Elliptic curve

An elliptic curve over a field $K$ is a smooth projective curve of genus one equipped with a specified $K$-rational point $O$. The chord-and-tangent law makes its rational points an abelian group with identity $O$.

For an elliptic curve $E/K$, regard $E(K^{\mathrm{sep}})$ as a discrete Galois module for the absolute Galois group $G_K$. Its continuous Galois cohomology has $H^0(K,E)=E(K)$; $H^1(K,E)$ classifies torsors under $E$. For an integer $n$ invertible in $K$, the Kummer exact sequence of an elliptic curve gives $0\to E(K)/nE(K)\to H^1(K,E[n])\to H^1(K,E)[n]\to 0$. If $[n]Q=P$, the first map sends $P$ to $\sigma \mapsto \sigma Q-Q$. Changing $Q$ changes the cocycle by a coboundary; this class vanishes exactly when $P$ has a division point over $K$.

For $d\in {\mathbb{Q}}^{\times }$, the smooth projective curve $X^3+Y^3+d{Z}^3=0$, with identity $O=(1:-1:0)$, is a genus-one elliptic curve. Its partial derivatives have no common projective zero. The invertible linear projective transformationsends it to ${\mathcal{Y}}^2\mathcal{Z}={\mathcal{X}}^3-432d^2{\mathcal{Z}}^3$, taking $O$ to $(0:1:0)$. Indeed $(X+Y{)}^3+3(X+Y)(X-Y{)}^2=4(X^3+Y^3)=-4d{Z}^3$. In affine coordinates $u=X/Z,v=Y/Z$, this gives $x=-12d/(u+v)$ and $y=-36d(u-v)/(u+v)$, with inverse $u=(y-36d)/(6x)$, $v=(-y-36d)/(6x)$. Thus diagonal sums of two cubes are naturally studied through elliptic-curve addition formula.

There are infinitely many rational pairs $(u,v)$ with $u^3+v^3=7$. The corresponding Weierstrass equation of an elliptic curve is $y^2=x^3-21168$, and $(u,v)=(2,-1)$ maps to $P=(84,756)$. The primes $5$ and $11$ are good, with point counts $6$ and $12$ because cubing is bijective in their residue fields. Reduction of torsion points on an elliptic curve bounds rational torsion by order dividing $6$. The flexes of a diagonal cubic exclude three-torsion for $d=-7$, while $d/2=-7/2$ is not a cube and hence excludes two-torsion. Thus $P$ has infinite order.

All nonzero multiples of $P$ map to distinct rational affine pairs, since the only rational point at infinity of the diagonal cubic is its identity. The inverse map is $u=(y+252)/(6x)$ and $v=(252-y)/(6x)$.

For $d\in {\mathbb{Q}}^{\times }$ the rational torsion subgroup of $X^3+Y^3+d{Z}^3=0$ is cyclic of order $3$ if $d$ is a cube, cyclic of order $2$ if $d/2$ is a cube, and trivial otherwise.

To bound torsion, use the Weierstrass equation of an elliptic curve $y^2=x^3-432d^2$. At every good prime $\ell \equiv 2(\mathrm{mod}3)$, cubing is a bijection on ${\mathbb{F}}_{\ell }$, so the number of points is $\ell +1$. Reduction injects the prime-to-$\ell$ torsion, since multiplication by an integer prime to $\ell$ is invertible in the formal group of an elliptic curve. For any prime $q>3$, the Dirichlet theorem on primes in arithmetic progressions supplies a good $\ell \equiv 2(\mathrm{mod}3)$, $\ell \equiv 1(\mathrm{mod}q)$, excluding $q$-torsion. Good primes $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good primes $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point is killed by $6$.

The flexes of a diagonal cubic give rational three-torsion exactly when $d$ is a cube, and then it has order $3$. Rational two-torsion requires $x^3=432d^2$, whose sole possible rational root exists exactly when $d/2$ is a cube. This equivalence follows by comparing prime valuations modulo $3$, including the valuation at $2$. The two conditions cannot hold together because $2$ is not a rational cube. The asserted classification follows.

Let $r^3=d\ne 0$ and let $\zeta$ be a primitive cube root of unity. The nine inflection points of a plane cubic on $X^3+Y^3+d{Z}^3=0$ areThe Hessian criterion for flexes of a plane cubic gives the Hessian determinant $216dXYZ$, so every flex lies on one of the three coordinate lines. Their intersections give precisely the displayed points. With $(1:-1:0)$ as identity, these are exactly the three-torsion points are flexes of a plane cubic configurations: the tangent divisor at a flex is $3P$, giving $3(P-O)=0$.

Over $\mathbb{Q}$ only the first point with $j=0$ is rational on $Z=0$. The other two coordinate lines each give one rational point precisely when $d$ is a rational cube. Consequently the rational three-torsion subgroup is $\mathbb{Z}/3\mathbb{Z}$ when $d$ is a cube and is trivial otherwise.

Fix $O\in E(k)$. On a genus-one smooth projective curve, every degree-one divisor class has a unique effective point representative by the Riemann-Roch theorem: its section space has dimension one, since the complementary canonical-divisor degree is negative. Thus $P\mapsto [P-O]$ is a bijection onto the degree-zero Picard group of a curve. Transporting divisor-class addition proves the group axioms, with identity $O$, and characterizes $P+Q$ by $P+Q\sim R+O$.

Let $E$ be a smooth plane cubic with an inflection point $O$ as identity. A point $P$ with $3(P-O)=0$ in Picard group has a line whose intersection divisor with $E$ is $3P$. Indeed, ${\mathcal{O}}_E(3P)\cong {\mathcal{O}}_E(3O)\cong {\mathcal{O}}_E(1)$. On this genus-one curve, the Riemann-Roch theorem gives $h^0({\mathcal{O}}_E(1))=3$: the dual term has negative degree. Restriction of ambient linear forms is injective and its source also has dimension three, so it is an isomorphism. The section vanishing at $3P$ therefore comes from a line. Conversely, such a line gives $3P\sim 3O$, so $P$ is a three-torsion point. This includes $O$; points of exact order three are the other such points.

Every complex elliptic curve, as a complex Lie group, is the quotient of $\mathbb{C}$ by a rank-two lattice. The coordinate functions are built from the Weierstrass elliptic function and its derivative. Holomorphic group homomorphisms lift to scalar multiplication on $\mathbb{C}$, and their endomorphism condition is $aL\subseteq L$. This turns complex multiplication into a statement about lattice stabilizers.

The Weierstrass elliptic function and its derivative give an isomorphism $z\mapsto (\wp (z),{\wp }^{\prime }(z))$ from a one-dimensional complex torus to a smooth plane cubic, with $0$ sent to its identity at infinity. The three nonzero two-torsion points of a complex torus are zeros of ${\wp }^{\prime }$. Its triple pole implies that these three zeros are simple. If two half-period values of the Weierstrass elliptic function coincided, $\wp -e$ would have at least four zeros counted with multiplicity, contradicting its double pole. The three cubic roots are therefore distinct, proving smoothness of an algebraic variety. The degree-two map $\wp$ has fibers $\{z,-z\}$, distinguished by ${\wp }^{\prime }$ away from the half-periods, proving bijectivity. A line pulls back to an elliptic function with a triple pole, and the zero-pole sum of an elliptic function says that its three intersections sum to zero on the torus. This gives exactly the chord-and-tangent group law, including repeated intersections interpreted by intersection multiplicity.

An elliptic curve in characteristic zero admits no nonconstant rational map of projective varieties from the projective line. Over $\mathbb{C}$, write its equation as $y^2={\prod }_{j=1}^3(x-e_j)$ with distinct $e_j$. Substituting $x=u/v$ with coprime polynomials gives $(y{v}^2{)}^2=v{\prod }_j(u-e_{j}v)$. These four pairwise coprime factors must all be squares. The polynomial pencil with four square members forces $u,v$ constant, ruling out a nonconstant rational parametrization of an algebraic curve.

The j-invariant is unchanged by admissible changes of Weierstrass equation of an elliptic curve. With $b_2=a_1^2+4a_2$ and $b_4=a_1{a}_3+2a_4$, set $c_4=b_2^2-24b_4$ and $j=c_4^3/\Delta$, where $\Delta$ is the elliptic-curve discriminant. For good reduction over a local field, an integral equation with unit discriminant shows that $j$ lies in the valuation ring. Thus a negative valuation of $j$ proves bad reduction, even if the displayed equation has not been proved minimal.

Let $E$ be a genus one curve over an algebraically closed field and choose $O\in E$. The mapis an isomorphism. For a degree-zero divisor $D$, Riemann--Roch gives a nonzero section of $D+O$, so $D\sim P-O$ for some point $P$. If $P-O\sim P^{\prime }-O$ with $P\ne P^{\prime }$, a rational function with divisor $P-P^{\prime }$ would give a degree-one map to the projective line, contradicting that $E$ has genus one. The Abel--Jacobi map transports addition in the divisor class group to the elliptic-curve group law.

Over an algebraically closed field of characteristic different from two, a degree-two map from a smooth genus one curve to the projective line has a deck involution. After choosing an origin, all its fiber divisors on an algebraic curve are linearly equivalent and have the same group sum $s$, by the Abel-Jacobi map of a genus-one curve. The involution is consequently $p\mapsto s-p$, including its fixed ramification points. Composing two such involutions gives a translation on an elliptic curve.

On a smooth plane cubic with identity point $O$, a line through points $P$ and $Q$, interpreted as the tangent when $P=Q$, meets the cubic in a third point $R$. The chord-and-tangent group law defines $P+Q$ as the third intersection of the line through $R$ and $O$ with the cubic.

Assume the elliptic-curve discriminant is nonzero. Then $b\ne 0$, and tangency at $P$ gives $2P=(-b,b(a-1))$ and $-2P=(-b,0)$. The line through $P,2P$ has slope $1-a$ and gives $3P=(1-a,a-b-1)$, $-3P=(1-a,(1-a{)}^2)$. Consequently $P$ has exact order four when $a=1$, exact order five when $a=b+1$, and exact order six when $b=(a-1)(2-a)$. The nonsingularity condition excludes the zero values of $b$ that would otherwise invalidate these torsion conclusions.

For a smooth genus one curve with origin $O$, the map $P\mapsto [(P)-(O)]$ identifies geometric points with the degree-zero Picard group of a curve. Riemann-Roch theorem gives a unique effective representative for each divisor class of degree one: its Riemann-Roch space has dimension one. A line cuts a plane cubic in a divisor linearly equivalent to $3(O)$ when $O$ is the Weierstrass flex. The two successive line intersections in the chord-and-tangent group law therefore give addition of the corresponding divisor classes. Associativity follows by injectivity of the identification and associativity of divisor-class addition, including tangent intersection multiplicities.

For a fixed point $Q$ on an elliptic curve, translation is the automorphism $P\mapsto P+Q$. Its inverse is translation by $-Q$. Translations let local computations of an invariant differential on an elliptic curve control every point of the curve.

For a singular projective Weierstrass cubic, its smooth locus of a variety contains the point at infinity and carries the chord-and-tangent group law. The singular point is excluded. Over an algebraic closure, a node gives the multiplicative algebraic group and a cusp gives the additive group. A node whose branches are not defined over the base field gives a nonsplit one-dimensional algebraic torus. This group is distinct from an elliptic curve, whose projective model is smooth.

On $y^2=x^3+a_2{x}^2+a_{4}x+a_6$, let $\lambda$ be the secant slope through distinct points $P,Q$, or the tangent slope when $P=Q$. Then

On $y^2=x^3+ax+b$ in characteristic of a field other than two, take the elementary symmetric polynomials $s_i$ of $x(P),x(Q),x(P+Q)$. If the line through $P,Q$ is $y=\lambda x+\mu$, its third intersection is $-(P+Q)$, so coefficient comparison gives $s_1={\lambda }^2$, $s_2=a-2\lambda \mu$, $s_3={\mu }^2-b$. Hence $(s_2-a{)}^2=4s_1(s_3+b)$. This is an identity of rational functions; tangent cases follow by specialization, and coordinates at infinity are handled through the projective addition map.

A Weierstrass equation presents an elliptic curve aswith nonzero discriminant, together with the point at infinity $O$.

Over an algebraically closed field of characteristic different from two, a cubic Weierstrass model with distinct roots becomes this form by sending two roots to zero and one, then scaling $Y$ by a square root of the resulting nonzero coefficient. The third root becomes $\lambda \notin \{0,1\}$. The point $[0:1:0]$ is a flex and the curve is a genus one curve.

For $y^2={\prod }_{j=1}^3(x-{\lambda }_j)$ with distinct roots in characteristic different from two, the unique point at infinity has local coordinates $u=X/Y$, $v=Z/Y$ satisfying $v=u^3+O(u^5)$. Thus $x=u/v$ and $y=1/v$ have poles two and three. Regular functions off infinity have the unique form $A(x)+yB(x)$; the even and odd leading pole orders cannot cancel. A basis of $L(n{P}_{\infty })$ consists of $x^j$ with $2j\le n$ and $y{x}^j$ with $3+2j\le n$. This gives dimension one at $n=0$ and dimension $n$ for $n\ge 1$ without Riemann-Roch theorem.

Let $E$ be a genus one curve and choose $O\in E$. Since a canonical divisor is linearly equivalent to zero, Riemann--Roch gives $\ell (nO)=n$ for $n\ge 1$. Choose $x\in L(2O)\setminus L(O)$ and $y\in L(3O)\setminus L(2O)$. Their pole orders at $O$ are two and three. The seven functionsare linearly dependent because $\ell (6O)=6$, and comparison of pole orders puts the relation in Weierstrass form. The complete linear system $|3O|$ is very ample and realizes this relation as a smooth plane cubic. In characteristic other than two and three, completing the square and translating $x$ produce a Short Weierstrass form.

The discriminant of a Weierstrass equation is a polynomial $\Delta$ in its coefficients whose nonvanishing is equivalent to the projective cubic being smooth. For $y^2=x^3+Ax+B$,

An admissible change of Weierstrass coordinates has the formwhere $u$ is nonzero. These are exactly the coordinate changes preserving Weierstrass form and the point at infinity.

Over a field of characteristic other than two or three, every elliptic curve has a short Weierstrass equation $y^2=x^3+Ax+B$ with $4A^3+27B^2\ne 0$.

On $y^2=x^3+ax+b$ in characteristic other than two, the differentialextends regularly and without zeros across the point at infinity. Translation by every point preserves it.

On a nonsingular equation $y^2=x^3+a_2{x}^2+a_{4}x+a_6$ in characteristic different from two, $\omega =dx/(2y)$ is regular and nowhere zero. For a chord of slope $h$ giving $P+Q=(X,Y)$, differentiating the line-intersection identity yields $dX/dx=Y/y$, so $dX/(2Y)=dx/(2y)$. Extension across the exceptional cases proves invariance under every translation on an elliptic curve. The addition map consequently satisfies ${\mu }^{\ast }\omega ={\mathrm{pr}}_1^{\ast }\omega +{\mathrm{pr}}_2^{\ast }\omega$, and $[n{]}^{\ast }\omega =n\omega$.

Minimality here concerns the discriminant valuation of an integral Weierstrass equation of an elliptic curve over a discretely valued field.

Over a discretely valued field, a minimal Weierstrass equation has integral coefficients and has least possible discriminant valuation among integral Weierstrass equations related by admissible changes of variables.

Reducing a minimal integral Weierstrass equation modulo the maximal ideal gives a projective cubic over the residue field. Reducing primitive projective coordinates defines the reduction map on rational points.

An elliptic curve has good reduction at a finite place when its reduced minimal Weierstrass equation is nonsingular, equivalently when its minimal discriminant has valuation zero.

An integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant defines a smooth proper elliptic curve over the valuation ring. Properness identifies points over the local field with integral sections. The group law, constructed by adding degree-zero divisor classes, is a morphism over the valuation ring, and base change to the residue field preserves it. Hence reducing primitive projective coordinates is a group homomorphism, including exceptional affine addition cases.

A good reduction in characteristic $p$ is supersingular when its geometric $p$-torsion is trivial, equivalently its formal group of an elliptic curve has height two. Good inert primes of a CM elliptic curve are supersingular. The ordinary bounded-measure and torsion-Selmer formulation must then be modified, for instance using signed local conditions in the cyclotomic tower.

A good reduction in characteristic $p$ is ordinary when the geometric $p$-torsion has $p$ points, equivalently its formal group of an elliptic curve has height one. For a CM elliptic curve, a good prime split in its CM field is ordinary. One of the two roots of $X^2-a_{p}X+p$ is a $p$-adic unit, enabling the bounded ordinary p-adic L-function.

If an elliptic curve has good reduction over a local field $K$, its integral projective model is smooth. Every point over the residue field lifts by the Hensel lemma, applied in a chart with a unit partial derivative. Thus reduction gives an exact sequence $0\to E_1(K)\to E(K)\to \tilde{E}(k)\to 0$. The kernel of reduction of an elliptic curve is the maximal ideal of the valuation ring, with the formal group of an elliptic curve as its group operation.

For an elliptic curve with good reduction over a local field, the kernel of reduction consists of the points reducing to the identity. The parameter $t=-x/y$ identifies it with the maximal ideal equipped with the formal group of an elliptic curve.

On an integral Weierstrass equation of an elliptic curve with good reduction at a prime $p$, a nonidentity rational point reducing to the identity cannot have both coordinates integral at $p$. Indeed, an affine integral pair reduces to an affine point, distinct from the point at infinity. This can prove nonintegrality of a large linear combination without computing its coordinates. The nonidentity condition is essential and can often be checked using the known torsion subgroup.

At a prime of good reduction of an elliptic curve, the reduction map is a group homomorphism on all local points. If $N$ annihilates the reduced finite group, every $NP$ reduces to the identity at infinity. A finite point with integral affine coordinates reduces to an affine point, so a nonidentity $NP$ cannot have integral affine coordinates. The identity has no affine coordinates and must be treated separately.

An elliptic curve has bad reduction at a finite place when its reduced minimal Weierstrass equation is singular, equivalently when its minimal discriminant has positive valuation.

Discriminants of isomorphic Weierstrass equations of an elliptic curve differ by a twelfth power of the scale parameter, so their valuations differ by a multiple of twelve. An integral model with good reduction has discriminant valuation zero. Thus a displayed integral model with discriminant valuation not divisible by twelve proves intrinsic bad reduction. More generally, a displayed discriminant valuation between zero and eleven is already minimal, since any smaller integral-model valuation would be nonnegative and differ by twelve. Positive valuations divisible by twelve alone do not establish bad reduction; rescaling may remove them.

For odd nonzero $D$, the intrinsic bad primes of $y^2=x^3-D^{2}x$ are exactly the displayed set. If $v_p(D)=2k$, the change $x=p^{2k}X$, $y=p^{3k}Y$ removes that even valuation from the parameter and gives unit discriminant at odd $p$. If the valuation is odd, after the same removal the discriminant valuation is six. Discriminant valuations of isomorphic Weierstrass models differ by multiples of twelve, so no integral model has unit discriminant. At two the original discriminant valuation is also six. Thus primes dividing a nonminimal displayed discriminant need not be intrinsic bad primes: $E_9$ is isomorphic to $E_1$ and has good reduction at three.

At a prime of good reduction with residue characteristic $p$, reduction is injective on the prime-to-$p$ torsion of an elliptic curve. Comparing reductions at two good primes therefore bounds rational torsion by finite point counts.

For an elliptic curve over $\mathbb{Q}$, reduction at a good prime $p$ is injective on the $\ell$-primary torsion subgroup when $\ell \ne p$: its kernel is a formal group, where multiplication by $\ell$ is bijective. Hence a rational point of order $\ell$ forces $\ell \mid \#\tilde{E}({\mathbb{F}}_p)$. If for every possible prime $\ell$ some good $p\ne \ell$ has a point count not divisible by $\ell$, rational torsion is trivial. In particular two distinct good primes $p,r$ with coprime counts, $p\nmid \#\tilde{E}({\mathbb{F}}_r)$ and $r\nmid \#\tilde{E}({\mathbb{F}}_p)$ suffice: the coprime counts exclude other primes, and the two cross conditions exclude $p,r$. This argument does not incorrectly apply prime-to-$p$ injectivity to $p$-power torsion.

Let an elliptic curve over $\mathbb{Q}$ have good reduction at distinct primes $p,r$, with point counts $N_p,N_r$. For a torsion point, its $\ell$-primary part injects into every reduction whose residue characteristic differs from $\ell$. Consequently its $\ell$-power order has exponent at most $\min (v_{\ell }(N_p),v_{\ell }(N_r))$ when $\ell \ne p,r$, at most $v_p(N_r)$ when $\ell =p$, and at most $v_r(N_p)$ when $\ell =r$. This precise form avoids assuming injectivity on residue-characteristic torsion. Counts five at two and seven at three force trivial rational torsion.

For good reduction of an elliptic curve over a complete local field of residue characteristic $p$, smoothness and the Hensel lemma make reduction surjective. Its kernel is a formal group on the maximal ideal, where multiplication by $q^m$, $q\ne p$, is bijective by prime-to-residue-characteristic multiplication on a formal group. Thus there is no nonzero $q$-power torsion in the kernel. Given a lift $P$ of a reduced point killed by $q^m$, choose the unique kernel point $R$ with $q^{m}R=q^{m}P$. Then $P-R$ is a torsion lift. This proves both surjectivity and uniqueness of torsion lifting. For finite residue field these groups are finite.

For an elliptic curve with good reduction over ${\mathbb{Q}}_2$, let $E_1$ be the reduction kernel and $E_2$ its formal subgroup with parameter in $4{\mathbb{Z}}_2$. The deep logarithm subgroup of a formal group makes $E_2$ torsion-free, while $[E_1:E_2]=2$. Thus the torsion of the reduction kernel has order at most $2$. Combined with the Hasse theorem for elliptic curves, rational torsion of order $16$ forces bad reduction at $2$.

For an elliptic curve $E:y^2=f(x)$ over a field of characteristic other than two and $d\in K^{\times }$, the curve $E^{(d)}:d{y}^2=f(x)$ is its quadratic twist by $d$. It becomes isomorphic to $E$ after adjoining $\sqrt{d}$; over a finite field its Frobenius trace is multiplied by the quadratic character of $d$.

A twist of an elliptic curve $E/K$ is an elliptic curve over $K$ that becomes isomorphic to $E$ over an algebraic closure of $K$. Twists are classified by the Galois-cohomology set $H^1(G_K,\mathrm{Aut}(E_{\overline{K}}))$.

An inflection point of a smooth plane cubic is a point where its tangent line has intersection multiplicity three. Once one inflection point is chosen as the identity, the nine geometric inflection points are exactly the 3-torsion points.

For a smooth homogeneous cubic in characteristic zero, its plane cubic flexes are precisely its intersections with the determinant of its Hessian matrix. Choose coordinates with a given smooth point $(0:0:1)$ and its tangent $y=0$. Up to a nonzero coefficient, the cubic starts as $y{z}^2+a{x}^{2}z+bxyz+c{y}^{2}z$ plus cubic terms in $x,y$. Its tangent contact has order three exactly when $a=0$, and the Hessian determinant at the point is a nonzero constant times $a$. This proves the criterion.

For $c\ne 0$ and $d^3\ne 27c$ in characteristic zero, the displayed plane cubic is smooth. Its Hessian determinant is $-6d^2(x^3+y^3+c{z}^3)+(216c-2d^3)xyz$. On the cubic this becomes $8(27c-d^3)xyz$. Hence all nine plane cubic flexes lie on the coordinate axes: $(1:-\zeta :0)$, $(-\sqrt[3]{c}\zeta :0:1)$ and $(0:-\sqrt[3]{c}\zeta :1)$ for ${\zeta }^3=1$.

A one-dimensional commutative formal group law over a ring $R$ is a series $F(X,Y)\in R[[X,Y]]$ satisfying the identity, commutativity, and associativity axioms. A homomorphism from $F$ to $G$ is a series $\theta (T)\in TR[[T]]$ satisfying $\theta (F(X,Y))=G(\theta (X),\theta (Y))$.

For a one-dimensional formal group law, write $f(T)={\partial }_{Y}F(T,0)$. Its invariant differential is $dT/f(T)$. Differentiating associativity gives ${\partial }_{Y}F(X,Y)f(Y)=f(F(X,Y))$, which expresses invariance under translation. In characteristic zero, integrating this differential constructs the formal logarithm.

An isomorphism from formal group law $F$ to $G$ over $R$ is a formal series $h(T)=a_{1}T+a_2{T}^2+\cdots$ with $a_1\in R^{\ast }$ and $h(F(X,Y))=G(h(X),h(Y))$. Coefficient recursion gives a compositional inverse over $R$. An isomorphism of formal series need not converge on an arbitrary prescribed analytic domain; convergence must be checked separately.

Every one-dimensional formal group law has a unique formal inverse $i_F(X)=-X+O(X^2)$ satisfying $F(X,i_F(X))=0$. Comparing coefficients recursively determines it degree by degree.

The formal additive group has law $F_a(X,Y)=X+Y$. Its multiplication-by-$n$ series is $[n{]}_a(X)=nX$.

The formal multiplicative group has law $F_m(X,Y)=X+Y+XY$, obtained by writing a multiplicative coordinate as $1+X$. Its multiplication-by-$n$ series is

Over an algebra over a field whose scalar field is $\mathbb{Q}$, the series $\exp (X)-1$ defines an isomorphism from the formal additive group to the formal multiplicative group, with inverse $\log (1+X)$.

A morphism $h(T)=uT+O(T^2)$ of one-dimensional formal group laws over a ring is an isomorphism when $u$ is a unit. Recursive coefficient comparison constructs its compositional inverse, and the morphism identity shows that this inverse is also a formal-group morphism.

For a one-dimensional formal group law $F$ over a ring $R$, the multiplication series satisfies $[n{]}_F(T)=nT+O(T^2)$. It is an isomorphism exactly when $n$ is a unit of $R$.

Let $R$ be a complete discrete valuation ring of residue characteristic of a field $p$, with maximal ideal $\mathfrak{m}$, and let $F$ be a one-dimensional formal group law over $R$. Its multiplication series is $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the invertible morphism criterion for formal group laws constructs its inverse with coefficients in $R$. Both formal power series converge on $\mathfrak{m}$, since their coefficients are integral and the powers of an element of $\mathfrak{m}$ tend to zero. Their compositions are the identity there. Thus multiplication by $n$ is bijective, including when $p=2$ and $n$ is odd. Every finite-order element of $F(\mathfrak{m})$ consequently has order of a group element equal to a prime power with prime $p$.

Over a characteristic-zero field, a one-dimensional commutative formal group law has a unique strict isomorphism to the additive formal group, called its formal logarithm. It satisfies ${\log }_F(F(X,Y))={\log }_F(X)+{\log }_F(Y)$.

For an integral one-dimensional formal group law over a finite extension of the p-adic field, the inverse of ${\partial }_{Y}F(T,0)$ has integral coefficients because its constant term is one. Integrating this invariant differential gives ${\log }_F(T)=T+{\sum }_{n\ge 2}{c}_n{T}^n$ with $n{c}_n$ integral. For $v_K(\pi )=1$ and $e=v_K(p)$ the displayed bound follows. The estimate $v_p(n)\le (n-1)/(p-1)$ makes the nonlinear part strictly contracting on ${\pi }^s{\mathcal{O}}_K$ for $s>e/(p-1)$, yielding the deep logarithm subgroup of a formal group.

Let $F$ be a formal group law over the valuation ring of a finite extension $K/{\mathbb{Q}}_p$, with $v_K(\pi )=1$ and $e=v_K(p)$. If $m>e/(p-1)$, its formal logarithm is an isomorphism $F({\pi }^m{\mathcal{O}}_K)\to ({\pi }^m{\mathcal{O}}_K,+)$. Indeed, its scaled series is $U+\pi R(U)$ with $R$ integral and convergent on ${\mathcal{O}}_K$; solving by contraction produces the formal group exponential. The subgroup has index $q^{m-1}$ in $F(\pi {\mathcal{O}}_K)$, where $q$ is the residue-field order.

Let $E/{\mathbb{Q}}_p$ have good reduction, and choose an integer $m>1/(p-1)$. Its filtration of elliptic-curve points over a local field has a torsion-free group $E_m$ identified by the formal logarithm with $p^m{\mathbb{Z}}_p$. The quotient group $G=E({\mathbb{Q}}_p)/E_m$ is finite. For $g\in G$ of order of a group element $N$, choose a lift $P$. There is a torsion point of an elliptic curve in this coset exactly when the displayed condition holds. Indeed, it is equivalent to the existence of $Q\in E_m$ with $[N]Q=[N]P$. Then $P-Q$ is killed by $N$; conversely every torsion representative has order $N$, since $E_m$ is torsion-free. The logarithmic division and the formal group exponential give the unique such $Q$. Changing the lift changes the logarithm by an element of $N{p}^m{\mathbb{Z}}_p$, so the criterion is well-defined. This finite-coset test includes residue-characteristic torsion; only the prime-to-p torsion lifts uniquely at good reduction are automatic.

For a one-dimensional commutative integral formal group law, the formal logarithm identifies $F(4{\mathbb{Z}}_2)$ with $4{\mathbb{Z}}_2$. This is an index-two subgroup of $F(2{\mathbb{Z}}_2)$. Choose a topological generator $h$ of the subgroup and an element $v$ outside it, and write $2v=ch$. If $c$ is odd, $v$ generates the whole group. If $c$ is even, $v-(c/2)h$ has order two and splits off a cyclic factor. The formal additive group and formal multiplicative group realize the two cases.

For odd $p$, a one-dimensional formal group law over ${\mathbb{Z}}_p$ has a formal logarithm ${\log }_F(T)=T+{\sum }_{j\ge 2}{c}_j{T}^j/j$ with $c_j\in {\mathbb{Z}}_p$. This follows by integrating its integral invariant differential. For $t\in p{\mathbb{Z}}_p\setminus \{0\}$, $v_p(c_j{t}^j/j)>v_p(t)$ for $j\ge 2$. Hence the series converges and $v_p({\log }_F(t))=v_p(t)$, so its kernel is zero. As it is a homomorphism into a characteristic-zero additive group, there is no nonzero torsion. In particular, at an odd prime of good reduction over $\mathbb{Q}$, reduction injects the whole rational torsion subgroup, including its $p$-primary part.

The formal group exponential is the compositional inverse of the formal logarithm. On a sufficiently small neighborhood over a characteristic-zero local field, both series converge and identify the formal group with the additive group.