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Apply the dual isogeny degree identity to $1-\varphi$. Its dual is $1-\hat{\varphi }$, soRearrangement givesThe right side is multiplication by an integer, proving the assertion. In particular the trace of an elliptic-curve endomorphism isThe argument includes $\varphi =0$ and $\varphi =1$ with the zero-map convention. Multiplying $\hat{\varphi }=[t]-\varphi$ by $\varphi$ also yields the useful characteristic relation

We use the basic dual isogeny identities in the endomorphism ring of an elliptic curve, extending the dual of the zero map to zero:Here $[n]$ denotes the multiplication-by-$n$ endomorphism, and the degree of the zero map is set to zero. To see the source of additivity, identify $E$ with its Picard group of degree-zero line bundles using the origin. Dualization is pullback on this group. Pullback of a degree-zero line bundle along the sum of two homomorphisms is the tensor product of its pullbacks along those homomorphisms, giving additivity; functoriality of pullback reverses composition. The last identity is the defining degree identity for a dual isogeny. Every nonzero elliptic-curve endomorphism is an isogeny.

The natural inclusion $\mathbb{Z}\hookrightarrow \mathrm{End}(E)$ remains injective even in positive characteristic: for nonzero $n$, the map $[n]$ has degree $n^2$ and is not zero. Therefore an equality of multiplication endomorphisms uniquely specifies its integer coefficient.

Move the nonidentity rational 2-torsion point to $T=(0,0)$ and complete the square. After a rational change of coordinates, the elliptic curve has an integral modelThe two-isogeny formula gives a partnerand the isogeny of elliptic curves and its dual isogeny areBoth extend at the exceptional points; their kernels are the identities and the distinguished 2-torsion points, and $\hat{\varphi }\varphi =[2]$.

The two-torsion square-class homomorphism is $\alpha (O)=1$, $\alpha (T)=[b]$, and $\alpha (x,y)=[x]$ otherwise, with values in ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$. Its kernel is $\hat{\varphi }{E}^{\prime }(\mathbb{Q})$. Define ${\alpha }^{\prime }$ on the partner in the same way. By the prime-support bound in two-isogeny descent, representatives are signed squarefree divisors $d$ of $b$, or of $b^{\prime }$ for the partner. A class is realized precisely when its quartic covering in a two-isogeny descent has a rational solution:For $UV\ne 0$, this comes from $x=d(U/V{)}^2$ and $y=dUN/V^3$; the cases $U=0$ or $V=0$ account for the distinguished point and the identity. Test this finite list using signs and local congruences, and exhibit rational solutions for the surviving classes. The local-to-global gap in isogeny descent matters: local solubility gives an upper bound from the isogeny Selmer group, so a rank determination requires matching actual classes or further descent, not merely declaring every locally soluble cover to have a rational point.

Once the two actual images are determined, the two-isogeny rank formula isFor completeness, put $A=E(\mathbb{Q})$ and $B=E^{\prime }(\mathbb{Q})$. The Mordell-Weil theorem gives $[A:2A]=2^r|A[2]|$. The index of a subgroup formula through $\hat{\varphi }B$ givesHere $\ker \hat{\varphi }\cap \varphi A=\varphi (A[2])$, since $\hat{\varphi }\varphi =[2]$, and this image has size $|A[2]|/2$. The denominator is therefore $4/|A[2]|$, proving the displayed rank of an elliptic curve formula.

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 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.

To compute the rank of an elliptic curve, translate the rational point of order two to $(0,0)$ and put the curve in the form $y^2=x^3+a{x}^2+bx$, with $a,b\in \mathbb{Z}$ after clearing denominators and $b(a^2-4b)\ne 0$. Its two-isogenous partner and the elliptic isogeny areThe maps extend over their exceptional points. The dual isogeny isThe two-torsion square-class homomorphism is $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\times }/{\mathbb{Q}}^{\times 2}$, defined by $\alpha (O)=1$, $\alpha ((0,0))=[b]$, and $\alpha ((x,y))=[x]$ otherwise. Define ${\alpha }^{\prime }$ similarly on $E^{\prime }$ with $b^{\prime }=a^2-4b$.

Here are the ingredients behind the descent procedure. For a generic chord $y=mx+c$, the cubic of intersection coordinates has root product $c^2$. As the third intersection is the negative of the group sum and negation leaves $x$ unchanged, this proves the square-class homomorphism property. For a line through $(0,0)$, the two nonzero intersection coordinates have product $b$, so the assigned class $[b]$ gives the same multiplicativity. A vertical line has two equal coordinates and contributes their square, and $O$ has identity class. These observations also cover tangencies and the exceptional sums. Its kernel is $\hat{\varphi }(E^{\prime }(\mathbb{Q}))$. Indeed, the displayed formula gives $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$. Conversely, if a point on $E$ has $x=r^2\ne 0$, the possible preimage coordinates satisfyIts elliptic-curve discriminant is $16(r^4+a{r}^2+b)=16y^2/r^2$, a rational square. A root $X$ and $Y=2rX$, with the sign chosen appropriately, give a preimage. The identity is always in the image. The point $(0,0)$ is in the image exactly when $b$ is a square: its nonzero preimage coordinates are $X=a\pm 2\sqrt{b}$, with $Y=0$. Thus the exceptional points satisfy precisely the same kernel statement. Interchanging the curves similarly gives $\ker {\alpha }^{\prime }=\varphi (E(\mathbb{Q}))$.

A prime not dividing $b$ can occur only with even valuation in an $x$-coordinate. If its valuation is positive, the factor $x^2+ax+b$ is a unit and $2v(y)=v(x)$; if negative, $x^2$ dominates that factor and $2v(y)=3v(x)$. Thus square classes in $\mathrm{im}\alpha$ have signed square-free representatives dividing $b$, giving finitely many candidates. To test a candidate $d$, put $x=d(u/v{)}^2$ and $y=duN/v^3$. The curve equation becomesA nontrivial rational solution, including the appropriate limiting points, is exactly a realization of the class. Clear denominators and arrange $\gcd (u,v)=1$ for congruence tests. Do the same for $E^{\prime }$ with $b^{\prime }$.

The factor in the two-isogeny rank formula can also be checked directly. Let $A=E(\mathbb{Q})$, $B=E^{\prime }(\mathbb{Q})$ and $t=\#A[2]$. The index of $\hat{\varphi }(B)$ in $A$ is $\#\mathrm{im}\alpha$, while the map from $B/\varphi (A)$ onto $\hat{\varphi }(B)/2A$ has kernelThe numerator has order two. Moreover $\ker \hat{\varphi }\cap \varphi (A)=\varphi (A[2])$, since $\hat{\varphi }\varphi P=2P$, and the latter has order $t/2$. Hence this quotient has order $4/t$, andThe Mordell-Weil theorem gives $[A:2A]=2^{r}t$, where $r$ is the rank. Cancelling $t$ yieldsIn practice, local congruence and real-solubility tests remove many candidate quartics, giving an upper bound, while rational point searches realize classes and give a lower bound. These often coincide. Local solubility alone is not a proof of a rational point: an everywhere locally soluble covering can still fail globally, so a remaining gap must be reported rather than treated as an exact rank.

For the curve at hand, $a=8$, $b=-7$, $b^{\prime }=92$, andThe first image is contained in $\{1,-1,7,-7\}$ and contains $1,-7$, the latter from the point $(0,0)$. To exclude $-1$, its quartic would require coprime $u,v$ withModulo $16$, an odd fourth power is $1$ and an even fourth power is $0$. If $u$ is odd and $v$ even the right side is $15$; if $u$ is even and $v$ odd it is $7$; if both are odd it is $14$. None is among the square residues $0,1,4,9$. Thus $-1$ is impossible. The image is a subgroup, so $7$ is also impossible: its product with the known class $-7$ would be $-1$. We obtain $\mathrm{im}\alpha =\{1,-7\}$.

For $E^{\prime }$, its quadratic factor is $X^2-16X+92=(X-8{)}^2+28>0$. Every nonzero rational $X$-coordinate is therefore positive. Prime support leaves only $1,2,23,46$ as possible classes; $1$ and $23=[92]$ are supplied by $O$ and the two-torsion point. The explicit point $(18,48)$ is on $E^{\prime }$ becauseIts class is $[18]=2$. Multiplying classes supplies $46$, so $\mathrm{im}{\alpha }^{\prime }=\{1,2,23,46\}$. The two-isogeny rank formula now provesAs an additional concrete point on the original curve, the dual isogeny sends $(18,48)$ to $(16/9,116/27)$. The descent calculation proves the rank; it does not by itself certify that this point generates the full free quotient.

An isogeny of elliptic curves $\varphi :E_1\to E_2$ over a field $K$ is a nonconstant morphism preserving the identity and addition. It is surjective and finite: its image is a closed connected subgroup, and a nonconstant morphism of smooth projective curves is finite. The degree of an isogeny is the degree of its induced extension of function fields. A separable isogeny has degree equal to the number of geometric kernel points; in general the degree is the length of the finite kernel scheme, incorporating inseparability. The degree of an isogeny multiplies under composition, and degree one means an isomorphism. In characteristic zero all elliptic isogenies are separable. In positive characteristic the Frobenius isogeny illustrates inseparability, while an elliptic isogeny of degree prime to the characteristic is separable.

The homomorphism group of elliptic curves includes the zero homomorphism, the constant identity map, and every nonzero member is an elliptic isogeny. The operations are pointwise addition and negation; for endomorphisms, composition also gives the endomorphism ring of an elliptic curve. We put $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We now prove the positive definiteness of degree on elliptic-curve homomorphisms in full.

Let $E=E_2$, and choose a Weierstrass model. Its coordinate $x$ has a double pole at $O$, and the generic fibre of $x$ consists of $P$ and $-P$. On $E\times E$ the rational function $x(P)-x(Q)$ therefore has Weil divisorwhere $\Delta =\{P=Q\}$, ${\Delta }_{-}=\{P=-Q\}$, $V=\{O\}\times E$, and $H=E\times \{O\}$. The generic zeros on the two diagonals have multiplicity one, and the generic poles on $V,H$ have multiplicity two; their intersections have codimension two and add no Weil divisor components. This argument also uses the general Weierstrass involution in characteristic $2$, where $-P$ need not mean changing just the sign of $y$.

For the sum and difference maps $a(P,Q)=P+Q$ and $s(P,Q)=P-Q$, the divisors $a^{\ast }(O)$ and $s^{\ast }(O)$ are ${\Delta }_{-}$ and $\Delta$. Consequently, for the degree-one line bundle $L={\mathcal{O}}_E(O)$,Pull back along $P\mapsto (\varphi (P),\psi (P))$ and take degrees of line bundles on $E_1$. A nonconstant map pulls back $L$ to degree equal to its degree, while a constant map pulls it back to a trivial bundle of degree zero. We obtain the degree parallelogram lawUsing the line-bundle identity is important: it still restricts correctly when $\varphi =\psi$ or $\varphi =-\psi$, when restriction of the displayed rational function would be identically zero.

Since $q(-\varphi )=q(\varphi )$, applying the identity to $n\varphi$ and $\varphi$ gives the second-difference recursionStarting from $q(0)=0$ and $q(\varphi )$ yields $q(n\varphi )=n^{2}q(\varphi )$ for all integers $n$. DefineIt is symmetric and odd in each variable. Applying the parallelogram identity to the four terms in its first expression givesInterchanging $u,w$ and using oddness also gives $B(u+w,v)-B(u-w,v)=2B(w,v)$. Adding proves $B(u+w,v)=B(u,v)+B(w,v)$; symmetry proves additivity in the other variable. Thus $q$ is a quadratic form with associated rational bilinear form $B$.

Every nonzero homomorphism has positive degree, so $q(\varphi )>0$ for $\varphi \ne 0$. The scaling identity also shows that the homomorphism group is a torsion-free group. Hence $q$ extends to its rational vector space. On the span of any finitely many rationally independent homomorphisms, the matrix of $B$ is rational. Positivity on nonzero rational vectors implies nonnegativity on real vectors by density. If the real matrix had a nontrivial kernel, its rational linear equations would have a nonzero rational solution, contradicting positivity. Thus its real extension is a positive-definite quadratic form. This establishes the requested positivity as well as the quadratic identities.

An important companion to degree is the dual isogeny. Under $E\cong {\mathrm{Pic}}^0(E)$ via $P\mapsto [P-O]$, the pullback ${\varphi }^{\ast }:{\mathrm{Pic}}^0(E_2)\to {\mathrm{Pic}}^0(E_1)$ defines $\hat{\varphi }:E_2\to E_1$. Divisor pushforward corresponds to $\varphi$, and pushforward after pullback multiplies divisor classes by $n=\mathrm{deg}\varphi$. Therefore $\varphi \hat{\varphi }=[n]$. Composing shows that $\hat{\varphi }\varphi -[n]$ takes values in the finite kernel of $\varphi$; a morphism from a connected smooth curve into this finite scheme must be constant, and its identity value is zero. ThusThe last equality follows from $\mathrm{deg}[n]=n^2$, already proved by applying $q(m\varphi )=m^{2}q(\varphi )$ to the identity map. The dual isogeny is unique, since the difference of two choices composed with $\varphi$ would vanish and $\varphi$ is surjective. It follows that being isogenous is an equivalence relation: duality supplies symmetry, and composition supplies transitivity. For a finite subgroup of geometric points of order prime to the characteristic, the quotient curve and its quotient map give an elliptic isogeny with that kernel; in general one uses finite subgroup schemes. This is the geometric source of explicit descent maps, including the two-isogeny formula in Question 4.

Finally elliptic isogenies preserve the rational rank of an abelian group over a number field. Their dual compositions are multiplication by $n$, so after tensoring the rational point groups with $\mathbb{Q}$ the elliptic isogeny becomes invertible. For a single curve, its endomorphism algebra in characteristic zero is $\mathbb{Q}$ or an imaginary quadratic field, giving ordinary integer multiplication or complex multiplication. In positive characteristic a supersingular curve can instead have a quaternionic endomorphism algebra. The positive degree form controls the sizes of these homomorphisms, while duality makes their arithmetic and geometric properties accessible.

A rational point of order $2$ can be moved to $(0,0)$, giving an integral modelIts two-isogeny descent partner and the two-isogeny formula areThe maps extend over the identity and $(0,0)$, have these two points as their respective kernels, and direct substitution and the doubling formula give $\hat{\varphi }\varphi =[2]$ and $\varphi \hat{\varphi }=[2]$.

Define the two-torsion square-class homomorphisms byand similarly ${\alpha }^{\prime }$ for $E^{\prime }$, using $b^{\prime }$. To see the homomorphism property, intersect a line $y=mx+c$ with $E$. Its three $x$-coordinates have product $c^2$, since the resulting monic cubic has constant term $-c^2$. The third point is the negative of the sum of the first two, and negation preserves $x$, so the square classes multiply as required. If the line passes through $(0,0)$, the other two coordinates have product $b$, giving exactly the prescribed exceptional value. Vertical lines and repeated intersections give the identity and tangent cases.

We also verify the kernel needed for descent:For a point on $E^{\prime }$, the first coordinate of its image under $\hat{\varphi }$ is $Y^2/(4X^2)$, a square. Conversely, if $P=(s^2,y)$ has square $x$-coordinate, solving $x(\hat{\varphi }(X,Y))=s^2$ givesIts discriminant is $16(s^4+a{s}^2+b)=(4y/s{)}^2$, so it has rational nonzero roots $X=a+2s^2\pm 2y/s$. Taking $Y=\pm 2sX$ gives a point on $E^{\prime }$; choose the signs so its image has second coordinate $y$. The exceptional point $(0,0)$ lies in the image exactly when $b$ is a square, as seen from the points with $Y=0$, $X=a\pm 2\sqrt{b}$. This proves the first kernel formula. Applying the same argument to $E^{\prime }$ and then rescaling the second partner by $(X,Y)\mapsto (X/4,Y/8)$ proves the second.

There are only finitely many candidate square classes. For a rational point write $x=M/e^2$ and $y=N/e^3$, with $\gcd (M,e)=1$. These denominator forms follow because a negative prime valuation of $x$ makes the leading cubic term the unique lowest-valuation term, forcing that valuation to be even. ThenIf a prime divides $M$ but not $b$, the second factor is a unit at that prime, so its valuation in $M$ is even. Thus every class in $\alpha (E)$ has a signed squarefree representative $d\mid b$. This is the prime-support bound in two-isogeny descent; the analogous candidates for ${\alpha }^{\prime }(E^{\prime })$ divide $b^{\prime }$.

The class $d$ is realized by a nonexceptional rational point precisely when its quartic covering in a two-isogeny descenthas a rational point with $UV\ne 0$. Indeed substitution $x=d(U/V{)}^2$, $y=dUN/V^3$ gives the original equation, and conversely the square-class condition gives these parameters. Clearing denominators permits $U,V,N$ integral with $\gcd (U,V)=1$. The zero-coordinate cases account for the already known identity and two-torsion classes.

For clarity, the two-isogeny rank formula contains a factor $4$, even when $E$ has full rational two-torsion. Put $t=\#E(\mathbb{Q})[2]$, which is $2$ or $4$, and let $r$ be the rank. From the two kernel formulas and $2E=\hat{\varphi }\varphi E$,Indeed the kernel of the induced map $E^{\prime }/\varphi E\to \hat{\varphi }{E}^{\prime }/2E$ is $\ker \hat{\varphi }/(\ker \hat{\varphi }\cap \varphi E)$. The intersection is $\varphi (E[2])$, of size $t/2$, so $\delta =4/t$. The Mordell-Weil theorem gives $[E:2E]=2^{r}t$. Therefore

This gives a practical rank procedure. Enumerate the finitely many signed squarefree divisors of $b,b^{\prime }$, and test their quartics over the reals and local fields at $2$ and primes dividing the relevant coefficients or elliptic-curve discriminant. These necessary local conditions give finite Selmer group of an elliptic curve upper bounds. Search the remaining coverings for rational points, or search the curves directly, to realize square classes and obtain lower bounds. One can also exhibit independent points and check independence by the canonical height pairing. When the upper and lower bounds agree, the rank is determined. Local solubility alone does not guarantee a rational point, so a persistent gap can reflect a nontrivial Tate–Shafarevich group rather than a missing congruence test.

Now take $a=0$, $b=-p^2$ for an odd prime $p$. The curve has four rational two-torsion points $O,(0,0),(p,0),(-p,0)$. Their images under $\alpha$ are $1,-1,p,-p$. The prime-support bound allows no other classes, soIts partner is $E_p^{\prime }:Y^2=X^3+4p^{2}X$. Any nonidentity rational point with $X\ne 0$ has $X>0$, because $X^2+4p^2>0$. Also the exceptional point has square class $4p^2=1$. Thus its image is contained in the four positive candidate classes $\{1,2,p,2p\}$. The rank bound for a prime congruent-number elliptic curve follows:

For an exact rank-one example, take $p=5$. The point $(X,Y)=(5,25)$ on $E_5^{\prime }:Y^2=X^3+100X$ realizes the class $5$. The five-adic obstructions for the congruent-number isogeny covers exclude the other two possible nontrivial classes. For $d=2$, the quartic isModulo $5$, if $5\nmid U$ the right side is $2$, a nonsquare. Hence $U=5U_1$ and $N=5N_1$; primitivity gives $5\nmid V$. Dividing by $25$ then gives $N_1^2=50U_1^4+2V^4\equiv 2(\mathrm{mod}5)$, again impossible. For $d=10$, the quartic is $N^2=10(U^4+V^4)$. At least one of $U,V$ is nonzero modulo $5$, and their fourth powers are $0$ or $1$, so their sum is $1$ or $2$ modulo $5$. The right side has five-adic valuation $1$, impossible for a square. Thus ${\alpha }^{\prime }(E_5^{\prime })=\{1,5\}$ exactly, andThe dual isogeny maps $(5,25)$ to $(25/4,-75/8)$ on $E_5$, giving an explicit rational point associated with this rank-one computation.

Let $t=q+1-\#E({\mathbb{F}}_q)$. The stated characteristic equation for the Frobenius isogeny isTake $\alpha ,\beta$ to be the complex roots of $T^2-tT+q$. They satisfy $\alpha +\beta =t$ and $\alpha \beta =q$. By the Hasse bound, $t^2\le 4q$. If the inequality is strict, the roots are complex conjugates and have squared modulus $q$. In the equality case both roots are the same real number $\pm \sqrt{q}$. Thus in every caseTo obtain every extension-field count without assuming distinct roots, set $t_n=\mathrm{tr}({\pi }^n)$ and $t_0=2$. The dual isogeny satisfies $\hat{\pi }\pi =[q]$ and $\pi +\hat{\pi }=[t]$. Consequently the sums ${\pi }^n+{\hat{\pi }}^n=[t_n]$ satisfyAlternatively, apply additivity of the trace of an elliptic-curve endomorphism directly to the characteristic equation multiplied by ${\pi }^{n-2}$. The sequence ${\alpha }^n+{\beta }^n$ has the same recurrence and initial values, so $t_n={\alpha }^n+{\beta }^n$ by induction. This is the Frobenius trace recurrence, valid also at a repeated root.

The fixed points of ${\pi }^n$ are $E({\mathbb{F}}_{q^n})$, and $1-{\pi }^n$ is again separable. Since $\mathrm{deg}{\pi }^n=q^n$, the trace-degree identity givesFactoring and using $\beta =q/\alpha$ yieldsThis is the elliptic-curve point count over a finite field formula and the Riemann hypothesis for an elliptic curve over a finite field. In particular the powers on both factors are necessary; they are damaged in the converted TeX but present in the original PDF.

Put $N(x)=x^3+x+2$, $H(x)=x^3+2x+1$ and $R(x)=N(x)/(x-1{)}^2$, all in characteristic three. Since $N(1)=1$, the numerator and denominator have no common factor, and $R$ has degree three. The map $x:E\to {\mathbb{P}}^1$ has degree two. Comparing the degrees in $x\circ \varphi =R\circ x$ gives $2\mathrm{deg}\varphi =3\cdot 2$, so $\mathrm{deg}\varphi =3$. This is the degree of an isogeny from its x-coordinate map.

In characteristic three, direct differentiation givesThe supplied $y$-coordinate is $H(x)y/(x-1{)}^3$, so the invariant differential on an elliptic curve satisfiesIn particular $\varphi$ is separable. Its dual isogeny satisfies $\hat{\varphi }\circ \varphi =[3]$. Since $[3{]}^{\ast }\omega =0$ and ${\varphi }^{\ast }\omega =\omega$, the scalar by which $\hat{\varphi }$ pulls back the differential is zero.

Pullback on the elliptic invariant differential is additive for sums of homomorphisms, by the addition identity proved in (a). ConsequentlyThe differential criterion for separability of an isogeny therefore gives separability exactly when $3\nmid m$, with $n$ arbitrary. Such a map is automatically nonzero. When $3\mid m$, every nonzero resulting map is inseparable; the zero map is not a separable isogeny. In fact the zero map occurs only at $(m,n)=(0,0)$: equality of the degrees of $[m]\varphi$ and $[n]\hat{\varphi }$ in a nontrivial vanishing relation would force $m=\pm n$, and then cancellation would force $\varphi =\pm \hat{\varphi }$, contradicting their different differential scalars.

First compute the Mordell-Weil group rank by two-isogeny descent. HereThe two-isogeny formula gives $\varphi :E\to E^{\prime }$. If $u=1+i$, then $u^4=-4$, andis an isomorphism over $K$. Its $x$-coordinate multiplier $u^2$ is a square in $K$.

Use the two-torsion square-class homomorphismFor any prime ideal of the Gaussian integers, if $v(x)>0$, then $x^2+i$ is a unit and $2v(y)=v(x)$; if $v(x)<0$, the term $x^3$ has strictly smallest valuation and $2v(y)=3v(x)$. Hence every valuation of $x$ is even. Since the Gaussian integers form a principal ideal domain, dividing by a square leaves a unit. Their units are $\{1,-1,i,-i\}$, whose square classes are $\{1,i\}$ because $-1=i^2$ and $i$ is not a square in $K$. For the latter assertion, $(a+bi{)}^2=i$ with $a,b\in \mathbb{Q}$ would imply $a^2=b^2$ and $2ab=1$, hence $a^2=1/2$, impossible for rational $a$. ThusBoth classes occur, at $O$ and $T$. This is the unit square-class bound for two-isogeny descent.

Define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly, with ${\alpha }^{\prime }((0,0))=[-4i]=[i]$. Since $\theta$ multiplies nonexceptional $x$-coordinates by a square, and preserves the exceptional classes as well, its image is also $\{1,[i]\}$. The standard kernel identities in two-isogeny descent areHere $\hat{\varphi }$ is the dual isogeny and $\hat{\varphi }\varphi =[2]$. These identities can be checked directly from the formulas: $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$, and conversely a square $x$-coordinate lets the quadratic equation for a preimage be solved using the curve equation. For example, if $x=r^2\ne 0$, the equation for $X$ is $X^2-(4x+2a)X+(a^2-4b)=0$, whose discriminant is $16(x^2+ax+b)=16(y/r{)}^2$; the $Y$-coordinate then follows from the dual formula. The exceptional points satisfy the same completed square-class criterion.

The two-isogeny index formula over a number field keeps track of a small kernel factor:In this case $\ker \hat{\varphi }=\{O,T^{\prime }\}$, where $T^{\prime }=(0,0)$, and $T^{\prime }\notin \varphi E(K)$ because ${\alpha }^{\prime }(T^{\prime })=[i]\ne 1$. Thus $\delta =2$, and the index is $2\cdot 2/2=2$. There is only one nonzero rational 2-torsion point on $E$: the other two would require $x^2=-i$, and $-i$ has the same nonsquare class as $i$. The Mordell-Weil theorem now givesso $r=0$.

It remains to identify all torsion, rather than merely the rank. The elliptic-curve discriminant is $-64i^3$, so the primes $(3)$ and $(2-i)$ have good reduction, with residue characteristics three and five. By the supplied point-count information their reduction groups have orders that are powers of two. The reduction of torsion points on an elliptic curve is injective on prime-to-residue-characteristic torsion. Every odd-primary torsion subgroup therefore injects into a group of two-power order at at least one of these two primes, and must be zero. All torsion is two-primary.

Finally, if a point had order four, its double would be $T$. The elliptic-curve addition formula givesFor $2P=T$ the denominator is nonzero, so $x^2=i$, impossible in $K$. A point of higher two-power order would have a multiple of order four, so it too is excluded. Thus the only torsion points are $O,T$. Together with rank zero,

Translate the rational 2-torsion point to $(0,0)$ and clear denominators to obtain the integral Weierstrass equation of an elliptic curve. Nonsingularity requires $b\ne 0$ and $b^{\prime }=a^2-4b\ne 0$. The two-isogeny descent usesand its dual isogenyThese formulas extend across their missing affine points as degree-two isogenies of elliptic curves, with kernels $\{O,(0,0)\}$ on their respective curves; substitution gives $\psi \varphi =[2]$ and $\varphi \psi =[2]$.