Jekyll2021-06-03T18:34:36+00:00https://oxyde.ga/feed.xmlOxideOxideThe class number formula2021-04-30T00:00:00+00:002021-04-30T00:00:00+00:00https://oxyde.ga/2021/04/30/Class-number<h1 id="introduction">Introduction</h1> <p>Factorisation is an essential idea in number theory. After learning the basics of abstract algebra, one learns that it’s possible to factorise an element of any commutative ring, not just integers. In an arbitrary ring however, a factorisation is not necessarily unique.</p> <p>There is a theorem which relates unique factorisation in a ring to the value of a $\zeta$ function. It also helps in determining whether the ring has unique factorisation if you can calculate the value of that $\zeta$ function.<sup id="fnref:approx" role="doc-noteref"><a href="#fn:approx" class="footnote" rel="footnote">1</a></sup></p> <p>I’ll describe the rings, then the $\zeta$ function, and then the connection between them.</p> <h1 id="fields">Fields</h1> <p>Let $K = \mathbb{Q}(\sqrt{m} )$ where $m &lt; 0$ is a square-free integer. It consists of numbers of the form $a+b\sqrt{m}$, where $a, b \in \mathbb{Q}$. The field $K$ is called an <em>imaginary quadratic field</em>, since it is a degree 2 extension of $\mathbb{Q}$ given by $K = \mathbb{Q}[x]/(x^2 - m)$.</p> <p>It also has a <em>ring of integers</em> $\mathcal{O}_K$, which is the analogue of $\mathbb{Z}$ for this field. In fact, $\mathcal{O}_K$ is the <a href="https://en.wikipedia.org/wiki/Integral_element#Integral_closure_in_algebraic_number_theory">integral closure</a> of $\mathbb{Z}$ in $K$.</p> <p>Unfortunately, the ring $\mathcal{O}_K$ is not always a unique factorisation domain.<sup id="fnref:lame" role="doc-noteref"><a href="#fn:lame" class="footnote" rel="footnote">2</a></sup> For example, in $\mathbb{Z}[\sqrt{-5}]$ we have $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$. The <em>class number</em> measures exactly how much unique factorisation fails in $\mathcal{O}_K$.<sup id="fnref:classnumber" role="doc-noteref"><a href="#fn:classnumber" class="footnote" rel="footnote">3</a></sup></p> <p>Now, let $N = \begin{cases} |m| &amp; \text{if } m \equiv 1 \mod{4} \\ 4|m| &amp; \text{if } m \equiv 2,3\mod{4} \end{cases}$</p> <p>The number $N$ is known as the <em>discriminant</em> of the number field. We can define a group homomorphism $\chi : (\mathbb{Z}/N\mathbb{Z})^\times \to {\pm 1}$ as follows: for any prime $p$ not dividing $m$, put $\chi(p) = \left( \frac{m}{p} \right)$ (This is the Legendre symbol, which outputs $1$ if $m$ has a square root in $\mathbb{F}_p$ and $-1$ otherwise). Using <a href="https://en.wikipedia.org/wiki/Quadratic_reciprocity">quadratic reciprocity</a>, we can extend to all integers coprime to $m$.</p> <h1 id="l-functions">L functions</h1> <p>We then form the series $L(s,\chi) = \sum_{n=1}^{\infty} \frac{\chi(n)}{n^s}.$ where $\chi(n)$ is $\chi(n \mod N)$ if $n$ is coprime to $N$, and $0$ otherwise. This type of series, which we can define for any homomorphism $\chi : (\mathbb{Z}/n\mathbb{Z})^\times \to \mathbb{C}^\times$, is known as an $L$-function<sup id="fnref:dirich" role="doc-noteref"><a href="#fn:dirich" class="footnote" rel="footnote">4</a></sup>, and is an extension of the Riemann $\zeta$ function.<sup id="fnref:riemann" role="doc-noteref"><a href="#fn:riemann" class="footnote" rel="footnote">5</a></sup></p> <p>For example, if $m = -1$, then $N=4$. We have $\chi(1 \mod 4) = 1$ and $\chi(3 \mod 4) = -1$. So, $L(s,\chi) = 1 - \frac{1}{3^s} + \frac{1}{5^s} - \frac{1}{7^s} + \dots$ And in particular, $L(1,\chi) = 1 - \frac13 +\frac15 - \frac17 + \dots = \frac\pi4$</p> <h1 id="the-theorem">The theorem</h1> <p>Here is the main theorem, called the class number formula for quadratic imaginary fields.</p> <blockquote> <p>For an imaginary quadratic field $K$, let $h_K$ be its class number and $w_K$ the number of roots of $1$ in $K$. Then: $h_K = \frac{w_K}{2} L(0,\chi) = \frac{w_K \sqrt{N} }{2\pi} L(1,\chi).$</p> </blockquote> <p>So, whether or not the ring of integers is a unique factorisation domain depends on the value of a certain $\zeta$ function.</p> <p>For example, let’s again take $m = -1$, so $K = \mathbb{Q}(\sqrt{-1} )$. The roots of unity in this field are $1, -1, \sqrt{-1}$, and $-\sqrt{-1}$, so $w_k = 4$. Using the theorem, we get: $h_K = \frac{w_K \sqrt{N}}{2\pi} L(1,\chi) = \frac{4 \cdot 2}{2\pi} L(1,\chi) = \frac{4}{\pi} \cdot L(1,\chi)$ Here $L(1,\chi) = 1 - \frac{1}{3} + \frac{1}{5} - \dots = \frac{\pi}{4}$, so $h_K = 1$. Therefore $\mathcal{O}_K = \mathbb{Z}[\sqrt{-1}]$ is a principal ideal domain. It is striking that we took a detour through analysis to prove an algebraic fact!</p> <hr /> <div class="footnotes" role="doc-endnotes"> <ol> <li id="fn:approx" role="doc-endnote"> <p>An exact value is not necessary, sometimes even an approximation suffices. <a href="#fnref:approx" class="reversefootnote" role="doc-backlink">&#8617;</a></p> </li> <li id="fn:lame" role="doc-endnote"> <p>If $\mathcal{O}_K$ were always a unique factorisation domain, then Fermat’s last theorem would have been proved almost 150 years earlier than it was! In 1847, <a href="https://en.wikipedia.org/wiki/Gabriel_Lam%C3%A9">Gabriel Lamé</a> gave a <a href="https://math.stackexchange.com/q/953462">wrong</a> <a href="https://gallica.bnf.fr/ark:/12148/bpt6k29812/f310.item">proof</a> that relied on $\mathbb{Z}[\zeta_n]$ being a unique factorisation domain, where $\zeta_n$ is a primitive $n$th root of unity. <a href="#fnref:lame" class="reversefootnote" role="doc-backlink">&#8617;</a></p> </li> <li id="fn:classnumber" role="doc-endnote"> <p>It is the order of a certain group known as the <a href="https://en.wikipedia.org/wiki/Ideal_class_group">ideal class group</a>. When the order is $1$, $\mathcal{O}_K$ is a unique factorisation domain. <a href="#fnref:classnumber" class="reversefootnote" role="doc-backlink">&#8617;</a></p> </li> <li id="fn:dirich" role="doc-endnote"> <p>It’s a <a href="https://en.wikipedia.org/wiki/Dirichlet_L-function">Dirichlet $L$-function</a> to be precise; there are many types of $L$-functions in number theory. <a href="#fnref:dirich" class="reversefootnote" role="doc-backlink">&#8617;</a></p> </li> <li id="fn:riemann" role="doc-endnote"> <p>The Riemann $\zeta$ function is an $L$-function with $N=2$, because $(\mathbb{Z}/2\mathbb{Z})^\times = {1}$ and the homomorphism must map $1$ to $1$. <a href="#fnref:riemann" class="reversefootnote" role="doc-backlink">&#8617;</a></p> </li> </ol> </div>OxideIntroduction Factorisation is an essential idea in number theory. After learning the basics of abstract algebra, one learns that it’s possible to factorise an element of any commutative ring, not just integers. In an arbitrary ring however, a factorisation is not necessarily unique.Grothendieck and Einstein on greatness1918-04-23T00:00:00+00:001918-04-23T00:00:00+00:00https://oxyde.ga/1918/04/23/Einstein<p>An excerpt of Albert Einstein’s 1918 address to the Physical Society in Berlin on the occasion of Planck’s 60th birthday (the emphasis is my addition):</p> <blockquote> <p>Nobody who has really gone deeply into the matter will deny that in practice the world of phenomena uniquely determines the theoretical system, in spite of the fact that there is no logical bridge between phenomena and their theoretical principles; this is what Leibnitz described so happily as a “<a href="https://en.wikipedia.org/wiki/Pre-established_harmony">pre-established harmony</a>.” Physicists often accuse epistemologists of not paying sufficient attention to this fact. Here, it seems to me, lie the roots of the controversy carried on some years ago between Mach and Planck.<sup id="fnref:mach" role="doc-noteref"><a href="#fn:mach" class="footnote" rel="footnote">1</a></sup></p> <p>The longing to behold this pre-established harmony is the source of the inexhaustible patience and perseverance with which Planck has devoted himself, as we see, to the most general problems of our science, refusing to let himself be diverted to more grateful and more easily attained ends. I have often heard <strong>colleagues try to attribute this attitude of his to extraordinary will-power and discipline – wrongly, in my opinion. The state of mind which enables a man to do work of this kind is akin to that of the religious worshiper or the lover; the daily effort comes from no deliberate intention or program, but straight from the heart.</strong></p> </blockquote> <p>Alexander Grothendieck had a similar view, expressed in the preface of <a href="https://en.wikipedia.org/wiki/Alexander_Grothendieck#Manuscripts_written_in_the_1980s"><em>Récoltes et Semailles</em></a>:</p> <blockquote> <p><strong>Il est vrai aussi que l’ambition la plus dévorante est impuissante à découvrir le moindre énoncé mathématique, ou à le démontrer</strong>—tout comme elle est impuissante (par exemple) à “faire bander” (au sens propre du terme). Qu’on soit femme ou homme, ce qui “faitbander” n’est nullement l’ambition, le désir de briller, d’exhiber une puissance, sexuelle en l’occurence—bien au contraire!</p> <p>Mais c’est la perception aiguë de quelque chose de fort,de très réel et de très délicat à la fois. On peut l’appeler “la beauté”, et c’est là un des mille visages de cette chose-là. D’être ambitieux n’empêche pas forcément de sentir parfois la beauté d’un être, ou d’une chose, d’accord. Mais ce qui est sûr, c’est que ce n’est <em>pas</em> l’ambition qui nous la fait sentir…</p> </blockquote> <p>Translation:</p> <blockquote> <p><strong>It is also the case that the most totally consuming ambition is powerless to make or to demonstrate the simplest mathematical discovery</strong>—even as it is powerless (for example) to “score” (in the vulgar sense). Whether one is male or female, that which allows one to ‘score’ is not ambition, the desire to shine, to exhibit one’s prowess, sexual in this case. Quite the contrary!</p> <p>What brings success in this case is the acute perception of the presence of something strong, very real and at the same time very delicate. Perhaps one can call it “beauty”, in its thousand-fold aspects. That someone is ambitious doesn’t mean that one cannot also feel the presence of beauty in them; but it is <em>not</em> the attribute of ambition which evokes this feeling…</p> </blockquote> <hr /> <div class="footnotes" role="doc-endnotes"> <ol> <li id="fn:mach" role="doc-endnote"> <p>Ernst Mach did not believe in the physical reality of atoms, in opposition to Planck. Even in 1910, 5 years since Einstein’s Brownian motion paper, Mach wrote:</p> <blockquote> <p>The essential difference between us concerns belief in the reality of atoms… My answer is simple: If belief in the reality of atoms is so crucial, then I renounce the physical way of thinking, I will not be a professional physicist, and I hand back my scientific reputation. In short, thank you so much for the community of believers, but for me freedom of thought comes first.</p> </blockquote> <p><a href="#fnref:mach" class="reversefootnote" role="doc-backlink">&#8617;</a></p> </li> </ol> </div>OxideAn excerpt of Albert Einstein’s 1918 address to the Physical Society in Berlin on the occasion of Planck’s 60th birthday (the emphasis is my addition):