#fermatslasttheorem

[mercurialblack]

Yo mama so fat she couldn't fit in Wiles' proof of Fermat's Last Theorem

[sciencetarot]

9 The Hermit: Fermat's Last Theorem Pierre de Fermat was a 17th-century French mathematician with a habit of taking notes in the margins of books he was reading. One such note, found after his death, announced that he had discovered a marvelous mathematical proof "which this margin is too narrow to contain." By not writing the solution, he left a tempting and frustrating offer for generations afterward. Many mathematicians struggled to produce Fermat's proof, consuming lifetimes. The proof was finally achieved in 1995 after centuries of persistent work done without assurance that a solution truly existed. The Hermit represents the focus and isolation of Fermat's studies, and of those who worked to solve his mystery. Many discoveries and insights of nature and of ourselves can be developed only in nurturing solitude, separated from distraction and fulfilled through thought and study.

[fermat4thewin]

Robert Light Duncan

aka: Bobby “Hajji” Duncan, Hajji Booby, Old Dawg Duncan, and so on.

[reflections-in-t]

『コンタクト』よりも『インターステラー』よりもサスペンスフルで斬新、タルコフスキーよりも美しく静謐な映像、そして『2001年宇宙の旅』よりも哲学的で円環的な奥行きのある世界観… . ファーストコンタクト系SF映画の最高峰に位置しそうな『メッセージ／Arrival』 . 退院後はじめて観た長編映画が今年最上の素晴らしい作品で、幸福です。 . . #Arrival #メッセージ #ドゥニヴィルヌーヴ #DenisVilleneuve #エイミーアダムス #AmyAdams #ジェレミーレナー #JeremyRenner #フォレストウィテカー #ForestWhitaker #ヘプタポッド #サピアウォーフの仮説 #SapirWhorfHypothesis #フェルマーの最小時間の #FermatsLastTheorem #フィボナッチ数列 #FibonacciNumber #あなたの人生の物語 #StoryOfYourLife #テッドチャン #TedChiang #プリズナーズ #Prisoners #複製された男 #Enemy #ボーダーライン #Sicario #BladeRunner2049 #AmazonVideo . .

[local-to-global-blog]

A note on how Wiles’ 3-5 Switch works.

The main theorem that Wiles proved actually could not prove Fermat’s Last Theorem or the Taniyama-Shimura Conjecture on its own. The main theorem states that if $E$ is a rational elliptic curve that is semistable at an odd prime $p$ and whose mod $p$ representation, $\overline{\rho}_{E,p}:G_{\mathbb{Q}}\rightarrow GL_2(\mathbb{F}_p)$, is irreducible and modular, then $E$ itself is modular. Proving this is the bulk of the work in Wiles’ proof. But this doesn’t say that any particular elliptic curve is modular, it just gives us a way to lift mod $p$ modularity to elliptic modularity. To finish off Fermat’s Last Theorem, he needed to use this to show that any sufficiently semistable elliptic curve was modular.

The trick is to use a theorem of Langlands and Tunnell about automorphic representations. The gist of their theorem is that the structure of $GL_2(\mathbb{F}_3)$ is simple enough to ensure that every mod 3 representation is, in fact, modular. Wiles idea is then to take a semistable elliptic curve and use this theorem to show that the mod 3 representation is modular, and then use his main result to conclude that the original elliptic curve is itself modular. But to apply his Main Result, we would have to require that the mod 3 representation be irreducible. This is not always the case. So Langlands and Tunnell will only work with elliptic curves whose mod 3 representation is irreducible. The theorem of Langlands and Tunnell is very complex and difficult, to circumvent it Wiles would seemingly have to prove an even deeper result than they, a modularity result for representations mod $p$ for $p>3$, which would be a feat on its own.

Instead, he came up with the 3-5 Switch.

The idea is to show that when the mod 3 representation is not irreducible, then the mod 5 representation will be irreducible. Moreover, we can temporarily work with an auxiliary elliptic curve that has the same mod 5 representation and is semistable at 5, but we can prove is modular, implying that the mod 5 representation of our original elliptic curve is semistable at 5, irreducible and modular, hence the curve itself is modular. This will prove it for enough elliptic curves to get Fermat’s Last Theorem.

Mod 3 Reducible implies Mod 5 Irreducible (for curves semistable at 3 and 5)

The first step in making this work is to show that if $E$ is an elliptic curve semistable at 3 and 5 whose mod 3 representation is not irreducible, then its mod 5 representation is irreducible.

We can picture the situation using the image for this post. The idea is that if the mod p representation is reducible, then the action of $G_{\mathbb{Q}}$ on the elliptic curve will have a cyclic subgroup of order $p$ that is $G_{\mathbb{Q}}$-invariant. So, if $E$ is reducible at 3 and 5, then it will have a cyclic subgroup of order 15 that is fixed under $G_{\mathbb{Q}}$. We then consider the modular curve $X_0(15)(\mathbb{Q})$ whose points correspond to pairs $(E,C_{15})$ of rational elliptic curves where $C_{15}$ is a cyclic subgroup of $E$ fixed by $G_{\mathbb{Q}}$. This modular curve is actually very concrete and we can do very explicit computations on it. Particularly, we find that there are only four rational points on it that are not cusps. Moreover, we can find the elliptic curves associated to each point and this will show that none of them are semistable at 5. This means that the curves we’re interested in, which we assume to be semistable at 5, cannot be represented by any of these points. It then follows that our elliptic curve must be irreducible mod 5.

Reducible Mod 3 and semistable at 3 and 5 implies Mod 5 is Modular

That was the easy part. Parameterize all possible counterexamples, show that our curve cannot be in the parameterization. Next we need to show that if $E$ is a rational elliptic curve, semistable at 3 and 5, reducible mod 3 and irreducible mod 5, then it is, in fact, modular mod 5. The idea is to use an auxiliary elliptic curve that is close enough to $E$, that we know is modular and then transfer this modularity onto $E$.

We look at a different modular curve,  $X(E,5)(\mathbb{Q})$.  The points of this will correspond to pairs $(E’,\varphi)$, where $E’$ is an elliptic curve and $\varphi$ is an isomorphism between the mod 5 representation of $E$ and the mod 5 representation of $E’$. The elliptic curve we started with is actually on this modular curve through the point $(E,\textrm{id})$.   In fact, $X(E,t)(\mathbb{Q})\simeq P^1(\mathbb{Q})$,  so it has infinitely many points. 

We note that if $(E’,\varphi)$ is 5-adically “close enough” to $(E,\textrm{id})$ then $E’$ will inherit many properties of $E$. Particularly, we will get that $E’$ is semistable at 3 and 5. Of note is the fact that there are infinitely many such $E’$ that are “close enough” to $(E,\textrm{id})$ on $P^1(\mathbb{Q})$. Our idea is then to pick an $(E’,\varphi)$ close enough to $(E,\textrm{id})$ so that we retain semistability, but that is also irreducible mod 3 and hence modular by Wiles main result.

How do we know that we can do this? Let’s parameterize all the troublesome points. Let $X(E,5;3)(\mathbb{Q})$ be all triples $(E’,\varphi, C_3)$ of rational elliptic curves with an isomorphism between mod 5 representations, but with the additional $G_{\mathbb{Q}}$-fixed subgroup $C_3$. This happens only when the mod 3 representation of $E’$ is reducible. Note that by just forgetting last entry into the triple we get a map  $X(E,5;3)(\mathbb{Q})\rightarrow X(E,5)(\mathbb{Q})$. We want to ensure that we can avoid the image of this when we choose $E’$.

The modular curve  $X(E,5;3)(\mathbb{Q})$ is fairly concrete and we can compute its genus as being bigger than one. Falting’s Theorem then says that this curve has only finitely many rational points. Hence the image in $X(E,5)(\mathbb{Q})$ is finite. Since we can choose from infinitely many points in $X(E,5)(\mathbb{Q})$, it follows that we can avoid these trouble points. 

This means that if $E$ is semistable at 3 and 5, reducible at 3 and irreducible at 5, then we can find an elliptic curve $E’$ that is semistable at 3 and 5, whose mod 5 representation is isomorphic to the mod 5 representation of $E$ and whose mod 3 representation is irreducible. The theorem of Langlands and Tunnell shows that the mod 3 representation is modular, and since we have semistability and irreducibility, Wiles’ main result tells us that $E’$ is modular. Hence the mod 5 representation of $E’$ is modular. Hence the mod 5 representation of $E$ is modular. Since, then $E$ is semistable at 5, and the mod 5 representation is irreducible and modular, Wiles’ main result, again, tells us that $E$ itself is modular. This then finally proves Fermat’s Last Theorem.

Many, many details are glossed over in this, but it is a really fun result and it’s good to have a broad overview of what is happening when trying to figure it out in more detail. Since you do have a broad layout of it, you can look into more of the finer points that have been glossed over and try to it out from This Online Lecture by Karl Rubin, which covers everything pretty satisfyingly.