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For odd $p$, let $y^2=f(x)$ be a nonsingular cubic Weierstrass equation of an elliptic curve over ${\mathbb{F}}_p$. Its Trace of Frobenius is $a_p=-{\sum }_x\chi (f(x))$, where $\chi$ is the Legendre symbol extended by $\chi (0)=0$. By the Euler criterion, reduce this sum to $f(x{)}^{(p-1)/2}$. The sum of a positive power $x^j$ over ${\mathbb{F}}_p$ is zero unless $p-1$ divides $j$, when it is $-1$. Since the polynomial has degree $3(p-1)/2<2(p-1)$, only its $x^{p-1}$ coefficient contributes.

The determinant of the columns $(1,\wp (z),{\wp }^{\prime }(z))$, $(1,\wp (w),{\wp }^{\prime }(w))$ and $(1,\wp (-z-w),{\wp }^{\prime }(-z-w))$ vanishes identically. Its apparent cubic poles at $z=0,-w$ have order at most two because the coefficient multiplying the cubic-pole entry vanishes linearly. Column coincidences give six zeros, unless $3w$ is a lattice point, when they give four zeros with one multiple zero. In the half-period case, the two apparent zero locations at poles are removable zeros: ${\wp }^{\prime }(w)=0$ and the two $1/z$ terms cancel, leaving $O(z)$. Thus the zero count exceeds possible pole order in every case. The argument principle proves the identity, which expresses collinearity of the three points on the Weierstrass equation of an elliptic curve.

Put a flex at $[0:1:0]$ with tangent line $Z=0$. The tangent section has a triple zero there, so $F(X,Y,0)=a{X}^3$ with $a\ne 0$. Smoothness gives $b\ne 0$. In characteristic different from two, replacing $Y$ by $Y+(cX+eZ)/(2b)$ completes the square and produces a Weierstrass equation of an elliptic curve using only projective linear coordinate changes.

For an elliptic curve over a local field with integral Minimal Weierstrass equation, let $E_0(K)$ consist of points with nonsingular reduction and let $E_1(K)$ be the kernel of $E_0(K)\to {\tilde{E}}_{\mathrm{ns}}(k)$. This definition also applies at a prime of bad reduction of an elliptic curve. The point at infinity is always smooth. With $t=-x/y$ and $w=-1/y$, the Weierstrass equation of an elliptic curve becomesThe unit linear coefficient in $w$ gives a unique integral formal power series $w(t)=t^3+O(t^4)$. Then $x=t/w$ and $y=-1/w$, so the nonidentity points of $E_1(K)$ have $v(x)=-2r$ and $v(y)=-3r$ for $r=v(t)>0$. The local parameter $t$ bijects $E_1(K)$ with ${\mathfrak{m}}_K$ and turns its addition into the formal group of an elliptic curve. At good reduction this agrees with the kernel of reduction of an elliptic curve.

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.

For a Weierstrass equation of an elliptic curve, use the local parameter $t=-x/y$ and put $w=-1/y$. The formal kernel of a minimal Weierstrass equation construction gives $w=t^3+a_1{t}^4+(a_1^2+a_2)t^5+(a_1^3+2a_1{a}_2+a_3)t^6+O(t^7)$. Substitution into the elliptic curve group law yields the displayed initial coefficients and the formal inverse $-t/(1-a_{1}t-a_{3}w(t))$. In characteristic zero, the invariant differential on an elliptic curve becomes $(1+a_{1}t+(a_1^2+a_2)t^2+O(t^3))dt$, giving the formal logarithm $t+a_1{t}^2/2+(a_1^2+a_2)t^3/3+O(t^4)$. The addition coefficients are integral polynomial expressions and hence also apply in positive characteristic of a field.

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.

At a point of a smooth algebraic curve, its local ring is a discrete valuation ring. A local parameter is a generator $t$ of its maximal ideal, equivalently an element of valuation one. It measures the order of a zero of a function or pole by expressing functions as powers of $t$ times units. Over the complex numbers its analytic counterpart is a local coordinate vanishing simply at the point. For an elliptic curve in a Weierstrass equation of an elliptic curve, $t=-x/y$ is the standard parameter at the identity.

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.

For a short Weierstrass equation of an elliptic curve with coefficients in the p-adic integers, a point with nonintegral $x$ has valuations $v_p(x)=-2m$ and $v_p(y)=-3m$ for some integer $m\ge 1$: the term $x^3$ strictly dominates $Ax+B$ in the equation, so $2v_p(y)=3v_p(x)$. Its formal parameter $T=-x/y$ therefore has valuation $m\ge 1$. These points, together with the identity, form the formal neighborhood $E_1({\mathbb{Q}}_p)$ of the identity, with parameter in $p{\mathbb{Z}}_p$ and the elliptic formal group law over the p-adic integers. For a point with $x\in {\mathbb{Z}}_p$, the equation gives $y\in {\mathbb{Z}}_p$ too.

The formal neighborhood has no nontrivial torsion. Multiplication by an integer coprime to $p$ has series $[m](T)=mT+O(T^2)$ and preserves $|T{|}_p$, so any torsion there must have $p$-power order. If $p\ge 3$, part (i) gives $|T{|}_p\ge p^{-1/(p-1)}>p^{-1}$, contradicting $T\in p{\mathbb{Z}}_p$.

The printed hypotheses also allow $p=2$. In that case, a nontrivial torsion point in the formal neighborhood would have order a power of two and hence yield a point of order two in that neighborhood. But every nonidentity point of order two on a short Weierstrass equation of an elliptic curve has $y=0$ and $x$ a root of the monic cubic over the p-adic integers $X^3+AX+B$. Such a root lies in the p-adic integers, so it cannot belong to the formal neighborhood, whose nonidentity points have nonintegral coordinates. This excludes the remaining case. This is the torsion-free formal subgroup for a short Weierstrass equation. Thus every affine torsion point satisfiesWith good reduction, the reduction homomorphism $E({\mathbb{Q}}_p)\to \tilde{E}({\mathbb{F}}_p)$ has kernel exactly this torsion-free formal neighborhood. Its restriction to the torsion subgroup is therefore injective, proving the required subgroup isomorphism. The short model over the p-adic integers is essential to the prime-two argument; no blanket torsion-injectivity assertion for arbitrary models is needed.

We prove that the only finite-order rational point is the origin: $\boxed{E(\mathbb{Q}{)}_{\mathrm{tors}}=\{O\}}$. The identity always has order one, so “no elements of finite order” must be understood as no nonidentity torsion points.

The given discriminant is a unit at $5$ and $7$, so the integral Weierstrass equation of an elliptic curve has good reduction at both primes. For an odd prime $p$, completing the square in $y^2+xy=f(x)$ shows that the number of $y$ above $x$ is determined byA nonzero quadratic residue gives two points, zero gives one, and a nonresidue gives none. Direct evaluation givesIncluding the origin gives $\#\tilde{E}({\mathbb{F}}_5)=10$ and $\#\tilde{E}({\mathbb{F}}_7)=13$.

At a good prime $p$, reduction is injective on torsion of order prime to $p$: a point in its kernel belongs to the formal group, where multiplication by that order is bijective by part(i), so a point killed by it must be zero. Suppose a nonidentity rational point has finite order. Taking a suitable multiple gives a point of prime order $\ell$. If $\ell \notin \{5,7\}$, reduction at both primes forces $\ell$ to divide both $10$ and $13$, impossible. If $\ell =5$, reduction at $7$ forces $5\mid 13$, impossible. If $\ell =7$, reduction at $5$ forces $7\mid 10$, also impossible. This exhausts all primes and proves the claim. The torsion exclusion from good reduction counts explicitly handles the two residual characteristics instead of incorrectly assuming prime-to-$p$ injectivity also controls $p$-torsion.

Let $O=(0:1:0)$ be the point at infinity of the Weierstrass equation of an elliptic curve. It is the identity element for the elliptic curve group law. For an affine rational point $P=(x,y)$, the inverse isIndeed, the two affine intersections with the vertical line have these ordinates, since their sum as roots of the quadratic in $y$ is $-a_{1}x-a_3$. This reflection is valid for a generalized Weierstrass equation of an elliptic curve in every characteristic of a field.

For distinct $P,Q$, draw their line; for $P=Q$, draw the tangent line. Let $R$ be its third intersection with the elliptic curve, counted with intersection multiplicity. Then the sum is the reflection of the third intersection:A vertical line has third intersection $O$, so $P+(-P)=O$. If a tangent has third intersection $O$, then $2P=O$. Define addition with $O$ by $P+O=P$; these conventions include all exceptional cases of the affine construction.

The construction is defined over the coefficient field: after two known intersections, the remaining root is rational over that field. It gives a commutative group operation. To see why it is associative, use the elliptic curve group law from Riemann-Roch: the map $P\mapsto [P-O]$ identifies the elliptic curve with its degree-zero divisor class group. A line has intersection divisor $P+Q+R$ linearly equivalent to $3O$, the divisor of the line at infinity. Thus $[P-O]+[Q-O]+[R-O]=0$. The geometric sum agrees with addition of divisor classes, which is associative.

Here the inverse in the elliptic curve group law is $(x,y)\mapsto (x,-y-1)$. For $P_0+P_1$, the line is $y=2-2x$. Substitution in the Weierstrass equation of an elliptic curve givesThe third intersection is $(3,-4)$, so reflection givesFor the difference, first take $-P_1=(1,-1)$. The line through $P_0$ and $-P_1$ is $y=2-3x$, and now the intersection polynomial is $x(x-1)(x-8)$. Its third point is $(8,-22)$, givingFinally, implicit differentiation gives $(2y+1)y^{\prime }=3x^2-7$. The tangent line at $P_1$ therefore has slope $-4$ and equation $y=4-4x$. Its intersection polynomial factors as $(x-1{)}^2(x-14)$. The double root records the intersection multiplicity of the tangent, and the remaining intersection is $(14,-52)$. Hence

Work near the identity $O$ of a generalized Weierstrass equation of an elliptic curve. On the projective chart where $Y\ne 0$, use the local parameter $t=-x/y$ and the auxiliary coordinate $w=-1/y$. Substitution givesThe coefficient of $w$ in the linearized equation at $(0,0)$ is one. Consequently, successive coefficient comparison determines a unique formal power series $w(t)$, without dividing by any coefficient of the equation. Its initial terms areRecover the affine coordinates as Laurent series $x=t/w(t)$ and $y=-1/w(t)$; in particular $x=t^{-2}+O(t^{-1})$ and $y=-t^{-3}+O(t^{-2})$.

The elliptic curve group law is a regular morphism near $(O,O)$. Applying the local parameter to the sum and passing to completed local rings producesOne can construct the series without any nonunit division. PutThe divided difference is an integral formal power series, including its tangent line specialization at $T_1=T_2$. The line $w=\lambda t+\nu$ meets the cubic in the two chosen points and a third point $R$. Substitution in the equation shows that its cubic and quadratic coefficients in $t$ are $-A,-B$, whereSince $\lambda$ has total degree at least two and $\nu$ at least three, $A$ is a unit. The sum of the roots therefore givesThe last expression reflects the third intersection according to the chord-and-tangent group law. Both denominators have constant term one. This proves integrality of the universal construction and also provides an explicit coefficient algorithm. Expanding it, the resulting formal group of an elliptic curve beginsThe integral coefficient recursion and the unit-denominator addition calculation work universally over $\mathbb{Z}[a_1,a_2,a_3,a_4,a_6]$; thus these coefficients specialize in every characteristic of a field. This is the initial coefficients of the elliptic formal group at infinity construction, rather than a definition requiring division by the elliptic-curve discriminant.

The identity, commutativity and associativity of the elliptic curve group law give the defining identities of a one-dimensional commutative formal group law:Reflection in the generalized Weierstrass equation of an elliptic curve gives its formal inverse explicitly:Successive addition gives the multiplication series $[n{]}_F(T)=nT+O(T^2)$.

In characteristic zero, an especially useful description comes from the invariant differential on an elliptic curve:Integrating with zero constant term constructs the formal logarithmTranslation invariance of the invariant differential on an elliptic curve implies $L(F(X,Y))=L(X)+L(Y)$. Thus the formal logarithm converts the formal group law into the formal additive group in characteristic zero. Expanding this identity to degree three also recovers the displayed coefficients of $F$; in positive characteristic the integral addition construction remains valid although the logarithmic denominators may not exist.

The Mordell-Weil theorem gives $E(\mathbb{Q})\cong E(\mathbb{Q}{)}_{\mathrm{tors}}\oplus {\mathbb{Z}}^r$. Move the rational point of order two to $(0,0)$ and clear denominators to obtain an integral Weierstrass equation of an elliptic curveThe two-isogeny formula gives $E^{\prime }:Y^2=X(X^2-2aX+a^2-4b)$ and dual isogenies $\varphi :E\to E^{\prime }$, $\hat{\varphi }:E^{\prime }\to E$ with $\hat{\varphi }\varphi =[2]$. Compute the torsion subgroup using reduction of torsion points on an elliptic curve at several good primes, followed by division polynomials of an elliptic curve to identify the possible points.

For the free part, use two-isogeny descent. The two-torsion square-class homomorphism isin ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$, with kernel $\hat{\varphi }{E}^{\prime }(\mathbb{Q})$; define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly. The prime-support bound in two-isogeny descent restricts the first image to signed square-free divisors of $b$ and the second to those of $a^2-4b$. For each candidate $d$, solve its two-isogeny descent quarticTest local solubility at the real place and the relevant p-adic fields, discard impossible classes, and search the surviving coverings for rational points. A point with $UV\ne 0$ gives $x=d(U/V{)}^2$, $y=dUN/V^3$. Include the identity and rational 2-torsion cases separately. Once both rational images are determined, the two-isogeny rank formula givesThe point searches give explicit candidates for a basis. Their canonical height pairing can certify independence; Mordell-Weil saturation and effective height bounds then certify that they generate the full free part. Rank alone does not certify generators of the whole Mordell-Weil group. If the rational images are not fully determined, the procedure instead supplies lower and upper bounds on the rank of an elliptic curve.

An elliptic curve over $\mathbb{Q}$ has good reduction at $p$ if it admits an integral Weierstrass equation of an elliptic curve over ${\mathbb{Z}}_p$ whose reduced projective cubic is a smooth algebraic curve over ${\mathbb{F}}_p$. Equivalently, its Minimal Weierstrass equation has a unit elliptic-curve discriminant, or its smooth proper model is an elliptic curve over the valuation ring. The existence of an appropriate integral model matters: a nonminimal equation can have discriminant divisible by $p$ even when the curve has good reduction. Equivalently,

The supplied integral equation has elliptic-curve discriminant $\Delta =-11$. It therefore has good reduction at every $p\ne 11$. At eleven its discriminant has valuation one. Under a change of Weierstrass equation of an elliptic curve, discriminant valuations differ by a multiple of twelve. A good integral model would have valuation zero, which is impossible. This is the discriminant valuation obstruction to good reduction; it also shows the displayed model is minimal at eleven. Hence

Choose an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. For $P\in E({\mathbb{Q}}_p)$, choose homogeneous coordinates $[X:Y:Z]$ in ${\mathbb{Z}}_p$ with at least one unit by multiplying all three coordinates by a common power of $p$. DefineTwo primitive integral representatives differ by a unit, so this is well-defined. Every point of the reduced cubic is a nonsingular point of an algebraic curve, and the identity reduces to $[0:1:0]$. This definition includes points whose affine coordinates are nonintegral.

To prove the good-reduction map respects elliptic addition, use the smooth proper integral model. The elliptic curve group law is a morphism on this model, constructed by addition of degree-zero divisor classes; it is defined even where an affine chord formula has a vanishing denominator. Passing to the residue field commutes with this morphism. Henceso reduction is a homomorphism of abelian groups.

For surjectivity, the simple-root Hensel lemma says that if $g\in {\mathbb{Z}}_p[t]$, $g(t_0)\equiv 0(\mathrm{mod}p)$ and $g^{\prime }(t_0)\not\equiv 0(\mathrm{mod}p)$, then there is a unique $t\in {\mathbb{Z}}_p$ with $t\equiv t_0(\mathrm{mod}p)$ and $g(t)=0$. The point at infinity lifts to $O$. At an affine point $(\overline{x},\overline{y})$ of the reduced equation $F(x,y)=0$, being a nonsingular point of an algebraic curve means at least one partial derivative $F_x,F_y$ is nonzero. If $F_y\ne 0$, fix any integral lift $x_0$ of $\overline{x}$ and apply the Hensel lemma to $g(t)=F(x_0,t)$ to lift $\overline{y}$. If $F_x\ne 0$, fix $y_0$ and lift $\overline{x}$ instead. This proves the surjectivity of good reduction over a local field and gives

An isogeny of elliptic curves over $k$ is a nonconstant morphism $\varphi :E_1\to E_2$ defined over $k$ and preserving the identity. Such a morphism is a group homomorphism: apply the Mumford rigidity lemma to $\varphi (P+Q)-\varphi (P)-\varphi (Q)$, whose restriction to either coordinate axis is zero. A nonconstant morphism of smooth projective curves is finite and surjective, so its degree of an isogeny is the positive integerDegrees multiply under composition. A separable isogeny has degree equal to the number of geometric points in its kernel of an isogeny. In positive characteristic of a field, inseparable isogenies also occur, for example the Frobenius isogeny; their total degree includes inseparable multiplicity, so counting geometric kernel points alone is insufficient. Every isogeny of degree $d$ has a dual isogeny $\hat{\varphi }$ satisfying $\hat{\varphi }\varphi =[d]$ and $\varphi \hat{\varphi }=[d]$.

Pointwise addition and negation make the homomorphism group of elliptic curves $H={\mathrm{Hom}}_k(E_1,E_2)$ an abelian group. Its zero is the constant identity map; every other element is an isogeny, because its nonconstant image is the whole target curve. Define $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We prove the positive definiteness of degree on elliptic-curve homomorphisms, including arbitrary characteristic.

Let $L={\mathcal{O}}_{E_2}(O)$, a line bundle associated to a divisor of degree one. On $E_2\times E_2$ write $m(P,Q)=P+Q$, $d(P,Q)=P-Q$ and let ${\pi }_i$ be the projections. The divisor proof of the degree parallelogram law givesHere is a direct justification. The degree-two Weierstrass coordinate $x$ has a double pole at $O$, and $x(P)=x(Q)$ precisely when $Q=P$ or $Q=-P$. Consequentlywhere $D_{+}$ is the diagonal and $D_{-}$ the graph of negation. Since $D_{+}=d^{-1}(O)$ and $D_{-}=m^{-1}(O)$, this principal divisor gives the displayed line bundle identity. This remains valid in characteristic two: the general Weierstrass equation of an elliptic curve of a smooth algebraic curve has a nontrivial generic involution $(x,y)\mapsto (x,-y-a_{1}x-a_3)$, so the two graphs still occur. This identity is also the symmetric-line-bundle form of the Theorem of the square.

Pull back the bundle identity along $(\varphi ,\psi ):E_1\to E_2\times E_2$ and take degrees of line bundles. For a nonzero map, $\mathrm{deg}({\varphi }^{\ast }L)=q(\varphi )$; for a zero map its pullback is trivial and has degree zero. Thus the degree parallelogram law isUsing bundles rather than pulling back the rational function itself is essential when $\varphi =\pm \psi$, where that function can vanish identically. Negation preserves degree, so $q(-\varphi )=q(\varphi )$. Applying the parallelogram identity to $n\varphi$ and $\varphi$ gives the quadratic degree recursion for elliptic multiplication, and induction from $q(0)=0$ yields $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$.

Define the bilinear degree pairing for elliptic-curve homomorphisms byIt is symmetric and odd in each argument. Two applications of the parallelogram law giveAdding proves $B(x+z,y)=B(x,y)+B(z,y)$, and symmetry proves additivity in the other argument. Also $B(\varphi ,\varphi )=q(\varphi )$. Therefore $q$ is an integer-valued quadratic form. The integral polarization is $2B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$; $B$ itself may be half-integral.

Finally $q(\varphi )>0$ for every nonzero $\varphi$, since an isogeny has positive finite degree. If $n\varphi =0$ with $n\ne 0$, the identity $q(n\varphi )=n^{2}q(\varphi )$ forces $\varphi =0$, so $H$ is a torsion-free abelian group. On any finite-dimensional rational span in $H\otimes \mathbb{Q}$, the pairing has a rational matrix and is positive on every nonzero rational vector. By density it is nonnegative on real vectors; if that rational matrix were singular, its kernel would contain a nonzero rational vector, contradicting strict positivity. Thus its real extension is a positive-definite quadratic form. This establishes the required positivity as well as the quadratic identity. For example, on the subgroup of multiplication maps $[n]$, $q([n])=n^2$; on the endomorphism ring of an elliptic curve with complex multiplication it is the imaginary-quadratic field norm, explaining the lattice geometry behind isogeny degrees.

The Weierstrass elliptic function and its derivative give the maponto the Weierstrass equation of an elliptic curve $Y^{2}Z=4X^3-g_{2}X{Z}^2-g_3{Z}^3$. Near zero, multiplying these coordinates by $z^3$ gives $[z+O(z^5):-2+O(z^4):z^3]$, so the map extends holomorphically and is an immersion at the origin.

The meromorphic map $\wp :E\to {\mathbb{P}}^1$ has degree two, since its only pole on the torus has order two. Its evenness implies $\wp (u)=\wp (v)$ exactly when $u\equiv v$ or $u\equiv -v$. Oddness of ${\wp }^{\prime }$ distinguishes these two alternatives unless they are the same point class. At each nonzero half-period, ${\wp }^{\prime }=0$. These three points exhaust its zeros, since its only pole has order three. The three half-period values are distinct: equality of two of them would give two distinct zeros each of order at least two for $\wp -e$, contradicting its total pole order two. They are roots of $4X^3-g_{2}X-g_3$, so the polynomial has distinct roots and the cubic is smooth. At a half-period $h$ with value $e$, ${\wp }^{\prime \prime }(h)=2{\prod }_{e^{\prime }\ne e}(e-e^{\prime })\ne 0$. Elsewhere ${\wp }^{\prime }$ is nonzero. The coordinate pair therefore has nonzero derivative at every finite point, and the map is injective. Compactness and the smooth cubic equation then identify it as an embedding onto the cubic.

The hyperplane section at infinity has divisor $3[O]$. Thus the intersection divisor $D_{\ell }$ of any line with this cubic is linearly equivalent to $3[O]$. Applying the principality criterion on a complex elliptic curve to $D_{\ell }-3[O]$ shows that its three points, counted with intersection multiplicity, have sum zero in $E$.

Conversely, if the degree-three divisor $D=[u]+[v]+[w]$ has point sum zero, Question 1 gives a meromorphic function with divisor $D-3[O]$. A meromorphic function with no poles away from $O$ and pole order at most three lies inTo prove this description, subtract multiples of ${\wp }^{\prime }$ and $\wp$ to cancel the cubic and quadratic Laurent principal parts. The remaining possible simple-pole coefficient is a residue, which must be zero because it is the only pole on a compact torus. The remaining holomorphic elliptic function is constant. Thus the constructed function is a linear form in the projective coordinates, and its line has intersection divisor exactly $D$. This proves the chord-and-tangent collinearity criterion is $u+v+w=0$ in $E$, with multiplicities understood.

For three pairwise distinct point classes away from $O$, ordinary collinearity is equivalent to linear dependence of the coordinate columns, soThe original PDF has this three-column determinant; the TeX's extra copied column is a transcription defect.

There is also a genuine implicit distinctness condition in the printed determinant assertion. If $u=v={\omega }_1/10$ and $w={\omega }_1/5$, all entries are finite and the determinant is zero because two columns coincide, but $u+v+w=2{\omega }_1/5\notin \Lambda$. Thus the unrestricted literal equivalence is false. To specify repeated intersections, use the confluent elliptic collinearity determinant: for a double point at $u$ replace the second column by $c^{\prime }(u)$, where $c(z)=(1,\wp (z),{\wp }^{\prime }(z){)}^t$; for a triple point use $c(u),c^{\prime }(u),c^{\prime \prime }(u)$. These impose tangency or inflection, and recover the correct conditions $2u+w\in \Lambda$ or $3u\in \Lambda$. If a point is $O$, use the homogeneous coordinates or a local homogeneous lift, since the displayed affine entries have poles there.