramification. Note that $S_1=S_v\otimes_{\mathbf{Q}_p}k$. For $b=2,3$, the specialization property in OA1 identifies the kernel of $S_b\to S_1$ with the image of $t:S_{b-1}\to S_b$.

The congruence bound. The goal is $\dim_k S_b\ge b$ when the first $b$ weighted logarithmic coefficients below tend to zero. For a Heegner point $y_{q_i}\in E(H_{q_i})$ of conductor $q_i$ and a fixed $p$-adic embedding above $v$, define

\mathscr L_i(t)=\sum_{g\in\mathop{\mathrm{Gal}}\nolimits(H_{q_i}/K)} (1+t)^{-r_i(g)}\log_v(y_{q_i}^{\,g}) =\sum_{j=0}^{b-1}\ell_{i,j}t^j\pmod{t^b}, \tag{4}

where $r_i(g)$ represents $\tau_i(g)$ in $\{0,\ldots,p^{n_i}-1\}$. At $t=0$, the Hecke trace relation gives $\ell_{i,0}=(a_{q_i}-2)\log_v y$, with $a_q=q+1-\#E(\mathbf{F}_q)$.

For the theta construction (OA1, §4), OA1 fixes a real quadratic field $F$, using the Goldfeld announcement, so that the additional quadratic-twist summand of the modified Selmer group over $L=KF$ vanishes. The auxiliary primes $q_i$ are chosen to split also in $F$. Let $A$ be a two-dimensional Hermitian space over $L$, definite at both real places, and $W$ a four-dimensional skew-Hermitian space whose unitary group has signature $(3,1)$ at both places. Base change and Jacquet–Langlands transfer attach to $E$ a fixed representation $\pi$ of $U(A)$. Fix a vector $\varphi\in\pi^\vee$. The associated theta family on $U(W)$ has the form

\Theta_{i,m}(g)=\int_{[U(A)]} \theta_{\Phi_{i,m}}(g,h)\,\varphi(h)\, (\mathcal C^m\nu_i)^{-1}(\det h)\,dh, \qquad [U(A)]=U(A)(F)\backslash U(A)(\mathbf A_F).

Here $\theta_{\Phi_{i,m}}$ is the theta kernel determined by a Schwartz function $\Phi_{i,m}$, and $\mathcal C$ is a fixed algebraic character. The index $i$ refers to the tame character $\nu_i$ and the local Schwartz function at $q_i$ (the other finite local data stay fixed). The character $\nu_i$ is pulled back to the norm-one torus of $L/F$ through $N_{L/K}$. The integer $m$ changes the algebraic twist and the archimedean Schwartz function, hence the weight.

Cuspidality requires vanishing of the constant Fourier–Jacobi coefficient at every boundary component. Each is a theta lift to a definite rank-two unitary group. With local intertwiners fixed, these lifts are controlled by one global invariant $S_{i,m}$. The Rallis and Waldspurger formulas give the period relation

A_{i,m}S_{i,m}^2=B_{i,m}P_{0,i,m}^2P_{1,i,m}^2

in the tame group algebra (OA1, Lemma 5.2). Here $A_{i,m},B_{i,m}$ collect the local factors, while $P_{0,i,m},P_{1,i,m}$ are toric periods for $E$ and an auxiliary quadratic twist. For $m_i\to\infty$ with $m_i\to0$ $p$-adically, OA1 relates $P_{0,i,m_i}$ to $\mathscr L_i(t)$ by a BDP-type formula.

Integral control and cuspidal congruences. In OA1, $\ell_{i,j}\to0$ for $j<b$ implies that every constant Fourier–Jacobi coefficient of $\Theta_i=\Theta_{i,m_i}$ tends to zero modulo $t^b$ (§5, Proposition 5.1). The integral estimates are uniform in conductor and weight: bounded denominators for the periods and $S_{i,m}$, bounded valuation of $A_{i,m}(0)\ne0$, and uniform bounds for the local maps expressing boundary restrictions in terms of $S_{i,m}$ (§4, Lemma 4.5 and Lemma 4.6). At the same time, a positive-index Fourier–Jacobi coefficient has valuation bounded above at $t=0$ in the same integral lattice (§6, Proposition 6.5). Thus the whole form does not vanish along with its boundary. Multiplication by a lift $h_i$ of a power of the Hasse invariant on the ordinary locus, followed by lifting to higher weight, gives (§7)

h_i\Theta_i\equiv f_i\pmod{p^{M_i},\ t^b}, \qquad M_i\to\infty,\quad f_i\text{ a classical cusp form on }U(W). \tag{5}

The surviving coefficient is used to preserve the cuspidal Hecke congruence over all of $R_b$.

The associated four-dimensional Galois representations are compared with a twist of $V$ and two characters. An off-diagonal $R_b$-module $D_b$ records extensions of one character by the $V$-block. Its dual gives classes satisfying the modified Selmer conditions, including splitting at $\bar v$, and the resulting bound is (§9, Proposition 9.7)

\ell_{i,j}\longrightarrow0\ (j<b) \quad\Longrightarrow\quad \mathop{\mathrm{Fitt}}\nolimits_{0,R_b}(D_b)=0,\qquad \mathop{\mathrm{Hom}}\nolimits_{R_b}(D_b,R_b)\hookrightarrow S_b, \qquad \dim_k S_b\ge b. \tag{6}

Suppose now, for a contradiction, that $y$ is torsion. If $\mathop{\mathrm{loc}}\nolimits_v(a)\ne0$, the trace relation gives $\ell_{i,0}=0$. Then (6) with $b=1$ contradicts $S_1=0$ from (2). It remains to treat the case of zero localization.

Zero localization and auxiliary primes. Suppose $\mathop{\mathrm{loc}}\nolimits_v(a)=0$, so $\dim_k S_1=2$. Here $E(\mathbf{Q})$ is finite, since a non-torsion rational point would have nonzero local logarithm. Thus the Selmer corank comes entirely from $\text{Ш}(E/\mathbf{Q})[p^\infty]$. The aim is to prove $\dim_k S_3\le2$.

A first-order lift of $x\in S_1$ is an element of $S_2$ reducing to $x$ at $t=0$. Global reciprocity gives two linear conditions on $x$ that must hold for such a lift to exist. In a basis $(a,h)$ of $S_1$, write $x=\alpha a+\beta h$. OA1 chooses the limiting auxiliary Frobenius $\gamma$ so that these conditions are

H_\gamma\binom{\alpha}{\beta}=0,\qquad H_\gamma=\begin{pmatrix}0&c_\gamma\\-c_\gamma&0\end{pmatrix}, \qquad c_\gamma\ne0. \tag{7}

A commutator calculation and Chebotarev supply such auxiliary primes (OA1, Lemmas 3.5–3.6). Since $H_\gamma$ is invertible, only $x=0$ can lift. Thus $S_2\to S_1$ is the zero map, as is $S_3\to S_1$, since it factors through $S_2$. Iterating the specialization property gives

S_b=\mathop{\mathrm{im}}\nolimits(t^{b-1}:S_1\longrightarrow S_b), \qquad tS_b=0,\qquad \dim_k S_b\le2 \quad(b=1,2,3). \tag{8}

Division of the Heegner class. To obtain the opposite bound $\dim_k S_3\ge3$ from (6), it suffices to show $\ell_{i,0},\ell_{i,1},\ell_{i,2}\to0$. OA1 does this by dividing the weighted Heegner Kummer class twice (OA1, Proposition 3.9). Its limit $\mathscr K_3$ over $R_3$ satisfies Kummer conditions at both primes over $p$.

The torsion-ness of $y$ gives $\mathscr K_3(0)=0$, hence $\mathscr K_3=t\mathscr K’$ with $\mathscr K’$ over $R_2$. The Kummer conditions place $\mathscr K'(0)$ in $H_f^1(K,V)\otimes k=ka$, which is fixed by complex conjugation: the negative eigenspace comes from the analytic-rank-zero twist $E^{(K)}$ and vanishes. On the other hand, the Heegner conjugation relation has sign $-w(E)=+1$ by $p$-parity, while conjugation sends $t$ to $-t$ modulo $t^2$. Comparing linear coefficients shows that $\mathscr K'(0)$ is anti-invariant, so it must vanish. Dividing once more gives

\mathscr K_3=t^2 z,\qquad z\in H_f^1(K,V)\otimes_{\mathbf{Q}_p}k=ka. \tag{9}

Since $z\in ka$ and $\mathop{\mathrm{loc}}\nolimits_v(a)=0$, the localization of $\mathscr K_3$ at $v$ vanishes through degree two. OA1’s uniform Kummer-to-logarithm estimate (Lemma 3.8) gives $\ell_{i,0},\ell_{i,1},\ell_{i,2}\to0$. Equation (6) then contradicts (8):

3\le\dim_k S_3\le2. \tag{10}

Thus $y$ is non-torsion, and Gross–Zagier and Kolyvagin give the rank-one conclusion in (1).

Relation to earlier work. The first half of the argument may be viewed as a tame analogue of Skinner’s converse strategy: a BDP-type formula links the vanishing of a Heegner logarithm to an automorphic congruence, from which one extracts nonzero Selmer classes. The passage from congruences to Galois extensions follows the approach of Skinner–Urban and Wan, but the variation here is through auxiliary ring-class characters away from $p$. (For an interesting precursor of tame congruences, see [Brumer–Stark].)

A key novelty is to combine this tame congruence method with division of the Heegner class. As in Sylvester, division recovers information invisible in the ordinary trace. In OA1, the torsion-ness assumption permits two $t$-divisions; conjugation and zero localization then give three logarithmic coefficients tending to zero. The BDP-type formula connects this vanishing to the theta congruence, giving the lower bound $\dim_k S_3\ge3$, while the auxiliary Frobenius conditions obstruct Selmer lifts and give $\dim_k S_3\le2$.

Acknowledgements. The author thanks Frank Calegari, Marco Sangiovanni Vincentelli, Chris Skinner and Ye Tian for helpful discussions.

Posted in Guest Post, Kartik Prasanna., Mathematics | 2 Comments

A Brief Digest of Some New OpenAI Results

Posted on October 8, 2026 by Persiflage

I thought it might be worth discussing some of the recent OpenAI results in my field, or at least in algebraic number theory. For each of them, I’d like to offer a few thoughts, if I have any, on what seems to be new and what constitutes the breakthrough. My hands are currently shot with RSI, so I am narrating this to ChatGPT and asking to convert it to html, so let’s see how that goes

Added: I heard that OpenAI has already withdrawn some of their papers on (partial results towards) the Hodge Conjecture due to mistakes being found. Apparently they found a sign error! That may be the best evidence of sentience yet!

Modularity of elliptic curves over imaginary quadratic fields.

This problem was almost already solved by Caraiani and Newton in their paper. They prove the modularity of elliptic curves over any imaginary quadratic field $F$ such that the elliptic curve $X_0(15)$ has rank zero over $F$. In particular, this applies to a positive proportion of imaginary quadratic fields.

The problem is what to do with elliptic curves which, for example, have reducible mod $3$ and mod $5$ representations. This is the problem that OpenAI overcame, by what I think is a very nice trick.

As already indicated in the comments on my last post, Bao Le Hung has understood this trick and has posted a corresponding paper on Hexagon Math. This argument, as Bao points out, seems quite robust, and is already enough to prove modularity of elliptic curves over all totally real fields and all CM fields not containing $\zeta_5$ (and no doubt more tricks could be applied to that case as well).

I view this result as definitely something a human could come up with. It’s a really nice trick rather than a fundamental breakthrough. I’m sure variations of it will be useful in other ways, and this is definitely worth exploring.
Fontaine–Mazur modularity at the prime $2$.

It’s a little bit hard for me to see exactly what the new idea is in this 70-plus-page paper. Superficially, I’m not sure I find this particularly surprising in light of recent results of Lue Pan and others over the past few years. My first thought is that the main difficulties at $p=2$ are technical. At the same time, I wouldn’t exactly say there’s a lot of money in proving this particular result, and so the few people most capable of possibly proving it have probably been thinking about other things. It is very clean, which is useful for applications.

That said, upon closer inspection this paper may turn out to have some really clever new trick, but I can’t see what that is on a superficial reading. I would actually say I’m surprised that this is a problem OpenAI would have tried to solve, except for the fact that it gets used as an input for Hilbert’s tenth problem.
Hilbert’s tenth problem over the rational numbers.

This one threw me for a loop. I remember that a number of people, perhaps starting with Mazur and others, thought about this problem in the context of Hilbert’s tenth problem over $\mathbf{Z}$. In particular, one of the issues became whether $\mathbf{Z}$ was existentially definable over $\mathbf{Q}$. All the talks I saw on this problem seemed to suggest that the answer to this was negative.

As for Hilbert’s tenth problem itself, it seemed plausible to me that it was actually solvable over the rational numbers. I don’t think I would have put money on that, however. I still certainly hope that Hilbert’s tenth problem has a positive answer over the rational numbers for algebraic curves, and I think that this should be a consequence of conjectures that we believe.

I hope to be able to encourage someone I know to write a guest blog post on this paper at some point. In fact, that might be true for a number of the other papers as well.
Artin’s conjecture on primitive roots.

When I first saw this, my impression was: well, of course Hooley’s argument should work if you knew the quasi-Riemann hypothesis for Dedekind $L$-functions. Except that OpenAI doesn’t prove the quasi-Riemann hypothesis for all Dedekind $L$-functions, just for Dirichlet $L$-functions. So instead, the argument is a little bit different, but also not at all surprising.

The cheapest way to try to prove Artin’s conjecture is as follows. For example, let’s concentrate on $2$ as the element one wants to be a primitive root. Take primes $p$ of the form $8k+3$. For each such prime, $2$ will be a quadratic nonresidue. But now let’s imagine that $p-1=2q$, where $q$ is prime. If $2$ is not a quadratic residue, the only way it could fail to be a primitive root is if it were a $q$-th power, in which case $2^2\equiv 1 \pmod p$, which is ridiculous unless $p=3$.

So the “only thing” you need to prove is the existence of infinitely many pairs of primes $p$ and $q$ with $p-1=2q$ and $p\equiv 3\pmod 8$. Of course, this immediately returns you to another quite difficult problem. On the other hand, this is not so far in spirit from the work on bounded gaps between primes by Zhang, Maynard, and others.

Instead, we could try to write $p-1=2aq$, where now, instead of insisting that $a=1$, we require that all the prime factors of $a$ are reasonably large, for example of size at least $\exp((\log p)^\delta)$ for some positive $\delta$. The exact details of the size restrictions will be a combination of what one needs in Hooley’s argument and what one can actually prove. Here $q$ is some large prime, itself of size at least some fixed power of $p$. I think it might literally be larger than $p^{9/10}$ in the actual paper.

The point is that because all the prime factors of $a$ are large, one doesn’t have to use GRH for Dirichlet $L$-functions to eliminate the possibility that $2$ is a cube, a fifth power, a seventh power, and so on. One has reduced the problem to one of the other regimes in Hooley’s argument, where one doesn’t need the Riemann hypothesis, but instead can use elementary sieving arguments.

Of course, one is now left with the difficult problem of finding primes $p=2aq+1$, where $a$ is not necessarily $1$, but has prime factors of some restrictive shape. How does one do that? Understanding the prime divisors of $p-1$ and finding primes of this form is certainly not easy. But this is where one is really choosing primes satisfying various congruences, and for this one can use the quasi-Riemann hypothesis for Dirichlet $L$-functions rather than for Dedekind $L$-functions. That, at least, is my first guess as to what is going on.

So again, this is an interesting type of argument which I don’t find particularly surprising. But no one was going to write this paper, which depends on the quasi-Riemann hypothesis for Dirichlet $L$-functions, without knowing anything about general Dedekind $L$-functions. And this brings us to the next problem.
The quasi-Riemann hypothesis.

WTF.

I think if you had asked me what was more likely, that AI would prove the Riemann hypothesis or that it would prove the quasi-Riemann hypothesis, I might honestly have thought that the first was more likely.

My belief, and I suspect this is shared by a certain number of other people, is that the actual proof of the Riemann hypothesis is not going to be an analytic proof at all, but rather something much more arithmetic, involving ideas that we don’t necessarily understand yet. I think that in many situations in number theory, analytic techniques are not the right way of thinking about the problem. They may give the first nontrivial results, but are ultimately insufficient for the final proof.

One example here is the Ramanujan conjecture concerning the size of the coefficients of $\Delta$. There is a bound of order $O(p^{6})$ first proved by Hardy. A significant improvement came from Rankin $O(p^{6-1/5})$, using what we would now call Rankin’s method, involving the symmetric square in automorphic terms, although Rankin certainly didn’t think about it this way. I also just learnt right now that Kloosterman had also improved Hardy’s result to $O(p^{6-1/8})$. These were analytic approaches. Of course, Langlands understood that one could also prove the Ramanujan bound from functoriality, but I don’t think of functoriality as a problem in analytic number theory either.

When it comes to the Riemann zeta function, there are many results going back a long time concerning zero-free regions. Perhaps my point of view was that these represented something close to the limits of analytic arguments, and that one would need a more algebraic perspective to make a truly significant improvement. Obviously, I was totally wrong, and something was clearly missed. But I have no idea of the structure of what is going on. I still believe, although perhaps with slightly less conviction, that improvements to these analytic arguments are not going to be sufficient to prove the full Riemann hypothesis.

Another thing that would be surprising is the idea that one might be able to prove the Riemann hypothesis for $\zeta(s)$ without proving the corresponding hypothesis for Dirichlet $L$-functions.

This surprising result puts us in an unusual situation. We now have a very strong intermediate result that no one ever imagined would be proved without also getting GRH. People have proved results assuming the generalized Riemann hypothesis, but no one has thought to prove results of the form: suppose you knew the quasi-Riemann hypothesis for Dirichlet $L$-functions, but nothing about Dedekind $L$-functions.

What this means is that there may be quite a few results which have previously been proved under GRH that might now be provable using only the quasi-Riemann hypothesis for Dirichlet $L$-functions. Artin’s primitive root conjecture is exactly one of these things. I would imagine that, once an expert knew what had been proved, they would see that this application was not surprising at all.

One of my favorite applications of the generalized Riemann hypothesis concerns the discriminant bounds of Stark and Odlyzko. But for all the applications I have in mind, which concern general number fields, one really needs the corresponding result for all Dedekind $L$-functions rather than merely Dirichlet $L$-functions. So we have no improvement on those bounds, at least as yet.
Results on BSD.

There are some results proving BSD under certain rank conditions. My first impression is that these are technical improvements on results already in the literature, and that they represent no genuine progress towards the higher-rank version of BSD. At this point I don’t have so much interest in looking at them myself, but I would be happy to have a guest blog post explaining anything that is new, or what the new trick or idea might be.
Catalan’s constant is irrational.

As we saw in the AI proof that $\zeta(5)$ is irrational, there are clearly a bunch of new ideas, or at least new optimizations, that we were not previously aware of. I feel that some of these papers are among the least well written, in the sense that certain optimizations are taking place without being motivated at all.

I am helping to run a conference at SLMath on irrationality in January, and one of our tasks will clearly be to try to understand what these new methods are, and whether they are simply new tricks or evidence of something more substantial. Of course, Apéry’s result from almost 50 years ago is still something we don’t really understand.

I don’t know to what extent this is true of the other papers, but it’s amusing to see this paper cite two crank papers, both of which give nonsense proofs that Catalan’s constant is rational, and which AI casts some shade on.

One thing I’m confused, or perhaps bemused, by is that I haven’t yet seen OpenAI give an alternative proof of the irrationality of $L(2,\chi_{-3})$, although it wouldn’t be too surprising if such a proof existed.
The irrationality measure of $\pi$ is $2$.

This is an amazing result. I’ve yet to have time to look at it at all, but Wadim Zudilin has already noted to me (perhaps unsurprisingly) the close similarity between the arguments here and Roth’s argument for algebraic irrationals. Again, this is something that I hope we shall address at our upcoming conference.

Broadly speaking, for the problems in the Langlands program that I am interested in, while OpenAI did an amazing job, nothing it did at this point seems at all superhuman in the way that the proof of the quasi-Riemann hypothesis seems, and perhaps other results in other fields seem, although I can’t comment on those.

Posted in Mathematics, Uncategorized | Tagged AI, Ana Caraiani, Artin Conjecture, BSD, Catalan's Constant, Fontaine-Mazur Conjecture, GRH, Hardy, Hilbert's 10th problem, Hooley, Irrationality, James Newton, OpenAI, Ramanujan Conjecture, Rankin, Rankin-Selberg, Riemann Hypothesis, Roth's Theorem | 14 Comments

The OpenAI problem dump

Posted on October 6, 2026 by Persiflage

As we await the latest AI-generated results (whether gurgling through the arXiv, appearing on 4chan, or coming straight from the horse’s mouth), I thought I would put on record some interesting open problems from my own field. I don’t mention these in any way to “move the goalposts”, but rather to be a little more explicit about what makes them interesting. In particular, each of these problems—which I consider significant—is only a very special case of a much larger problem, and even solving that larger problem would not be the final goal. I had wanted to say a little more about each question, but I learnt that the OpenAI problem dump is coming today, so let’s write down some problems in advance, in no particular order.

Your exam begins now. To get full marks you must answer the full question, but partial credit is available. Outside materials including the total sum of human mathematics may be used. Each question is worth 10 points. Hints are given to some of the problems. Good Luck!

Lehmer’s Problem I: Does there exist a prime $p$ such that $\tau(p)=0$, where

\[q \prod_{n=1}^{\infty} (1-q^n)^{24} = \sum_{n=1}^{\infty} \tau(n) q^n?\]

Although this is a classical problem, it’s hard to understand how interesting it is. Famously, we know that $\tau(p)/p^{11/2} \in [-2,2]$ by Deligne, and we now know that these values are distributed according to the Sato–Tate measure. On probabilistic grounds, $\tau(p)=0$ should occur only finitely often; indeed, even $\tau(p) \equiv 0 \bmod p^2$ should occur only finitely often. So after checking the first billion primes, the answer is presumably no. But the point is not so much the answer itself as what the theory of modular forms can tell us about it. We don’t even know what type of question this is! One can relate it to a question about how many ways $p$ can be written as a sum of $24$ squares, but that hardly seems promising. Much more promising, to me, is the fact that the associated Galois representation

\[\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{Q}_p)\]

has large global image, while $\tau(p)=0$ would force it to be locally induced at $p$. Now the question starts to connect to other conjectures, such as Greenberg’s conjecture, which says that, in this setting, if the Galois representation were locally split, then $\rho$ would have to be CM (still open!). But this relationship is undercut by the fact that there are examples of modular forms $f$ with $a_p=0$ that are not CM, most notably when $f=f_E$ has weight $2$ and $E$ is non-CM but supersingular at $p \ge 5$. Now weight $k=2$ is different on probabilistic grounds, but there are also examples with $a_p=0$ in weights $k=4$ and $k=6$.

One can also ask about the analogue for higher-weight forms of level $N=1$. Now the question starts to relate to Maeda’s conjecture, which predicts that, for a normalized cuspidal eigenform of weight $k$, the field $K$ generated by its coefficients has degree equal to the full dimension $d$ of the space of cusp forms. Stronger versions predict not only that the Galois group of the Galois closure is $S_d$, but also that $K$ is generated by any single coefficient $a_p$ of $f$. The latter implies that $a_p \ne 0$ as soon as the dimension of the space is greater than one. Again, it’s totally unclear whether the conjecture is false, true for probabilistic reasons, or true for some good reason.
Lehmer’s Problem II: Is the Mahler measure of a nonzero algebraic integer that is not a root of unity bounded away from $1$? Progress here would be really interesting. One aspect of this problem is that I don’t think we have a convincing heuristic either way. There has been some nice progress in this general area recently (by Dimitrov, Smith, Orloski–Sardari…), although these methods do not seem to have much to say about Lehmer’s problem. A proof would probably be more interesting than a family of counterexamples, but who knows.
Artin’s conjecture for $2$-dimensional Galois representations. Naturally, one can restrict to the case of even representations with projective image $A_5$. If one can answer this, then one can also ask about the full Artin conjecture for any nontrivial irreducible Artin representation of $G_{\mathbf{Q}}$.
The modularity of elliptic curves over general number fields. The case of imaginary quadratic fields seems very much like something that could be done, although I don’t know how to do it. But general number fields are more interesting. Here one can also ask about—and might have to solve—the problem of constructing Galois representations for automorphic forms of algebraic type, as explained in Buzzard–Gee. One can further ask for the modularity of all compatible systems of Galois representations, which encompasses both this question and the previous one on Artin’s conjecture.
Leopoldt’s conjecture. It used to be said that you were not a real algebraic number theorist until you had proved Leopoldt’s conjecture (usually incorrectly). More generally, for geometric Galois representations $V$, there are conjectures about $H^1(\mathbf{Q},V)$ and the dimension of its image in $H^1(\mathbf{Q}_p,V)$, including Janssen’s conjecture.
The algebraic independence of $\zeta(2),\zeta(3),\zeta(5),\zeta(7),\ldots$. More generally, the Grothendieck period conjecture (even just for mixed Tate motives!).
The existence of infinitely many ordinary primes for any cuspidal modular eigenform. More generally, a Newton = Hodge theorem for a density-one set of primes for any motive $M$ over any number field $F$ whose Sato–Tate group is connected.
The BSD conjecture in its full form, in arbitrary rank. The BSD conjecture elliptic curves over any number field. Then Beilinson’s conjecture…
A disproof of Vandiver’s conjecture. It’s surely false (?!), so this is more a question of finding some really clever way to compute $K_{12}(\mathbf{Z}),K_{16}(\mathbf{Z}),K_{20}(\mathbf{Z}),\ldots$ until the order of one of these groups has an interesting prime factor. More generally, prove that there are infinitely many primes $p$ for which it fails. At the same time, prove there are infinitely many regular primes.
Last but not least, RH, or GRH, or GRH for $L(s,M)$ attached to any irreducible motive $M$.

One could go on and on, for extra credit: Schanuel’s conjecture. Given an algebraic curve $X/\mathbf{Q}$, is there an algorithm to determine $X(\mathbf{Q})$? Are there infinitely many Mersenne primes and finitely many Fermat primes? What comes after the Langlands conjectures? …
I don’t know how well ChatGPT will do, but 10/100 would be extremely impressive to say the least. I would certainly guess that Vandivier’s conjecture is the easiest to disprove since it is “only” a computation. But it might be quite a difficult computation, I’m not sure. And even knowing ALL of these wouldn’t answer many questions I am still interested in.

Update: The results are in here. As BLH says, modularity of elliptic curves over imaginary quadratic fields is in, for 2 points. The quasi-Riemann hypotheses (!?) for 3 points (which proves Artin’s conjecture, but not the Artin’s conjecture I gave above!). So that’s on the one hand 5/100 but on the other hand still OMFG Jaw-droppingly amazing (and of course who knows how well it would have done had it been given the list of questions…)

Posted in Mathematics | Tagged Alex Smith, Beilinson, Bryce Orloski, Janssen, Kevin Buzzard, Lehmer's Conjecture, Leopoldt Conjecture, Naser Sardari, Toby Gee, Vandiver's Conjecture, Vesselin Dimitrov | 15 Comments

Journals

Posted on October 5, 2026 by Persiflage

In my recent post, I listed four broad topics which I think our profession should urgently try to address. Here I want to say something about journals.

Whenever I have tried to write one of these posts, I always get stuck on the issue of incentives. There are many things I would like to see, but there is not much point saying what I want to happen without providing any mechanism for us to get there. There are two broad categories of incentives: professional, namely what criteria we use to hire people, give people tenure, give people prizes, etc.; and personal, the inherent drivers that motivate us. The former we have some control over, the latter is a little more slippery.

A lot has been written about the beauty of understanding in the last few months. However, even a cursory examination of the arXiv shows where the money is. There has been a mad and honestly undignified rush to be “the first” to prove one result or another. There is a very wide spectrum of people using AI, and I am definitely not trying to imply that everyone is using it badly. I have seen some excellent uses of AI to prove interesting things, and I’m excited by the math that is to come!

But who could blame people for rushing? I have the luxury of taking this time to use AI to explore interesting new math without needing every experiment to turn into a publication. I also have a good tenured job. For younger people, more papers in better journals could mean a better job and a better life. We cannot continue to reward the production of papers in exactly the same way and then be surprised when people organize their work around producing more papers.

This brings me to the Leiden declaration, which calls for transparent disclosure of the use of automated tools. It is already creaking at the seams. Its spirit is not being observed: we are seeing everything from vague admissions of “assistance” to complete denial. The point of a disclosure should be to tell the reader exactly what happened, not to leave them guessing how much is concealed inside weasel words like “assistance.” I’m not saying that every use of AI should come with the full prompts, but it would be nice to see them now and then!

I am really uncomfortable with a default middle ground where AI is being used in an ambiguous way. That is one of the reasons I found the disclosure discussed in this post so refreshing. It told us plainly where the proofs came from. We should be encouraging that candour, not making people feel that they have to minimize the machine’s contribution to make the paper professionally acceptable. Of course, papers that are genuine collaborations between man and machine exist, and naturally these cases are more subtle, which only makes the point we should have more practice being clear.

One of the few levers we have to affect how mathematicians act is through hiring and through journals. And what are journals doing? As far as I can see, far too little. This cannot be business as usual. Editorial boards have to decide now what kinds of papers they want to publish, and what those publications are supposed to contribute.

I do not think this is simply a question of allowing or forbidding AI. We should be willing to reconsider what constitutes an interesting mathematical contribution. We should also be willing to consider new forms of papers, whether using AI or not. Being more demanding about what is worth publishing is perfectly consistent with being less constrained by traditional ideas about who must have produced it.

For example, let’s consider this new paper. For context, as part of my thesis, I proved that there did not exist any non-zero semistable abelian variety $A/\mathbf{Z}[1/N]$ when $N=6$ or $N=10$ assuming GRH. This paper removes the dependence on GRH.

To make a point, I have listed the authors as Astra and Fable, and instead the paper has a “Human acknowledgment” section. Suppose (for arguments sake) I thought this was interesting enough that it should be part of the mathematics literature in some way. Would putting my name at the top, and relegating the machines to an acknowledgment, make it a better paper? Would writing it under my name and saying that AI “contributed to some of the arguments” make it more or less transparent who did what?

There was a comment on my earlier post asking why Constantin (or someone else in this position) should list themself as an author at all if the mathematics was not due to them. Here is one institutional obstacle: the arXiv does not allow AI tools to be listed as authors. Not only would the paper above not be considered by any journal in its current form, but I can’t even post it on the arXiV.

Naturally, the reason to insist on papers having human Authors is to put responsibility onto humans for what they post. That is a real issue, but a brief glance at the arXiV should convince you that a human listed as an author is hardly a guarantee of anything. Why must responsibility and credit for discovering the mathematics be bundled together? Someone could be responsible for presenting a paper, explaining what has been checked, answering questions, and dealing with corrections without claiming to have discovered its arguments. The best AI disclosure is one which honestly says what the person presenting it actually did.

(The arXiv does, however, now insist on a full English version of every submission. At least we have got rid of the French; remember Agincourt!)

You may well ask what I am doing about this, given my editorial duties.A&NT (Algebra and Number Theory) is a journal with plenty of prestige, especially for a specialist journal, while also being relatively new and so perhaps less constrained by “tradition.” I strongly believe that it is a perfect journal to experiment with new forms of papers, so what better place than A&NT?

I emailed the board asking, as a thought experiment, whether each editor would accept a paper of the kind recently discussed here if it were submitted and passed the referee process. There were many who would. But the discussion made it clear that we had substantially different views about how the journal should respond to AI. I don’t think there is any point running a journal right now as if AI doesn’t exist, and if it was just going to be business as usual, there isn’t much sense for me to remain on the board.

So I resigned.

The point is not that journals should publish more AI papers, or fewer AI papers. It is that they should make some decisions about what is worth publishing, and be willing to experiment with how it is presented. Papers that have a nice idea which didn’t work. Papers with speculative ideas that the author has no idea how to prove. Papers with good expositions, papers which only have conjectures, the list goes on. Unfortunately, right now, the claim that there is “so much more to mathematics than theorems” is not exactly what one might conclude by looking at any recent journals. What we absolutely should not be doing is encouraging everyone to rush out conventional-looking papers while refusing to consider experiments that make it harder to pretend that nothing has changed. If mathematical understanding is what we want to defend, then it ought also to be what we reward.

And yet, I am not pessimistic. The pace of change is so fast that what is obvious to me now (and has been obvious to others for far longer) may simply be obvious to everyone in a years time.

Added: Link to Hexagon Math.

Posted in Mathematics | Tagged AI, Astra, Fable, John Stallings, Thesis | 15 Comments

AI disclosures done right

Posted on September 27, 2026 by Persiflage

Another shiny new AI proof, this time a proof by Constantin Kogler of the Odlyzko–Poonen conjecture which says that a polynomial with coefficients chosen randomly from $\{0,1\}$ is irreducible. What I want to emphasize on this occasion is the exemplary AI disclosure statement. To quote just the first line, “The proofs are due to GPT-6 Astra”. I feel that in the next few months we are going to have an avalanche of new papers attributing “assistance” from AI of varying levels of honesty; I’ve already noted it will be hard for mathematicians to admit even to themselves what they actually contributed. Thus I think it is important to normalize the contributions of papers such as these which give full credit to AI. What goes unmentioned is the fact that the author clearly put in some genuine effort to craft a very nicely written paper. (The author also got Astra to formalize the proof in Lean.) I can’t speak for everyone, but I can say personally that when I see a paper such as this, it gives me great delight, and gives me no AI-related angst at all. I want to commend Constantin for his efforts! The next question is: how would a journal respond to this paper as a submission? I have some thoughts…

Posted in Mathematics | Tagged AI, Andrew Odlyzko, Bjorn Poonen, Constantin Kogler | 9 Comments

Roots of Unity

Posted on September 26, 2026 by Persiflage

This post is a follow-up on a previous post on the abelian house.

For an algebraic integer $\alpha$, the house $\overline{|\alpha|}$ is the absolute value of the largest conjugate of $\alpha.$ Raphael Robinson made a number of conjectures concerning cyclotomic integers with small house, some of which were solved a long time ago by Jones, Cassels, and Schinzel, and one much more recently by Frederick Robinson (no relation!) and Michael Wurtz as part of an REU, discussed in this post. My description there was actually somewhat inaccurate; the work of Robinson-Wurtz did not fully resolve all of Robinson’s conjecture and indeed one remained outstanding. In one or two email conversations I had with Kiran Kedlaya, it certainly seemed plausible that these methods could be pushed forward to resolve the remaining conjecture, though some work was certainly going to be required. As always,the issue is passing from “effective in principle” to “effective in practice”. I’m happy to say that this conjecture has now been resolved! In a recent paper of Jitendra Bajpai, Srijan Das, Kiran Kedlaya, Nam Le, Meghan Lee, Antoine Leudière, and Jorge Mello, the authors build and improve upon the previous methods to prove Robinson’s conjecture and more. (Added: this was “recent” when the first draft of this blog post was written in 2025 which seems infinitely long ago; and in fact just yesterday Kiran sent me the sequel!) Many of the arguments are similar to those of previous papers, but in this type of problem a small improvement can make the difference between a feasible and an intractable problem. One problem they raise as a future project is to compute the first limit point of houses of algebraic integers that is not itself a house of an algebraic integer, a topic that was the main point of this post. They remark in passing that they found a significantly smaller limit point than the one I had found, namely

\[\frac{25}{4} = 6.25 < \alpha = \displaystyle{\frac{97 + 26 \sqrt{13}}{27} = 7.064604\ldots}\] With some trepidation I checked to see how confidently I had predicted that $\alpha$ was the smallest such point. Here is the relevant excerpt:

Moreover, I think it quite likely (and quite provable, perhaps with a certain amount of computational effort) that [$\alpha$] is the smallest number in $\overline{{{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}} \setminus {{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$.

Not too bad! I didn’t formally conjecture it, or (worse) claim it to be true, so I consider that a success.

Posted in Mathematics | Tagged Antoine Leudière, Cassels, itendra Bajpai, Jitendra Bajpai, Jones, Jorge Mello, Kiran Kedlaya, Meghan Lee, Nam Le, Robinson, Roots of Unity, Srijan Das, The Abelian House | Leave a comment

Are business schools intellectually bankrupt? (Part Two)

Posted on September 24, 2026 by Persiflage

Nicholas Polson, a professor of Finance at Booth (Chicago’s business school) in the news for authoring 258 papers in 2026. It turns out that I had an interesting interaction with Polson in this past which I never blogged about. Instead, I just found this draft of a blog post from 2017; I thought it might be a light diversion.

From 2017:

There are essentially three types of people who claim proofs of the Riemann Hypothesis.

First, there are the cranks. The crank often writes something which is utterly incoherent — possibly invoking something from physics, possibly also simultaneously proving RH and Fermat’s Last Theorem at the same time. Alternatively, there are cranks who have some basic knowledge of mathematical formalism, and manage to scribble down something which at least shares some of the grammatical structure of a mathematical argument. My mental image of a crank used to be a retired 60 year old engineer, but, like so many other things, the internet has expanded my horizons, and nowadays cranks come in many different flavours.

Second, there are the amateurs with a sufficient amount of hubris that they somehow believe they can make a contribution a problem which — after a century of study — is clearly very deep and almost everyone believes will require fundamentally new ideas to crack. These people are often successful in their own careers — including Nobel prize winners in Chemistry (or was that a proof of Fermat?). This appears to be related to a (no doubt well-known) variation of the Dunning–Kruger effect — people with a high level of competence in one domain mistakenly overrate their competence in another (in this case number theory). This is usually a very bad mistake when it comes to higher mathematics. (Although don’t imagine that the shoe is never on the other foot — try to imagine for a moment otherwise smart mathematicians pontificating about biology.)

Third, there is the mentally ill. (Obviously there is often a non-trivial intersection between some or all of these classes.)

It is really not worth my while bothering to look at any purported proof of RH — it’s fairly clear that attempting to interact with any character of type one, two, or three above is not really worth one’s time. However, recent circumstances have unfortunately brought me to do so. It began when a probabilist was asked to review a short paper for MathSciNet in a Brazilian probability journal. The reviewer then noticed that the main claim of the paper was a proof of the Riemann hypothesis. Naturally he was confused! The author of the paper was a professor at the University of Chicago. Not of the mathematics department, of course, nor of the statistics department, but of the business school! (For more on business schools, see this post.) The reviewer decided to consult Reddit, who suggested forwarding it to me:

Seeing as the author is at U Chicago (but not in the math department), just email Calegari and inform him that a “colleague” of his has somehow managed to get a proof of RH published in a low-tier probability journal. He’ll deal with it (or get someone to), that’s a professional embarrassment for U Chicago.

Well, of course, I do whatever Reddit tells me to do (/s)

I place the argument — to the extent that I can tell — in the second class. Half an hour of study was sufficient to determine a hole big enough to drive a lorry through. It was the type of mistake that I might have made (but didn’t) when I was 15. There’s a certain amount of probabilistic window dressing in the paper which is syntactically related to real mathematics but tangential to RH. Once this window dressing is removed, the argument is essentially as follows:

Take the function $ 1/\zeta(s)$, then take its inverse Mellin transform.
Assume without comment properties of this inverse Mellin transform which require RH.
Deduce RH by considering the Mellin transform again.

In fact, the good news is that the argument can be upgraded in a smaller number of pages so that it proves that all the zeros of the function $(1/4-s)(3/4-s)$ lie on the critical line. Now that would be a spectacular result!

Let’s prove it! We can start at around equation (2.35) in the paper, where the function

\[ \displaystyle{G(x) = \sum_{k=1}^{\infty} \frac{ (-1)^{k+1}}{\Gamma(k)} \frac{\xi(1/2) \xi(k + 3/2)}{\xi(k + 1/2)} x^k,}\]

is defined, with corresponding Laplace transform

\[\displaystyle{m_G(s) = \int_{0}^{\infty} x^{s-1} G(x) dx.}\]

Here $\xi(s)$ is the part of the zeta function coming from the Hadamard factorization (i.e. without the Gamma factors). Now let’s imagine instead that $\xi(s)$ is just the function $3/4 – s$.
A standard calcuation (for either the $\xi$ coming from Riemann or from $3/4 – s)$ gives

\[ \displaystyle{m_{G}(s) = \int_{0}^{\infty} x^{s-1} G(x) dx = \frac{\xi(3/2 – s) \Gamma(1+s) \xi(1/2)}{\xi(1/2 – s)}}.\]

In particular, this is a purely “formal” calculation (Ramanujan Master Theorem Style). The crux of the argument, the part where something “gets done” and one gets access to $\xi$ in the critical strip is where he “eliminates” $m_{G}(s)$ by using the identity

\[\displaystyle{
\frac{1}{\xi(1/2 – s)} = \frac{m_G(s)}{\xi(1/2)\Gamma(1+s) \xi(3/2 – s)}.}\]

He really wants to use this identity (see the line before (2.45), where variables have been changed slightly) in the range

\[ \displaystyle{s \in (-1/2,0).}\]

After all, he wants to understand $\xi$ — even if just on the real line — in the critical strip. In order to get this, you need to know something about the growth of $G(x)$ as $x$ goes to infinity, because you want the Mellin transform to be well defined. In order to get convergence for real negative s close to zero (which he uses), you certainly want to assume that

\[\displaystyle{G(x) = O(x^{\epsilon})}.\]

So let’s assume exactly this. And now let me prove that $3/4 – s$ has no zeroes for $1/2 \le \mathrm{Re}(s)\le 1$ which is nonsense. From our equation above, $m_G(s)$ is now well defined for $\mathrm{Re}(s)$ in $(-1/2,0)$. But then the RHS is well defined in this range, so the LHS has no poles, so $\xi(1/2 – s)$ has no zeroes for $-1/2 \le \mathrm{Re}(s) \le 0$, or $\xi(s)$ has no zeroes for $1/2 \le \mathrm{Re}(s) \le 1$.. Done!

So what is wrong with this argument? Obviously one actually has to understand the growth of $G(x)$. In the case of $\xi(s) = r – s$, one can compute $G(x)$ explicitly, and at least away from $1/2$ where symmetry forces $G(x)$ to vanish one can compute it explicitly in terms of incomplete Gamma functions and get

\[\displaystyle{G(x) \sim x^{\mathrm{Re}(s) – 1/2}}.\]

Of course, this is no surprise, since this exactly eliminates the contradiction. So now let us return to the paper. We have the function $G(x)$, and we need to say something
about its growth at infinity. In order to prove RH we need to show that it grows slower than any power of $x$. So what is going to happen? Well, we are going to have that

\[\displaystyle{G(x) = O(x^{\rho – 1/2 + \epsilon})},\]

where $\rho$ is the supremum over the real parts of all the non-trivial zeros of $\zeta(s)$. So, in order to prove RH, one only needs to prove … the Riemann Hypothesis!

Note that understanding the growth of functions like $G(x)$ and their link to RH is not new. In fact, already over 100 years ago, Riesz proves the following. Let

\[ \displaystyle{ F(x) = \sum_{n=1}^{\infty} (-1)^n \frac{x^n}{\zeta(2n) \Gamma(n)}.}\]

Then

\[ \displaystyle{F(x) = O(x^{1/4 + \epsilon})}\]

is equivalent to RH. What I have sketched above is basically already a moral explanation of this argument — poles of a function imply growth of the inverse Mellin transform and vice versa. Indeed, Grosswald (in the paper cited by Polson!) proves that the rate of growth (up to $\epsilon$) of $F(x)$ is exactly $x^{\theta/2}$ where $\theta$ is the supremum of the real part of zeros of $\zeta(s)$.

Well, now at least we know the answer to “what does it take to get you to look at my proof of RH?” The answer: you have to have tenure at Chicago, and you have to have a published proof of RH.

One month later: This, at least, was the original story as of a month ago. But there was a twist. Greg Lawler and I actually contacted the author of this paper. Communications via email were not particularly successful. He actually produced a second purported proof (!?) which was worst than the first — basically writing down integrals for a complex parameter s related to $1/\zeta(s)$ paying no attention to the domains of applicability, and then using (in effect) precisely facts about convergence of these integrals which require being careful about the domain of applicability. But then we met in person, and he was very polite, and seemed to realize that both approaches were flawed. The original published paper was retracted by the author, and balance was restored. Success! Or at least I thought so, until I just found out that he recently updated his second paper with more of the same claims! (A little effort — more than it is worth — shows that simply by changing some 1/2s to 1/3rds or 2/3rds one can prove that $\zeta(s)$ doesn’t have any zeroes at all.)

So where does this lead us with respect to the question in the title? I guess in the context where producing banal observations about human behavior that have been repeatedly observed by others gets you a (not really a) Nobel Prize, paying someone $400000 a year (Note: guesstimate) to produce quisquilian proofs of the Riemann Hypothesis sounds perfectly sane. C’est la vie.

Posted in Uncategorized | Tagged (not really a) Nobel Prize, Andrew Gelman, Business School, Cranks, GRH, retraction watch, Riemann Hypothesis | 3 Comments

zeta(5) is irrational

Posted on September 24, 2026 by Persiflage

So one day after an amazing result by humans, the computers have struck back! This time, a proof that $\zeta(5)$ is irrational. The paper I saw gives me the impression of being almost entirely AI generated (if so, it then comes with a very dishonest disclosure), but never mind, thanks for the compute!

I spent some effort trying to see what was going on. Since the argument has now been verified in Lean, it seemed more useful to try to identify the key ideas and where they could be traced, and where they were new.

One way I tried to understand this was to insist that ChatGPT try to find in the literature where similar ideas had been used before. I am sure that I did not do a great job with the literature, but I did my best in the several hours I had available. I sure hope Jacob Tsimerman’s suggestion that AI might reach superhuman exposition will come soon. The discussion below connects the arguments to the work of Prévost (not cited in the paper), but there may well be other sources that ChatGPT learnt these ideas from. Certainly Hankel determinants are also considered by Zudilin, and I suspect that he would do a better job unearthing the ideas behind this paper (some of which are no doubt in his own papers) than I have done. Later on I quote from ChatGPT directly on what it thinks.

Put $D_0(t)=1$ and $D_m(t)=\prod_{j=1}^m(t+j^2)$ for $m\geq1$. Fix an integer $k\geq2$, and let $H_n^{(k)} =\sum_{v=1}^n v^{-k}$ with $H_0^{(k)}=0$, and define the Bernoulli numbers as usual by $u/(e^u-1)=\sum_{m\geq0}B_mu^m/m!$, so $B_1=-1/2$.

Definition. Let $t$ and $X$ be independent indeterminates. Define the $\mathbf{Q}$-vector space

\[\begin{gathered} \mathcal R=\bigcup_{m\geq0}\left\{\frac{P(t)}{D_m(t)}:P(t)\in\mathbf{Q}[t]\right\}\subset\mathbf{Q}(t),\\ \mathbf{Q}[X]_{\leq1}=\{aX+b:a,b\in\mathbf{Q}\}. \end{gathered}\]

Thus $\mathcal R$ consists of the rational functions of $t$ whose finite poles, if any, are simple and belong to $\{-1^2,-2^2,\ldots\}$. There is no restriction on the polynomial part. For the fixed integer $k\geq2$, define the $\mathbf{Q}$-linear map $\mu_{k,X}:\mathcal R\longrightarrow\mathbf{Q}[X]_{\leq1}$ by

\[\begin{aligned}\mu_{k,X}(t^e)&=(-1)^eB_{2e+2}\frac{(2e+k)!}{(k-1)!(2e+2)!}&& (e\geq0),\\ \mu_{k,X}\left(\frac1{t+j^2}\right)&=j^{k-1}(X-H_j^{(k)})-\frac1{k-1}+\frac1{2j}&& (j\geq1).\end{aligned}\]

Write $\mu_k:\mathbf{Q}[t]\to\mathbf{Q}$ for its restriction to polynomials, and abbreviate $\mu_{k,X}((t+j^2)^{-1})$ to $\phi_{k,j}(X)$. The fact that this is well-defined follows from the partial fraction expansion. If one writes $R\in\mathcal R$ as $R(t)=P(t)+\sum_{j=1}^m c_j/(t+j^2)$, with $P(t)=\sum_e p_et^e\in\mathbf{Q}[t]$ and $c_j\in\mathbf{Q}$, then

\[\begin{aligned} \mu_{k,X}(R)={}&\left(\sum_{j=1}^m c_jj^{k-1}\right)X+\sum_e p_e\mu_k(t^e)\\ &+\sum_{j=1}^m c_j\left(-j^{k-1}H_j^{(k)}-\frac1{k-1}+\frac1{2j}\right). \end{aligned}\]

For a real number $\xi$, let $\operatorname{ev}_\xi:\mathbf{Q}[X]\to\mathbf{R}$ be evaluation at $X=\xi$, and define $\mu_{k,\xi}=\operatorname{ev}_\xi\circ\mu_{k,X}$. Thus $\mu_{k,\xi}(R)$ is a real number. For any field $F$ containing $\mathbf{Q}$, the same basis and formulas define an $F$-linear map on $\mathcal R_F=\bigcup_m D_m(t)^{-1}F[t]$, with values in $F[X]_{\leq1}$. We keep the notation $\mu_{k,X}$ for this extension and $\mu_k$ for its restriction to $F[t]$. When $F=\mathbf{R}$, we can specialize $X=\xi\in\mathbf{R}$ as above. In particular, this specifies the map on rational functions with real coefficients.

The formulas of Euler and Hermite

Let $f(y)=(e^{2\pi y}-1)^{-1}$ for $y>0$, and put

\[\begin{aligned} w_k(y)&=\frac{2(-1)^{k-1}y^k}{(k-1)!}f^{(k-1)}(y)\\ &=\frac{2(2\pi)^{k-1}y^k}{(k-1)!}\sum_{l\geq1}l^{k-1}e^{-2\pi l y}. \end{aligned}\]

The density $w_k(y)$ is positive for $y>0$, tends to $1/\pi$ as $y\to0$, and satisfies $w_k(y)=O_k(y^ke^{-2\pi y})$ at infinity. Euler’s and Hermite’s formulas give the following integral representation.

Proposition. For $R\in\mathcal R_{\mathbf{R}}$,

\[\mu_{k,\zeta(k)}(R)=\int_0^\infty R(y^2)w_k(y)\mathrm{d} y.\]

For $a>0$, write $\zeta(k,a)=\sum_{v\geq0}(v+a)^{-k}$. The corresponding transform identity is

\[\int_0^\infty\frac{w_k(y)}{y^2+a^2}\mathrm{d} y=a^{k-1}\zeta(k,a)-\frac1{2a}-\frac1{k-1}.\]

The transform identity is the integer-weight specialization of Prévost’s Stieltjes representation [3] (Theorem 1).

Hankel Determinants

Choose integers $0\leq N<K$, $r\geq1$, and $h\geq1$. Let $\mathcal P_h=\mathbf{Q}[t]_{<h}$ be the $h$-dimensional vector space of polynomials of degree less than $h$. Multiplication by $W(t)=D_N(t)^r/D_K(t)$ sends every polynomial to $\mathcal R$. Therefore

\[\beta_{k,X}:\mathcal P_h\times\mathcal P_h\longrightarrow\mathbf{Q}[X]_{\leq1},\qquad \beta_{k,X}(p,q)=\mu_{k,X}(Wpq)\]

is a well-defined symmetric bilinear map. Its matrix in the basis $1,t,\ldots,t^{h-1}$ and its determinant are

\[\begin{gathered} G_k(X)=\left[\mu_{k,X}\left(\frac{D_N(t)^rt^{i+j}}{D_K(t)}\right)\right]_{0\leq i,j<h},\\ \Delta_k(X)=\det G_k(X)\in\mathbf{Q}[X]. \end{gathered}\]

Such a matrix is called Hankel because its entries depend on $i+j$. (The notation suppresses $K,N,r,h$.) Since the entries have degree at most one in $X$, the determinant has degree at most $h$.

Lemma. After scalar extension to $\mathbf{R}$ and specialization $X=\zeta(k)$, the bilinear map $\beta_{k,\zeta(k)}$ is an inner product. For every nonzero real polynomial $q$ of degree less than $h$,

\[\beta_{k,\zeta(k)}(q,q)=\int_0^\infty W(y^2)q(y^2)^2w_k(y)\mathrm{d} y>0.\]

Thus $G_k(\zeta(k))$ is a positive-definite Gram matrix, meaning the matrix of pairwise inner products of a linearly independent list. In particular $\Delta_k(\zeta(k))>0$.

Allowing for a general rational $W$, Prévost’s arguments for $\zeta(2)$ and $\zeta(3)$ fit this same Hankel construction, using the one-pole choices $W_{2,n}(t)=1/(t+(2n+1)^2)$ and $W_{3,n}(t)=t/(t+(n+1)^2)$, respectively, with matrices of size $n+1$ [3]. These determinants are affine polynomials in $X$, and their roots are the corresponding rational approximations to $\zeta(k)$: a change of polynomial basis expresses each determinant as a rational factor times a Padé approximation error. Prévost’s coefficient and error estimates allow these determinants to be rescaled into integer linear forms whose nonzero values at $\zeta(2)$ or $\zeta(3)$ tend to zero, proving irrationality. On the other hand, in this new argument, the single pole of $W$ is replaced by the $K-N$ poles of $D_N(t)^r/D_K(t)$, producing a determinant of degree $K-N$ in $X$.

Proposition (Degree of the determinant). Take $h=K-N$. Then $\Delta_k$ has degree $h$, with

\[[X^h]\Delta_k(X)=(-1)^{h(h-1)/2}\prod_{j=N+1}^Kj^{k-1}D_N(-j^2)^{r-1}\ne0.\]

For a nonzero polynomial $\Delta\in\mathbf{Q}[X]$, a positive rational number $c$ with $c\Delta\in\mathbf{Z}[X]$ will be called a normalizing factor. For example, one can multiply by the least common multiple of the coefficient denominators. One can also divide out a common factor of the resulting integer coefficients; such division is important in this construction. The required estimate concerns the value of the normalized polynomial, including both effects.

Lemma. Let $\xi\in\mathbf{R}$. Suppose $P_n\in\mathbf{Z}[X]$, $\deg P_n\leq Cn$, and $0<P_n(\xi)\leq\exp(-cn^2+o(n^2))$, where $C,c>0$ are fixed. Then $\xi$ is irrational.

The strategy

The goal is to apply the integer-polynomial lemma with $\xi=\zeta(k)$ and with $P_n$ a positive rational multiple of the determinant $\Delta_k(X)$ in the Hankel matrix. “All” that has been done is to adjust Prévost’s positive measure by $D_N^r/D_K$ for suitable parameters $N$ and $K$. Increasing the numerator power actually makes the determinant larger, but it turns out that one obtains extra cancellations in the coefficient denominators that compensate. Establishing those cancellations uniformly is required for the proof, of course, but finding the right construction is the key step. Now it might seem as though this new factor has come out of nowhere, but even this is not actually new, and the rows of this matrix can be identified with well-poised hypergeometric integrals considered by Zudilin. Once the required arithmetic estimates are proven, one deduces the irrationality of $\zeta(k)$ with $k=2,3,4,5$, but not (at least directly) for larger $k$.

Prévost’s moments and the introduction of the zeta value

We now attempt some mathematical archaeology. Prévost’s Theorem 1 in [3] gives the Hurwitz-zeta representation above. Write $\mathrm{d}\nu_k(t)=w_k(\sqrt t)\,\mathrm{d}t/(2\sqrt t)$ for $t>0$. Its polynomial moments are

\[m_j:=\int_0^\infty t^j\mathrm{d}\nu_k(t)=\mu_k(t^j)=(-1)^jB_{2j+2}\frac{(2j+k)!}{(k-1)!(2j+2)!}\in\mathbf{Q}.\]

These are rational for every integer $k\geq2$. Their Hankel determinants compute rational orthogonal polynomials. Put $H_s=\det[m_{i+j}]_{0\leq i,j<s}$, with $H_0=1$. The determinant formula in Prévost’s equation (2.11), written for this moment sequence, is

\[p_s(z)=\frac{1}{H_s} \det\begin{pmatrix} m_0&m_1&\cdots&m_s\\ m_1&m_2&\cdots&m_{s+1}\\ \vdots&\vdots&&\vdots\\ m_{s-1}&m_s&\cdots&m_{2s-1}\\ 1&z&\cdots&z^s \end{pmatrix}.\]

The polynomial is monic of degree $s$ and satisfies $\int t^j p_s(t)\mathrm{d}\nu_k(t)=0$ for $j<s$; these are its orthogonality conditions. Both $H_s$ and the coefficients of $p_s$ are rational. The zeta value enters through a subsequent evaluation of a transform, as follows.

Define $\mathcal F_k(z)=\int_0^\infty(z+t)^{-1}\mathrm{d}\nu_k(t)$ for $z>0$. At a positive integer $a$, we get

\[\mathcal F_k(a^2)=a^{k-1}\bigl(\zeta(k)-H_a^{(k)}\bigr)-\frac1{k-1}+\frac1{2a}=\phi_{k,a}(\zeta(k)).\]

Consequently a rational approximation to $\mathcal F_k(a^2)$ gives a rational approximation to $\zeta(k)$. A Padé approximation matches finitely many coefficients of the asymptotic expansion $\sum_{j\geq0}(-1)^jm_jz^{-j-1}$, as $z\to+\infty$, by a rational function. The coefficients of that rational function are determined by the moment equations above. Prévost–Rivoal give effective approximation bounds when the Padé degree and $a$ increase together [4].

Replacing $\mathrm{d}\nu_k(t)$ by $\mathrm{d}\nu_k(t)/(t+a^2)$, polynomial division gives the affine moment

\[s_{a,j}(X):=\mu_{k,X}\left(\frac{t^j}{t+a^2}\right) =\sum_{l=0}^{j-1}(-a^2)^l m_{j-1-l}+(-a^2)^j\phi_{k,a}(X),\]

where the sum is empty at $j=0$. At $X=\zeta(k)$ this is exactly the $j$th moment of the corresponding measure.

The measure in the new argument is the following explicit modification of the same $\nu_k$:

\[\mathrm{d}\sigma_{k,K,N,r}(t)=W(t)\mathrm{d}\nu_k(t),\qquad W(t)=\frac{D_N(t)^r}{D_K(t)} =\frac{\displaystyle\prod_{a=1}^N(t+a^2)^{r-1}} {\displaystyle\prod_{j=N+1}^K(t+j^2)}.\]

Here $0\leq N<K$ and $r\geq1$ are integers. All factors are positive on the integration interval $t\geq0$. For fixed $K,N,r$ the moments are $\mu_{k,\zeta(k)}(Wt^j)$, and their Hankel determinant of size $h=K-N$ is $\Delta_k(\zeta(k))$. These are the moments used throughout the argument.

ChatGPT tells me: Uvarov studies the effect of precisely this type of rational multiplication on orthogonal polynomials [7]. Krattenthaler’s Theorem 1 and Proposition 13 express the modified Hankel determinant through the original $p_s$ and their Cauchy transforms $C_s(y)=\int p_s(t)/(y-t)\mathrm{d}\nu_k(t)$, for $y<0$ [8]. For the displayed modifier, each numerator node $-a^2$ gives $r-1$ rows $p_s^{(d)}(-a^2)/d!$ for $0\leq d<r-1$, and each denominator node gives a Cauchy-transform row. Thus the measure considered here belongs to an established family of rational modifications. The starting measure and remainder approximation come from Prévost and Prévost–Rivoal [2], [3], [4]. The exact algebra for multiplying that measure by $D_N^r/D_K$ is supplied by Uvarov’s rational modifications and Krattenthaler’s determinant identity [7], [8]. Zudilin then supplies the irrationality criterion for a positive moment determinant with sufficiently small coefficient denominators [5]. These results identify the starting approximation problem, the modified determinant, and the final integer-polynomial argument, respectively. The closest precedent for improving the coefficient normalization of the matrix is Brown’s Sections 7.4–7.5 [6]. There one first clears row denominators, then removes column common factors and uses congruences between rows or columns to reduce the total multiplier. Here the concrete mechanism uses congruences between square nodes, while also accounting for the polynomial parts in the partial-fraction expansions. Applying $\mu_k$ to those polynomial parts can introduce Bernoulli denominators. The required quantitative input is that the assigned divisibility holds simultaneously for every mixed pairing, with these contributions included.

To come back to the argument, one can (uniformly, for $k=2,3,4,5$) take $K=12n$, $N=n$, $r=6$, and $h=11n$, and write $\Delta_{k,n}$ for the corresponding determinant. (This is different from what is done in the Lean verification, but whatever.) Set

\[S_n=\frac{(K!)^{2h}4^{h-1}}{(N!)^{12h}\prod_{i=1}^{h-1}((2i)!)^2},\qquad F_{k,n}(X)=S_n\Delta_{k,n}(X).\]

One then has to estimate this at the real place and then control the denominators. In particular:

Theorem: $\zeta(k)$ is irrational for $k \in \{2,3,4,5\}$. For all sufficiently large $n$,

\[\begin{aligned}P_{k,n} & \in\mathbf{Z}[X], \\
\deg P_{k,n} & =11n \\ P_{k,n}(\zeta(k))& \le e^{-\gamma_k n^2} \\ (\gamma_2,\gamma_3,\gamma_4,\gamma_5)& =(30,96,6,3/2).\end{aligned}\]

References

[1] F. Beukers, A note on the irrationality of $\zeta(2)$ and $\zeta(3)$, Bull. London Math. Soc. 11 (1979), 268–272. doi:10.1112/blms/11.3.268.

[2] M. Prévost, A new proof of the irrationality of $\zeta(2)$ and $\zeta(3)$ using Padé approximants, J. Comput. Appl. Math. 67 (1996), 219–235. doi:10.1016/0377-0427(95)00019-4.

[3] M. Prévost, Remainder Padé approximants for the Hurwitz zeta function, Results Math. 74 (2019), article 51; arXiv:1709.05389v1.

[4] M. Prévost and T. Rivoal, Diagonal convergence of the remainder Padé approximants for the Hurwitz zeta function, J. Number Theory 222 (2021), 346–361. doi:10.1016/j.jnt.2020.10.019.

[5] W. Zudilin, A determinantal approach to irrationality, Constr. Approx. 45 (2017), 301–310. arXiv:1507.05697v2; doi:10.1007/s00365-016-9333-7.

[6] F. Brown, Mellin transforms, transfinite diameter and rational approximations of integrals, 2026, arXiv:2604.20741v1.

[7] V. B. Uvarov, The connection between systems of polynomials that are orthogonal with respect to different distribution functions, U.S.S.R. Comput. Math. Math. Phys. 9, no. 6 (1969), 25–36. doi:10.1016/0041-5553(69)90124-4.

[8] C. Krattenthaler, A determinant identity for moments of orthogonal polynomials that implies Uvarov’s formula for the orthogonal polynomials of rationally related densities, 2021. arXiv:2103.03969v1.

Posted in Mathematics | Tagged \zeta(5), Francis Brown, Irrationality, Marc Prévost, Tanguy Rivoal, Wadim Zudilin | 4 Comments

Real quadratic fields and finite quantum dilogarithms I

Posted on September 22, 2026 by Persiflage

Danylo Radchenko and Campbell Wheeler have posted an extraordinary new paper in which they prove that Stark units for real quadratic fields are algebraic numbers. (Not yet a Shimura reciprocity law that shows these generate abelian extensions but hey, this is only part I.) This, to me, is arguably the most significant mathematical result of the past month. There is a lot of rich mathematics going on here, and I apologize for not having the time to even summarize it. But I didn’t want it to go unmentioned!

Update: There is now a “part II” as it were, posted here.

Posted in Mathematics | Tagged Campbell Wheeler, Class Field Theory, Danylo Radchenko, Dilogarithm, Reciprocity, Shimura, Shintani, Stark | 8 Comments

An even PSL_2(F_23) Galois extension

Posted on September 20, 2026 by Persiflage

I was wrong!

I have said on a number of occasions that my favourite family of groups for the inverse Galois problem is $G=\mathrm{SL}_2(\mathbf{F}_p)$. This is still true, and the reason is still the same; the fact that $G$ has no non-trivial non-central involutions means that any such extension must be totally real up to twist, and thus it cannot come (for example) as the mod-$p$ reduction of some strongly compatible system (assuming $p > 5$). For a similar reason, the group $\mathrm{PSL}_2(\mathbf{F}_p)$ should also present difficulties under the additional assumption that the field is totally real, and I used this to argue that a totally real extension with Galois group $\mathrm{PSL}_2(\mathbf{F}_{23})$ would be more interesting to me than any other Galois closure of a degree $24$ field. It was then a delight to receive an email from Eray Karabiyik (a recent student of Zywina from Cornell, soon to be a postdoc at HIMIS, Shenzhen) giving me an explicit degree $24$ polynomial with all roots real whose splitting field had Galois group $\mathrm{PSL}_2(\mathbf{F}_{23})$!

Here is an introduction to his construction. The goal is to construct (more or less) a representation

\[\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_{23})\]

which is even, has image containing $\mathrm{SL}_2(\mathbf{F}_{23})$, and whose determinant lands in the squares $(\mathbf{F}_{23}^{\times})^2$. Its projectivization then has image exactly $\mathrm{PSL}_2(\mathbf{F}_{23})$. As mentioned above, this should not come from a regular rank $2$ (with coefficients) motive. But there is no such parity obstruction to its appearing as a constituent of the mod-$23$ reduction of a rank $4$ motive! So, for example, one could hope to find an abelian surface $A$ with

\[A[23]^{\mathrm{ss}} \simeq \rho \oplus \rho^{\vee}(1)\]

where $\rho$ is even and $(1)$ denotes the mod-$23$ cyclotomic twist. Suppose for now that $A$ and a principal polarization are defined over $\mathbf{Q}$.

One way to find $A$ so that $A[p]$ breaks up is for its geometric endomorphism ring to be the ring of integers $\mathcal{O}_K$ of a real quadratic field $K$. If this action is defined over $\mathbf{Q}$, however, then one just gets (once again) rank $2$ compatible systems which are odd. Suppose instead that the action of $K$ is only defined over a quadratic field $E$. Now, for a prime $p$ that splits in $K$, $A[p]$ decomposes over $E$ into two factors which are interchanged by the action of $\mathrm{Gal}(E/\mathbf{Q})$. On the other hand, at an inert prime, one gets a two-dimensional representation of $G_E$ over $\mathbf{F}_{p^2}$, with the non-trivial coset of $G_E$ acting semilinearly through the field automorphism, and generically with large image.

But if $p$ is ramified in $K$, with $(p)=\pi^2$, then the subspace $A[\pi]$ does give a representation $\rho:G_{\mathbf{Q}}\rightarrow\mathrm{GL}_2(\mathbf{F}_p)$. Writing $\omega$ for the mod-$p$ cyclotomic character and $\chi$ for the quadratic character of $E$, one obtains

\[0\longrightarrow\rho\longrightarrow A[p]\longrightarrow\rho\otimes\chi\longrightarrow0,\qquad \det\rho=\omega\chi.\]

Thus $\rho$ is even iff $E$ is imaginary. The condition that the determinant be a square is precisely $\chi=\omega^{(p-1)/2}$ with $p=23$. So one is indeed in with a chance assuming:

$K$ is a real quadratic field with discriminant divisible by $23$,
$E=\mathbf{Q}(\sqrt{-23})$.

In this case $\det\rho=\omega^{12}$ and $\rho\otimes\chi\simeq\rho^{\vee}(1)$, as desired. Of course, large image still has to be checked.

To find surfaces $A$ with endomorphisms by $K$, it helps to have access to the corresponding Hilbert modular surface. One can also insert $E$ into the story by suitable twisting. This is analogous to how one can look for $\mathbf{Q}$-curves defined over a quadratic field $E$, admitting an isogeny of degree $q$ to their conjugate, by twisting $X_0(q)$ by the Atkin–Lehner involution and the character of $E$. The quotient $X_0^+(q)$ forgets which conjugate one started with.

Here Eray takes $K=\mathbf{Q}(\sqrt{23})$, for which Elkies and Kumar give an explicit model of $Y_-(92)$. There are two relevant involutions: one exchanges the two RM embeddings, while the other switches the two principal polarization classes on the generic unpolarized RM surface. The latter correspond to totally positive units modulo squares (the fundamental unit $\epsilon$ is totally positive): starting with a principal polarization $\lambda$ and RM embedding $\iota$, the other class is represented by $\lambda\circ\iota(\epsilon)$. Both leave the unpolarized surface unchanged, corresponding to “$q=1$”. Eray searches on the appropriate non-K3 involution quotient, twisted by $E$ using the remaining involution. If I understood him, this turns out to be (birational to) a genus two fibration over $\mathbf{P}^1$. He then finds a point which gives rise to such an $A$.

Now I have simplified Eray’s example, because his $A$ is defined over an auxiliary quadratic field $H=\mathbf{Q}(\sqrt{18193})$, with the RM defined over $EH$. This is presumably either related to field of moduli versus field of definition issues at low level, or (somewhat relatedly) lifting rational points from the quotient by the involution back to the cover. It affects some of the discussion of Galois representations above, but (in favorable situations) is no longer visible when one considers projective representations.

As $p$ gets larger, one will either have to make $K$ larger (so that $p$ is ramified) or take a non-maximal order like $\mathbf{Z} + p \mathcal{O}_K$. In either case, the corresponding Hilbert modular surfaces (and all their quotients by involutions) will have general type for large enough $p$, and my guess is that rational points away from the special loci on the relevant Hilbert modular surfaces and their twisted quotients will become increasingly scarce and most likely eventually empty. So I suspect this is unlikely to produce totally real $\mathrm{PGL}_2(\mathbf{F}_p)$ or $\mathrm{SL}_2(\mathbf{F}_p)$ extensions for all (or even infinitely many) $p$, but it’s a very nice example nonetheless.

Posted in Mathematics, Uncategorized | Tagged Abhinav Kumar, David Zywina, Eray Karabiyik, Even Galois Representations, Hilbert modular surfaces, I was wrong, Inverse Galois Problem, Noam Elkies | Leave a comment

BSD and the p-converse (Guest Post: Ashay Burungale)

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A Brief Digest of Some New OpenAI Results

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AI disclosures done right

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Are business schools intellectually bankrupt? (Part Two)

zeta(5) is irrational

The formulas of Euler and Hermite

Hankel Determinants

The strategy

Prévost’s moments and the introduction of the zeta value

References

Real quadratic fields and finite quantum dilogarithms I

An even PSL_2(F_23) Galois extension

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