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The piecewise linear functions $h_n$, for distinct integer indices, are linearly independent over the integers. Restrict a finite integer relation to $y=1$. At $x=n$, only $h_n$ has a jump of one in its slope; hence the coefficient of that function must be zero. This proves that the underlying group of the two-generated free Abelian lattice-ordered group has infinite rank of an abelian group.

A concrete model of the two-generated free Abelian lattice-ordered group is the lattice subgroup of real-valued functions on ${\mathbb{R}}^2$ generated by the coordinate projections $X(x,y)=x$, $Y(x,y)=y$. Its elements are finite joins of finite meets of integer linear forms $ax+by$. Thus they are continuous positively homogeneous piecewise linear functions, with pointwise addition and lattice operations. This is the piecewise integer-linear representation of a free Abelian lattice-ordered group.

For the construction, first form the term algebra on two symbols using addition, negation, zero, join and meet, then identify terms that agree in every Abelian lattice-ordered group. Evaluation under any assignment of the two generators factors through this quotient, proving its universal property. To obtain the concrete function model, use the identity-detection theorem that the real line generates the variety of Abelian lattice-ordered groups. Its proof passes a failed identity to a prime ordered quotient, chooses its finitely many active linear pieces, and uses the rational-inequality alternative underlying Farkas' lemma to find a real witness. This makes evaluation on ${\mathbb{R}}^2$ injective as well as surjective onto the generated function group.

Its underlying group is not ${\mathbb{Z}}^n$ for any finite $n$. The integer-linear hinge functions have infinite independent rank: $h_k(x,y)=(x-ky)\vee 0$ are integer-linearly independent for distinct integers $k$. On $y=1$, only $h_k$ has a slope jump at $x=k$, so each coefficient in a finite relation must vanish. The underlying group therefore has infinite rank of an abelian group despite having two lattice generators.

No: there are at most three, including zero and the whole group. A two-generator ordered Abelian group is torsion-free and has rank of an abelian group at most two. A convex subgroup of an ordered group is pure: if $ng\in C$ with $n>0$, then $g$ lies between zero and $ng$ or their negatives, so $g\in C$. Therefore a proper inclusion $C<D$ of convex subgroups increases rational rank: an element of $D\setminus C$ cannot have a nonzero integer multiple in $C$. Convex subgroups form a chain, so a rank-two group permits at most two strict inclusions. This is the bound for convex subgroups of a finite-rank ordered Abelian group. It is sharp: lexicographically ordered group ${\mathbb{Z}}^2$ has exactly $0$, $\{0\}\times \mathbb{Z}$, and ${\mathbb{Z}}^2$.

The Fundamental theorem of finitely generated abelian groups says that every finitely generated abelian group is isomorphic toThe rank of an abelian group $r$ and the invariant factors $d_j$ are uniquely determined, with the finite summand allowed to be absent. Equivalently, one can decompose the torsion subgroup into cyclic groups of prime-power orders. The infinite cyclic part and torsion part must both be included in the statement.

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.

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 $\varphi \psi =[2]$.

Define homomorphisms to the square-class group of a field ${\mathbb{Q}}^{\times }/({\mathbb{Q}}^{\times }{)}^2$ byFor ordinary intersections with a line, the product of the three first coordinates is the square of its intercept, proving the homomorphism identity; tangent cases use the same product with multiplicities. For $T=(0,0)$, direct addition gives $x(P+T)=b/x(P)$, so $\alpha (P+T)=[b]\alpha (P)$. Also $\alpha (P)\alpha (-P)=1$ and $\alpha (T{)}^2=1$, which handle inverse points and $T+T=O$. These verify the exceptional values as well. Their kernels are $\psi E^{\prime }(\mathbb{Q})$ and $\varphi E(\mathbb{Q})$. One can check the first assertion directly: if $x=s^2\ne 0$, solving for the first coordinate of a preimage under $\psi$ giveswith rational corresponding ordinate. Conversely, on $E^{\prime }$ one has ${\psi }_x=Y^2/(4X^2)$, a square. The exceptional point $(0,0)$ has a rational preimage exactly when $b$ is a square. The other kernel assertion follows identically, since the double companion curve is isomorphic to $E$ by scaling its coordinates by four and eight.

We now justify the rank factor in the square-class index formula for two-isogeny descent, rather than forgetting a torsion correction. Put $A=E(\mathbb{Q})$, $B=E^{\prime }(\mathbb{Q})$, andThe preimage of $\psi \varphi A$ under $\psi$ is $\varphi A+\ker \psi$, soThe Mordell-Weil theorem gives $[A:2A]=2^r\#A[2]$, where $r$ is the rank of an abelian group of $A$. If $b^{\prime }$ is a square, $\#A[2]=4$ and $(0,0)\in \varphi A$, so $\delta =1$. If it is not a square, $\#A[2]=2$ and $\delta =2$. Hence $\delta \#A[2]=4$ in both cases, and

To bound these two images, let $\ell \nmid b$ and take a point with $x\ne 0$. If $v_{\ell }(x)>0$, then $x^2+ax+b$ is a unit, so $2v_{\ell }(y)=v_{\ell }(x)$ is even. If $v_{\ell }(x)<0$, the leading term $x^3$ uniquely has least valuation in the equation, so $2v_{\ell }(y)=3v_{\ell }(x)$ and again $v_{\ell }(x)$ is even. Thus every image class has a signed square-free integer representative supported on the prime divisors of $b$. The exceptional class $[b]$ also has that property. There are at most $2^{\nu (b)+1}$ such classes. Applying the same reasoning to $E^{\prime }$ gives $\#{\alpha }^{\prime }(B)\le 2^{\nu (b^{\prime })+1}$. The prime-support bound in two-isogeny descent is thereforeHere $\nu$ counts the distinct prime divisors of the absolute value of a nonzero integer; $\nu (1)=\nu (-1)=0$. The nonsingularity hypotheses exclude the otherwise undefined case $\nu (0)$.

For the unheaded practical procedure, enumerate signed square-free divisors $d$ of $b$, and of $b^{\prime }$ on the companion curve. A class $d$ is represented by a point of $E$ exactly when its quartic covering in a two-isogeny descenthas a rational solution with $(U,V)\ne (0,0)$. For $UV\ne 0$, reconstruct $x=d{U}^2/V^2$, $y=dUW/V^3$; substitution verifies the equivalence. Solutions with $V=0$ or $U=0$ account for the identity class or the class of $(0,0)$ respectively. Clearing denominators allows integral $U,V,W$ with $\gcd (U,V)=1$.

Test these finitely many coverings over the real numbers and local fields, especially at two and the primes dividing $b{b}^{\prime }$, to exclude impossible classes. Search the survivors for rational points, close the witnessed classes under multiplication, and use the index formula once both images are determined. Equivalently, local solubility gives a descent upper bound and independent rational points, certified for example by their canonical height of an elliptic curve pairing, give a lower bound. When the bounds coincide the rank is determined. This procedure often succeeds, but local solubility alone does not prove global solubility: a nonzero Tate–Shafarevich group can leave surviving coverings without rational points. In that case the computation gives a rigorous bound rather than a falsely certified exact rank.

The given homeomorphisms and homotopy invariance of homology givebecause $n\ge 2$, while $M\simeq S^{n-1}$ is path connected and $L$ is the disjoint union of two copies of $S^{n-1}$. HenceThe degree-one and degree-zero part of the Mayer–Vietoris sequence is thereforeTaking the rank of an abelian group throughout this short exact sequence givesso $\mathrm{rank}{H}_0(N)=2$. Since the zeroth homology group is freely generated by the path components,

Present a finitely generated abelian group $G$ as the cokernel of an integer matrix $A:{\mathbb{Z}}^m\to {\mathbb{Z}}^n$. Changing $A$ by the invertible row and column operations used in its Smith normal form does not change the isomorphism type of the cokernel. Ifthenwhere factors with $d_i=1$ may be omitted. This is the Fundamental theorem of finitely generated abelian groups; the rank of an abelian group $r$ and invariant factors $d_i>1$ are unique.

For an elliptic curve over a number field $K$, the Mordell-Weil theorem writes $E(K)$ as a finite torsion subgroup plus a free abelian group of finite rank of an abelian group $r$. This integer is the rank of the elliptic curve over $K$. It can increase under field extension. A two-isogeny descent supplies rank bounds from rational square-class images; the canonical height pairing can certify independence of rational points.

For $E:y^2=x(x^2+ax+b)$ and $E^{\prime }:y^2=x(x^2-2ax+a^2-4b)$ over $\mathbb{Q}$, define the square-class homomorphisms by $\alpha (x,y)=[x]$, $\alpha (O)=1$, $\alpha ((0,0))=[b]$, and similarly ${\alpha }^{\prime }$ with $b^{\prime }=a^2-4b$. Their kernels are the images of the dual degree-two isogenies of elliptic curves. If $r$ is the rank of an abelian group of $E(\mathbb{Q})$, then $2^r=\#\alpha (E)\#{\alpha }^{\prime }(E^{\prime })/4$. To check the factor four, put $\delta =[\ker \hat{\varphi }:\ker \hat{\varphi }\cap \varphi E]$. The index of $2E$ in $E$ is $\#\alpha (E)\#{\alpha }^{\prime }(E^{\prime })/\delta$. If $b^{\prime }$ is a square, $\delta =1$ and $\#E(\mathbb{Q})[2]=4$; otherwise $\delta =2$ and $\#E(\mathbb{Q})[2]=2$. In either case $\delta \#E(\mathbb{Q})[2]=4$.

For the two-isogeny formula $\varphi :E\to E^{\prime }$ over a number field, the two-torsion square-class homomorphisms have kernels $\hat{\varphi }{E}^{\prime }(K)$ and $\varphi E(K)$. ConsequentlyIndeed $2E(K)=\hat{\varphi }\varphi E(K)$, and the surjection from $E^{\prime }(K)/\varphi E(K)$ to $\hat{\varphi }{E}^{\prime }(K)/2E(K)$ has kernel $\ker \hat{\varphi }/(\ker \hat{\varphi }\cap \varphi E(K))$. Combined with the Mordell-Weil theorem, the index is $2^r\#E(K)[2]$, where $r$ is the rank of an abelian group. This formula prevents an erroneous extra factor of two in an isogeny descent.