Baker’s Theorem and the Class Number Problem (Part 3)

Disclaimer: This post is the third of a set of three(ish) about Baker’s Theorem and the Class Number Problem. It is intended to be informative and not too rigorous. Find part one here and part two here!

So, we’ve introduced both the Class Number One Problem and Baker’s Theorem, and now we’re going to try and tie them all together. I’m going to miss out some details, which I’ll write up in an addendum/part 4 for anyone who’s interested in the ins and outs rather than an overview of the proof. I refer heavily back to the first and second posts in this series, but I’ll briefly talk about where we are. My main references for this post are Baker and Wustholz’s “Logarithmic Forms and Diophantine Geometry” and Baker’s paper “On the Class Number of Imaginary Quadratic Fields“.

First of all, we have Baker’s Theorem. I quote the theorem Baker himself proved; we’ve had improvements since, but we only need a fairly weak version for our uses. For (as far as I’m aware) the strongest result to date, see this result from Matveev. What follows is the version of Baker’s Theorem we will use.

Baker’s Theorem Let $\alpha_{1},\, \dots,\, \alpha_{n} \in \mathbb{A} \backslash \{0,\, 1\}$ be such that $\log \alpha_{1},\, \dots,\, \log\alpha_{n}$ are linearly independent over $\mathbb{Q}$ . Let $\beta_{0},\,\beta_{1},\, \dots ,\,\beta_{n} \in \mathbb{A}$ , not all 0. Set $\Lambda := \beta_{0} + \beta_{1} \log \alpha_{1} + \dots + \beta_{n} \log \alpha_{n}$ and let $B= \max \{h(\beta_{0}),\, h(\beta_{1}),\dots,\, h(\beta_{n})\}$ , where $h(x)$ denotes the maximum of the absolute values of coefficients of the minimal polynomial of $x$ . Assume that $\Lambda \neq 0 .$ Then there exists an effectively computable constant $c:=c(n,\,\alpha_{1},\, \dots ,\, \alpha_{n})$ such that $\lvert \Lambda \rvert \geq (eB)^{-c}$ .

We are trying to show that there are finitely many imaginary quadratic fields with class number 1. I’m going to assume a knowledge of $L$ -series (I’ll talk more about these in the upcoming addendum). Throughout I will write $\mathbb{Q}(\sqrt{-d})$ to denote an imaginary quadratic field. With some consideration, we can show that if $\mathbb{Q}(\sqrt{-d})$ has class number 1, then the discriminant of this number field is $-d$ (consider fundamental discriminants and $-d \mod 4$ ).

By the time we’ve got to the point Baker proved his Theorem, we’d made some progress with the problem. Dickson (and others who improved his methods) used the connection between quadratic forms and quadratic fields to show that there were at least 9 imaginary quadratic fields with class number 1, and that other than these 9 there were no more down to $-d = -500,000,000$ . Heilbronn and Linfoot managed to show that there was maximally one further imaginary quadratic field with class number one. If this field existed, Heilbronn and Linfoot were able to show that this would prove that the Generalised Riemann Hypothesis would not be true (see further Siegel zeros, which have a wider connection to class numbers of fields and can be used to give lower bounds for class numbers). I might talk more about these proofs in the addendum, because there’s some really clever maths in there,

So, to summarise, we’ve got to a point where we have found 9 imaginary quadratic fields with class number 1, and we know there’s maximally one more to find, if it exists. Now let’s show it doesn’t, following Baker’s argument. Before we start, I’m going to give the explicit class number formula for quadratic fields. First, give a quadratic field $K$ of discriminant $D$ , we define the character $\chi (n) = \left( \frac{D}{n}\right)$ , where the right hand side of this expression is the Kronecker symbol. We then define the Dirichlet L-function to be $L \left(s,\,\chi\right) =\prod_{p} \left( 1-\frac{\chi(p)}{p^{s}}\right)^{-1}$ where we take the product over the primes. The one last thing I want to comment on this (for now) is that $\zeta_{K}(s) = \zeta(s)L(s, \chi)$ , where $\zeta_{K}$ is the zeta function for number field $K$ .

For a real quadratic field $K$ with discriminant $D_{K}$ we find that $h_{K}=\frac{\sqrt{\left| D_{K} \right|}}{2 \log \epsilon}L \left(s,\,\chi\right)$ where $\epsilon$ is the fundamental unit of $\mathcal{O}_{K}$ . For an imaginary quadratic field $K$ with $D_{K}$ and $m$ roots of unity in the field, the explicit class number formula gives us that $h_{K}=\frac{m \sqrt{\left| D_{K} \right|}}{2 \pi}L(1, \chi)$ . These are specific examples of the more general explicit class number formula.

We now are in a position to get started.

We let $-d < 0$ and $k>0$ be the discriminants of $K=\mathbb{Q}(\sqrt{-d})$ and $L=\mathbb{Q}(\sqrt{k})$ respectively, where $gcd(k,d)=1$ . We define the characters $\chi (n)=(\frac{k}{n})$ and $\chi' (n)=(\frac{-d}{n})$ via the Kronecker symbols. Then, for any $s>1$ , $L(s, \chi) L(s, \chi \chi')=\frac{1}{2}\sum_{f}\sum_{x,y \neq 0,0}\chi(f) f^{-s},$ where $f(x,y)=ax^{2}+bxy+cy^{2}$ with $a,\, b \in \mathbb{Z}$ is summed over the set of reduced quadratic forms with discriminant equal to the discriminant of $K$ . Eventually, we assume $K$ has class number one, which will greatly simplify this as we will only have one quadratic form remaining. This is the consequence of some nice maths that I’ll put in the addendum! First, we consider these $L$ -series and manipulate them and our class number formulas to get to a position where we can apply Baker’s Theorem. I omit lots of the working, if you’re interested in seeing the working drop me a message on twitter and I’ll send it over! The next few paragraphs are mostly about manipulating series and the like, so can be skimmed until we get to the point where we can apply Baker’s Theorem.

Playing with the $L$ -series, we can find that $L(s, \,\chi)L(s, \, \chi\chi')= \prod_{p}\bigg(\frac{p^{2s}}{p^{2s-1}}\bigg)\prod_{p|k}\bigg(\frac{p^{2s}-1}{p^{2s}}\bigg)\sum_{f}\frac{\chi(a)}{a}+\sum_{f}\sum_{r=-\infty}^{\infty}A_{r}e^{\frac{\pi irb}{ka}}$ , where we get to the last line by expressing the sum of $f$ as a Fourier series, with $A_{r}$ being the coefficients of this sum. We then take the limit as $s$ tends to 1 to find that $L(1, \chi) L(1, \chi \chi') = \frac{\pi^{2}}{6}\prod_{p|k}\bigg(1-\frac{1}{p^{2}}\bigg)\sum_{f}\frac{\chi(a)}{a}+\sum_{f}\sum_{r=-\infty}^{\infty}A_{r}e^{\frac{\pi irb}{ka}}$ . Calculating and bounding the terms in the Fourier Series we find that for $r \neq 0$ , $|A_{r}| \leq 2r'e^{\frac{-r'}{ka}}$ , where $r'=\frac{\pi |r|}{\sqrt{d}}$ , and $A_{0}=0$ apart from when $k$ is a power of prime $p$ , when $A_{0}=\frac{-2 \pi}{k \sqrt{d}}\chi (a) \log p$ .

We now assume that $K= \mathbb{Q}(\sqrt{-d})$ has class number one. Then, by assumption, see the next addendum post, $d$ is prime and $d \equiv 3 \mod 4$ . We also take $k \equiv 1 \mod 4$ . Further, as we assume $K$ has class number 1, there is just one reduced quadratic form, $x^{2}+xy+\frac{1+d}{4}y^{2}$ with discriminant equal to $D_{K}$ . Thus we see that $\chi(a)=1$ , $a=1$ so $\sum_{f} \frac{\chi(a)}{a}=1$ . Further, as there is only one reduced quadratic form (there is only one $f$ ) we can easily see that $\sum_{f}\sum_{r=-\infty}^{\infty}A_{r}e^{\frac{\pi irb}{(ka)}}=\sum_{r=-\infty}^{\infty}A_{r}e^{\frac{\pi ir}{k}}$ .

We use this to see that under the assumption that $\mathbb{Q}(\sqrt{-d})$ has class number one, $L(1, \chi) L(1, \chi \chi')=\frac{\pi^{2}}{6}\prod_{p|k}\bigg(1-\frac{1}{p^{2}}\bigg)+\sum_{r=-\infty}^{\infty}A_{r}e^{\frac{\pi irb}{k}}.$

Finally, by considering the explicit class formula, we can see that $L(1, \chi)L(1, \chi \chi')=\frac{2 \pi \log (\epsilon_{L}) h(k) h'(k)}{k \sqrt{d}},$ where $\epsilon_{L}$ is a fundamental unit of $L$ and $h'(k)$ is the class number of $\Q(\sqrt{-dk})$.

Putting this all together, we see that $\frac{2 \pi \log \epsilon_{k} h(k) h'(k)}{k \sqrt{d}}=\frac{\pi^{2}}{6}\prod_{p|k}\bigg(1-\frac{1}{p^{2}}\bigg)+\sum_{r=-\infty}^{\infty}A_{r}e^{\frac{\pi ir}{k}}.$ Let’s rearrange this, to get that $2 \log \epsilon_{k} h(k) h'(k)-\frac{\pi k \sqrt{d}}{6}\prod_{p|k}\bigg(1-\frac{1}{p^{2}}\bigg)=\frac{k \sqrt{d}}{\pi}\sum_{r=-\infty}^{\infty}A_{r}e^{\frac{\pi ir}{k}}.$ I promise we’re almost there now!

Again, I will omit the calculations, but if we take the absolute value of each side of the above, we find the value of $\frac{k \sqrt{d}}{\pi}\sum_{r=-\infty}^{\infty}\lvert A_{r} \rvert,$ where we have already bounded the $A_{r}$ by considering the Fourier Series. We find that $\frac{k \sqrt{d}}{\pi}\sum_{r=-\infty}^{\infty}|A_{r}| \leq 36 k e^{-\frac{\pi \sqrt{d}}{k}}$ , and hence it can be seen that $|2 \log (\epsilon_{k}) h(k) h'(k)-\frac{\pi k \sqrt{d}}{6}\prod_{p|k}\bigg(1-\frac{1}{p^{2}}\bigg)| \leq 36 k e^{-\frac{\pi \sqrt{d}}{k}}.$

We now pick some clever values of $k$ ; if we take $k=21, 33$ , then the respective fields have class number one, and thus we can find $| \log (\epsilon_{21}) h'(21)-\frac{32}{21}\pi \sqrt{d}| \leq 378 e^{-\frac{\pi \sqrt{d}}{21}}$ and $| \log (\epsilon_{33}) h'(33)-\frac{80}{33}\pi \sqrt{d}| \leq 594 e^{-\frac{\pi \sqrt{d}}{33}}.$ This looks more promising! We have a linear form in logarithms, we can start to see how Baker’s Theorem can be used!

We combine the above to see that
$\lvert 35\log (\epsilon_{21}) h'(21)-22 \log (\epsilon_{33}) h'(33)\rvert<57e^{\frac{-\pi \sqrt{d}}{100}}.$

We can use the class number formulas to give $h'(21)$ and $h'(33)$ in terms of $d$ .

This puts us in a position to use Baker’s theorem to say that
$\lvert 35\log (\epsilon_{21}) h'(21)-22 \log (\epsilon_{33}) h'(33)\rvert > d^{-C}$ where $C$ is an effectively computable constant. By Dixon’s arguments mentioned above, we can assume $d$ is greater than the heights of $\epsilon_{21}, \epsilon_{33}$ .

Using this, we can find a lower bound on $-d$ . Using Baker’s Theorem as stated, we get a bound of around $-10^{20}$ . Explicitly, what we have managed to show is that if there is another imaginary quadratic field with class number 1, $K=\mathbb{Q}(\sqrt{-d})$ , then $d< 10^{20}$ . We can then use a computer to check fields down to this bound, and we only find the 9 imaginary quadratics found by Dixon.

Some general comments I’d feel remiss to not mention (though there are LOTS of comments needed to make this all fully rigorous, see my next addendum post!): 21 and 33 are chosen in the proof as they are small enough to make calculations manageable. You could take any two positive quadratic fields with class number one and the argument would work. Further, I’ve attributed this all to Baker; his original proof actually used 3 logarithmic terms. Stark pointed out the method could be reduced to two logarithmic terms.

Heegner actually proved the class number one problem (ish) before Baker, but used modular forms, and his proof wasn’t fully understood at the time (or quite complete, Stark filled some gaps). This method is beautiful, ands doesn’t rely on the manipulations we’ve used above. However, Baker’s method holds a special place in my heart; a lot of my research is around applications of Baker’s Theorem, and this is where I first met it.

So coming up, an addendum where I talk about some cool maths to do with this but I thought would take away from the flow of the proof. In the meantime, any questions, comments etc, please drop me a tweet!

N.B. A lot of this material comes from my master’s project, which was supervised by Dr. Martin Orr, to whom I owe many thanks.

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