Title: The 10th Discriminant and Tensor Powers of $\mathbb{Z}$

“We plan to discuss very shortly certain achievements and disappointments of the $\mathbb{F}_1$-approach. In addition, we will consider a possibility to apply noncommutative tensor powers of $\mathbb{Z}$ to the Riemann Hypothesis.”

Here’s his talk, and part of the comments section:

Smirnov urged us to pay attention to a 1933 result by Max Deuring in Imaginäre quadratische Zahlkörper mit der Klassenzahl 1:

“If there are infinitely many imaginary quadratic fields with class number one, then the RH follows.”

Of course, we now know that there are exactly nine such fields (whence there is no ‘tenth discriminant’ as in the title of the talk), and one can deduce anything from a false statement.

Deuring’s argument, of course, was different:

The zeta function $\zeta_{\mathbb{Q} \sqrt{-d}}(s)$ of a quadratic field $\mathbb{Q}\sqrt{-d}$, counts the number of ideals $\mathfrak{a}$ in the ring of integers of norm $n$, that is
\[
\sum_n \#(\mathfrak{a}:N(\mathfrak{a})=n) n^{-s} \]
It is equal to $\zeta(s). L(s,\chi_d)$ where $\zeta(s)$ is the usual Riemann function and $L(s,\chi_d)$ the $L$-function of the character $\chi_d(n) = (\frac{-4d}{n})$.

Now, if the class number of $\mathbb{Q}\sqrt{-d}$ is one (that is, its ring of integers is a principal ideal domain) then Deuring was able to relate $\zeta_{\mathbb{Q} \sqrt{-d}}(s)$ to $\zeta(2s)$ with an error term, depending on $d$, and if we could run $d \rightarrow \infty$ the error term vanishes.

So, if there were infinitely many imaginary quadratic fields with class number one we would have the equality
\[
\zeta(s) . \underset{\rightarrow}{lim}~L(s,\chi_d) = \zeta(2s) \]
Now, take a complex number $s \not=1$ with real part strictly greater that $\frac {1}{2}$, then $\zeta(2s) \not= 0$. But then, from the equality, it follows that $\zeta(s) \not= 0$, which is the RH.

To extend (a version of) the Deuring-argument to the $\mathbb{F}_1$-world, Smirnov wants to have many examples of commutative rings $A$ whose multiplicative monoid $A^{\times}$ is isomorphic to $\mathbb{Z}^{\times}$, the multiplicative monoid of the integers.

What properties must $A$ have?

Well, it can only have two units, it must be a unique factorisation domain, and have countably many irreducible elements. For example, $\mathbb{F}_3[x_1,\dots,x_n]$ will do!

(Note to self: contemplate the fact that all such rings share the same arithmetic site.)

Each such ring $A$ becomes a $\mathbb{Z}$-module by defining a new addition $+_{new}$ on it via
\[
a +_{new} b = \sigma^{-1}(\sigma(a) +_{\mathbb{Z}} \sigma(b)) \]
where $\sigma : A^{\times} \rightarrow \mathbb{Z}^{\times}$ is the isomorphism of multiplicative monoids, and on the right hand side we have the usual addition on $\mathbb{Z}$.

But then, any pair $(A,A’)$ of such rings will give us a module over the ring $\mathbb{Z} \boxtimes_{\mathbb{Z}^{\times}} \mathbb{Z}$.

It was not so clear to me what this ring is (if you know, please drop a comment), but I guess it must be a commutative ring having all these properties, and being a quotient of the ring $\mathbb{Z} \boxtimes_{\mathbb{F}_1} \mathbb{Z}$, the coordinate ring of the elusive arithmetic plane
\[
\mathbf{Spec}(\mathbb{Z}) \times_{\mathbf{Spec}(\mathbb{F}_1)} \mathbf{Spec}(\mathbb{Z}) \]

Smirnov’s hope is that someone can use a Deuring-type argument to prove:

“If $\mathbb{Z} \boxtimes_{\mathbb{Z}^{\times}} \mathbb{Z}$ is ‘sufficiently complicated’, then the RH follows.”

If you want to attend the seminar when it happens, please register for the seminar’s mailing list.

This year’s goal is to obtain a somewhat definite verdict on the field-with-one-element hype.

In short, the plan is to outline Smirnov’s approach to the ABC-conjecture using geometry over $\mathbb{F}_1$, to describe Borger’s idea for such an $\mathbb{F}_1$-geometry and to test it on elusive objects such as $\mathbb{P}^1_{\mathbb{F}_1} \times_{\mathbb{F}_1} \mathsf{Spec}(\mathbb{Z})$ (relevant in Smirnov’s paper) and $\mathsf{Spec}(\mathbb{Z}) \times_{\mathbb{F}_1} \mathsf{Spec}(\mathbb{Z})$ (relevant to the Riemann hypothesis).

We did start with an historic overview, using recently surfaced material such as the Smirnov letters. Next, we did recall some standard material on the geometry of smooth projective curves over finite fields, their genus leading up to the Hurwitz formula relating the genera in a cover of curves.

Using this formula, a version of the classical ABC-conjecture in number theory can be proved quite easily for curves.

By analogy, Smirnov tried to prove the original ABC-conjecture by viewing $\mathsf{Spec}(\mathbb{Z})$ as a ‘curve’ over $\mathbb{F}_1$. Using the connection between the geometric points of the projective line over the finite field $\mathbb{F}_p$ and roots of unity of order coprime to $p$, we identify $\mathbb{P}^1_{\mathbb{F}_1}$ with the set of all roots of unity together with $\{ [0],[\infty] \}$. Next, we describe the schematic points of the ‘curve’ $\mathsf{Spec}(\mathbb{Z})$ and explain why one should take as the degree of the ‘point’ $(p)$ (for a prime number $p$) the non-sensical value $log(p)$.

To me, the fun starts with Smirnov’s proposal to associate to any rational number $q = \tfrac{a}{b} \in \mathbb{Q} – \{ \pm 1 \}$ a cover of curves

$q~:~\mathsf{Spec}(\mathbb{Z}) \rightarrow \mathbb{P}^1_{\mathbb{F}_1}$

by mapping primes dividing $a$ to $[0]$, primes dividing $b$ to $[\infty]$, sending the real valuation to $[0]$ or $[\infty]$ depending onw whether or not $b > a$ and finally sending a prime $p$ not involved in $a$ or $b$ to $[n]$ where $n$ is the order of the unit $\overline{a}.\overline{b}^{-1}$ in the finite cyclic group $\mathbb{F}_p^*$. Somewhat surprisingly, it does follow from Zsigmondy’s theorem that this is indeed a finite cover for most values of $q$. A noteworthy exception being the map for $q=2$ (which fails to be a cover at $[6]$) and of which Pieter Belmans did draw this beautiful graph

True believers in $\mathbb{F}_1$ might conclude from this graph that there should only be finitely many Mersenne primes… Further, the full ABC-conjecture would follow from a natural version of the Hurwitz formula for such covers.

(to be continued)