Set up FLT example
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example/02_first_reduction.typ
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example/02_first_reduction.typ
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#import "common.typ": *
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= First reductions of the problem <first-reduction>
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== Overview
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The proof of Fermat's Last Theorem is by contradiction. We assume that we have a
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counterexample $a^n + b^n = c^n$, and manipulate it until it satisfies the
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axioms of a "Frey package". From the Frey package we build a Frey curve --- an
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elliptic curve defined over the rationals. We then look at a certain
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representation of a Galois group coming from this elliptic curve, and finally
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using two very deep and independent theorems (one due to Mazur, the other due to
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Wiles) we show that this representation is neither reducible or irreducible, a
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contradiction.
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== Reduction to $n >= 5$ and prime
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#lemma(
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lean: <FermatLastTheorem.of_odd_primes>,
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)[
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If there is a counterexample to Fermat's Last Theorem, then there is a
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counterexample $a^p + b^p = c^p$ with an odd prime $p$.
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][
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Note: this proof is #link(
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"https://leanprover-community.github.io/mathlib4_docs/Mathlib/NumberTheory/FLT/Four.html#FermatLastTheorem.of_odd_primes",
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)[in mathlib already];; we run through it for completeness' sake.
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Say $a^n + b^n = c^n$ is a counterexample to Fermat's Last Theorem. Every
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positive integer is either a power of 2 or has an odd prime factor. If
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$n = k p$ has an odd prime factor $p$ then $(a^k)^p + (b^k)^p = (c^k)^p$ is
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the counterexample we seek. It remains to deal with the case where $n$ is a
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power of 2, so let's assume this. We have $3 <= n$ by assumption, so $n = 4 k$
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must be a multiple of 4, and thus $(a^k)^4 + (b^k)^4 = (c^k)^4$, giving us a
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counterexample to Fermat's Last Theorem for $n = 4$. However an old result of
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Fermat himself (proved as #link(
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"https://leanprover-community.github.io/mathlib4_docs/Mathlib/NumberTheory/FLT/Four.html#fermatLastTheoremFour",
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)[`fermatLastTheoremFour`] in mathlib) says that $x^4 + y^4 = z^4$ has no
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solutions in positive integers.
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]
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Euler proved Fermat's Last Theorem for $p = 3$;
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#lemma(
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lean: <fermatLastTheoremThree>,
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)[
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There are no solutions in positive integers to $a^3 + b^3 = c^3$.
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][
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The proof in mathlib was formalized by a team from the "Lean For the Curious
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Mathematician" conference held in Luminy in March 2024 (its dependency graph
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can be visualised #link(
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"https://pitmonticone.github.io/FLT3/blueprint/dep_graph_document.html",
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)[here]).
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]
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#corollary(
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lean: <FLT.of_p_ge_5>,
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uses: (<fermatLastTheoremThree>, <FermatLastTheorem.of_odd_primes>),
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)[
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If there is a counterexample to Fermat's Last Theorem, then there is a
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counterexample $a^p + b^p = c^p$ with $p$ prime and $p >= 5$.
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][
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Follows from the previous two lemmas.
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]
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== Frey packages
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For convenience we make the following definition.
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#definition(
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lean: <FLT.FreyPackage>,
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)[
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A _Frey package_ $(a, b, c, p)$ is three pairwise-coprime nonzero integers
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$a$, $b$, $c$, with $a eq.triple 3 thick (mod 4)$ and
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$b eq.triple 0 thick (mod 2)$, and a prime $p >= 5$, such that
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$a^p + b^p = c^p$.
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]
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Our next reduction is as follows:
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#lemma(
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lean: <FLT.FreyPackage.of_not_FermatLastTheorem_p_ge_5>,
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uses: (<FLT.FreyPackage>,),
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)[
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If Fermat's Last Theorem is false for $p >= 5$ and prime, then there exists a
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Frey package.
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][
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Suppose we have a counterexample $a^p + b^p = c^p$ for the given $p$; we now
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build a Frey package from this data.
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If the greatest common divisor of $a, b, c$ is $d$ then $a^p + b^p = c^p$
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implies $(a/d)^p + (b/d)^p = (c/d)^p$. Dividing through, we can thus assume
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that no prime divides all of $a, b, c$. Under this assumption we must have
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that $a, b, c$ are pairwise coprime, as if some prime divides two of the
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integers $a, b, c$ then by $a^p + b^p = c^p$ and unique factorization it must
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divide all three of them. In particular we may assume that not all of
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$a, b, c$
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are even, and now reducing modulo 2 shows that precisely one of them must be
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even.
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Next we show that we can find a counterexample with $b$ even. If $a$ is the
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even one then we can just switch $a$ and $b$. If $c$ is the even one then we
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can replace $c$ by $-b$ and $b$ by $-c$ (using that $p$ is odd).
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The last thing to ensure is that $a$ is 3 mod 4. Because $b$ is even, we know
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that $a$ is odd, so it is either 1 or 3 mod 4. If $a$ is 3 mod 4 then we are
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home; if however $a$ is 1 mod 4 we replace $a, b, c$ by their negatives and
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this is the Frey package we seek.
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]
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== Galois representations and elliptic curves
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To continue, we need some of the theory of elliptic curves over $QQ$. So let
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$f(X)$ denote any monic cubic polynomial with rational coefficients and whose
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three complex roots are distinct, and let us consider the equation
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$E : Y^2 = f(X)$, which defines a curve in the $(X, Y)$ plane. This curve (or
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strictly speaking its projectivisation) is a so-called elliptic curve (or an
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elliptic curve over $QQ$ if we want to keep track of the field where the
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coefficients of $f(X)$ lie). More generally if $k$ is any field then there is a
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concept of an elliptic curve over $k$, again defined by a (slightly more
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general) plane cubic curve $F(X, Y) = 0$.
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If $E$ is an elliptic curve over a field $k$, and if $K$ is any field which is a
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$k$-algebra, then we write $E(K)$ for the set of solutions to $y^2 = f(x)$with
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$x, y in K$, together with an additional "point at infinity". It is an
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extraordinary fact, and not at all obvious, that $E(K)$ naturally has the
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structure of an additive abelian group, with the point at infinity being the
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zero element (the identity). Fortunately this fact is already in mathlib. We
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shall use $+$ to denote the group law. This group structure has the property
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that three distinct points $P, Q, R in K^2$ which are in $E(K)$ will sum to zero
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if and only if they are collinear.
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The group structure behaves well under change of field.
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#lemma(
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lean: <WeierstrassCurve.Points.map>,
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)[
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If $E$ is an elliptic curve over a field $k$, and if $K$ and $L$ are two
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fields which are $k$-algebras, and if $f : K -> L$ is a $k$-algebra
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homomorphism, the map from $E(K)$ to $E(L)$ induced by $f$ is an additive
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group homomorphism.
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][
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The equations defining the group law are ratios of polynomials with
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coefficients in $k$, and such things behave well under $k$-algebra
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homomorphisms.
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]
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This construction is functorial (it sends the identity to the identity, and
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compositions to compositions).
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#lemma(
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lean: <WeierstrassCurve.Points.map_id>,
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uses: (<WeierstrassCurve.Points.map>,),
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)[
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The group homomorphism $E(K) -> E(K)$ induced by the identity map $K -> K$ is
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the identity group homomorphism.
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][
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An easy calculation.
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]
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#lemma(
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lean: <WeierstrassCurve.Points.map_comp>,
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uses: (<WeierstrassCurve.Points.map>,),
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)[
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If $K -> L -> M$ are $k$-algebra homomorphisms then the group homomorphism
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$E(K) -> E(M)$
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induced by the map $K -> M$ is the composite of the map $E(K) -> E(L)$ induced
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by $K -> L$ and the map $E(L) -> E(M)$ induced by the map $L -> M$.
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][
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Another easy calculation.
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]
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Thus if $f : K -> L$ is an isomorphism of fields, the induced map $E(K) -> E(L)$
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is an isomorphism of groups, with the inverse isomorphism being the map
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$E(L) -> E(K)$ induced by $f^-1$. This construction thus gives us an action of
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the multiplicative group $"Aut"_k(K)$ of $k$-automorphisms of the field $K$ on
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the additive abelian group $E(K)$.
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#definition(
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lean: <WeierstrassCurve.galoisRepresentation>,
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uses: (<WeierstrassCurve.Points.map_id>, <WeierstrassCurve.Points.map_comp>),
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)[
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If $E$ is an elliptic curve over a field $k$ and $K$ is a field and a
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$k$-algebra, then the group of $k$-automorphisms of $K$ acts on the additive
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abelian group $E(K)$.
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]
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In particular, if $overline(QQ)$ denotes an algebraic closure of the rationals
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(for example, the algebraic numbers in $CC$) and if
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$"Gal"(overline(QQ) slash QQ)$ denotes the group of field isomorphisms
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$overline(QQ) -> overline(QQ)$, then for any elliptic curve $E$ over $QQ$ we
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have an action of $"Gal"(overline(QQ) slash QQ)$ on the additive abelian group
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$E(overline(QQ))$.
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We need a variant of this construction where we only consider the $n$-torsion of
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the curve, for $n$ a positive integer. Recall that if $A$ is any additive
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abelian group, and if $n$ is a positive integer, then we can consider the
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subgroup $A[n]$ of elements $a$ such that $n a = 0$. If a group $G$ acts on $A$
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via additive group isomorphisms, then there will be an induced action of $G$ on
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$A[n]$.
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#definition(
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lean: <WeierstrassCurve.torsionGaloisRepresentation>,
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uses: (<WeierstrassCurve.galoisRepresentation>,),
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)[
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If $E$ is an elliptic curve over a field $k$ and $K$ is a field and a
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$k$-algebra, and if $n$ is a natural number, then the group of
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$k$-automorphisms of $K$ acts on the additive abelian group $E(K)[n]$ of
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$n$-torsion points on the curve.
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]
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If furthermore $n = p$ is prime, then $A[p]$ is naturally a vector space over
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the field $ZZ slash p ZZ$, and thus it inherits the structure of a mod $p$
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representation of $G$. Applying this to the above situation, we deduce that if
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$E$ is an elliptic curve over $QQ$ then $"Gal"(overline(QQ) slash QQ)$ acts on
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$E(overline(QQ))[p]$ and this is the _mod $p$ Galois representation_ attached to
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the curve $E$.
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In the next section we apply this theory to an elliptic curve coming from a
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counterexample to Fermat's Last theorem.
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