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Choose homogeneous polynomials $F,G$ defining the two projective plane curves. Their absence of a common irreducible component makes them coprime. For a point $O=[1:0:0]$ outside both curves, they have degrees $n,m$ in $X$ and nonzero constant leading coefficients. The resultantis therefore nonzero. Indeed, a vanishing resultant over $k(Y,Z)$ would give a common factor there, and Gauss's lemma would give a common component of the original curves.

The homogeneity of $F,G$ makes $R$ a homogeneous polynomial of degree $nm$: scaling $(Y,Z)$ by $a$ scales their $X$-roots by $a$, and the product of the $nm$ differences of roots scales by $a^{nm}$. Every point in the intersection projects from $O$ to a zero of $R$ in the projective line. Each such projection line contains only finitely many points of either curve, since it passes through $O$ outside them. Thus the intersection is finite already.

We may now choose $O$ also outside the finitely many joining lines of distinct intersection points. Projection is then injective on the intersection. A nonzero degree-$nm$ homogeneous polynomial on the projective line has at most $nm$ distinct zeros, so the resultant bound for intersections of plane curves givesThis derives the required cardinality bound without assuming the full intersection-multiplicity statement of Bézout's theorem.

For genuine degrees three and two, the resultant is the Sylvester matrix determinantEquivalently, if $r_1,r_2,r_3$ are the roots of $p$, counted with multiplicity, then $R(p,q)=a^2{\prod }_{i}q(r_i)$. Thus $R(p,q)\ne 0$ means that the polynomials have no common root, or equivalently are coprime over $\mathbb{C}$. One can also see this from the matrix: its singularity gives a nonzero relation $Ap+Bq=0$ with $\mathrm{deg}A<2$, $\mathrm{deg}B<3$. If $p,q$ were coprime, $p\mid B$ would force $B=0$, then $A=0$, a contradiction. Conversely, a common factor supplies such a relation after dividing $p,q$ by that factor. When a leading coefficient is zero, use the resultant for the actual polynomial degrees rather than assuming the displayed fixed-degree criterion unchanged.

An algebraic addition theorem for a meromorphic function $f$ means a nonzero polynomial $P(U,V,W)$ withidentically where the values are finite. A local relation extends meromorphically to the other regular arguments.

For a polynomial $f$ of positive degree $d$, let $z_i$ be the $d$ roots of $f(z)-U$ and $w_j$ the $d$ roots of $f(w)-V$. FormIts coefficients are separately symmetric polynomials in the two sets of roots. The Fundamental theorem of symmetric polynomials and the fixed nonzero leading coefficient of $f$ show they are polynomials in $U,V$. The expression is monic of degree $d^2$ in $W$, so it is nonzero. If $U=f(z)$, $V=f(w)$, one factor vanishes at $W=f(z+w)$. This proves the addition relation, including at exceptional multiple-root values by polynomial identity. A constant polynomial instead has the relation $W-f(0)=0$.

For the elliptic case, let $E=\mathbb{C}/\Lambda$. A nonconstant elliptic function $f$ defines a finite morphism $E\to {\mathbb{P}}^1$, so the function field of $E$ is a finite algebraic extension of $\mathbb{C}(f)$. The elliptic function-field decomposition expresses it as $\mathbb{C}(\wp ,{\wp }^{\prime })$, withThe Weierstrass addition formula makes translation algebraic:Differentiating and using the cubic differential equation also makes ${\wp }^{\prime }(z+w)$ rational in the four separate values. Therefore $f(z+w)$ lies in a finite algebraic extension of $\mathbb{C}(f(z),f(w))$. Its algebraic equation, after clearing denominators, is the required polynomial relation. This sketches why every elliptic function, not only $\wp$, has an algebraic addition theorem. Constants are already covered.

For the final assertion, consider $f(z)=e^{e^z}$, an entire function. If $c$ is a period, then $e^{e^z(e^c-1)}=1$ for every $z$. The continuous exponent must be a constant integer multiple of $2\pi i$; its derivative forces $e^c=1$. Hence its periods are exactly $2\pi i\mathbb{Z}$, making it a simply periodic function.

Suppose it had an algebraic addition theorem $P$. There are only finitely many values $V_0$ for which $P(U,V_0,W)$ is identically zero: a nonzero coefficient polynomial in $V$ already bounds this exceptional set. Choose a positive irrational $\alpha$ with $e^{\alpha }$ outside this set, and set $c=\log \alpha$. Specializing $w=c$ gives a nonzero polynomial $Q(U,W)=P(U,e^{\alpha },W)$ and, with $t=e^z$,All exponents with nonzero coefficients are distinct, by irrationality of $\alpha$. Divide by the exponential with largest real exponent and let $t\to +\infty$; its nonzero coefficient would have to tend to zero. This contradiction proves that simple periodicity does not imply an algebraic addition theorem. It is the simply periodic entire function without an algebraic addition theorem example.

The polynomial discriminant satisfiesA polynomial has no repeated roots over the algebraic closure exactly when it is coprime to its derivative. Consequently the assumed squarefree reduction has nonzero resultant with its derivative, so $p\nmid \mathrm{disc}(f)$. On the other hand the power basis is a basis of the order in a number field $\mathbb{Z}[\alpha ]$, and the discriminant-index formula for an integral lattice givesHere the field discriminant $d_L$ is an integer. ThereforeThis is squarefree reduction excludes a prime from the power-basis index; its argument establishes the index condition before applying the Kummer-Dedekind theorem.

Irreducibility modulo two implies irreducibility over $\mathbb{Q}$ by Gauss lemma for polynomials, so the Galois group embeds transitively in $S_4$. The discriminant is $229$, hence primes two and three are unramified. The factor-degree theorem says that square-free reduction with factor degrees $d_1,\dots ,d_r$ provides an element of the Galois group with cycle lengths $d_1,\dots ,d_r$. The reductions already found therefore give a 4-cycle and a 3-cycle.

The group order divides 24 and is divisible by four and three, so is 12 or 24. An order-12 subgroup of $S_4$ has index two and must be $A_4$: in its quotient every transposition has the same nontrivial image, since transpositions are conjugate and generate $S_4$, making the quotient map the sign. But a 4-cycle is odd. Therefore the order is 24 andThe discriminant can also be obtained from $f^{\prime }=x^2(4x+3)$ and the resultant, giving $256-27=229$.

Heights measure arithmetic size, and their finiteness and growth properties turn finite divisibility information into finite generation. For an elliptic curve $E$ over a number field $K$, the resulting Mordell-Weil theorem isThe distinction between this statement and the Weak Mordell-Weil theorem, which only asserts that $E(K)/mE(K)$ is finite, is exactly where heights enter.

For a rational point of the projective line, choose coprime integers $a,b$ and define the logarithmic projective heightFor a fixed number field, its absolute logarithmic version iswith the usual normalized absolute values. The product formula makes this independent of scaling $(a,b)$, and the local degree formula makes it independent of enlarging the number field. Heights are nonnegative and are useful precisely because they control numerator and denominator together, unlike an ordinary real absolute value.

The Northcott theorem states that algebraic points of bounded degree and bounded height form a finite set. Over $\mathbb{Q}$ this is immediate from the coprime numerator-denominator description. In general, bounded degree and height bound the coefficients of the primitive minimal polynomial: their elementary symmetric functions are bounded by its Mahler measure, which is the appropriate power of the multiplicative height. There are only finitely many such integer polynomials.

On a Weierstrass equation of an elliptic curve, define the naive logarithmic height $h_x(P)=h(x(P))$ and $h_x(O)=0$. Since an $x$-coordinate has at most two points above it, Northcott theorem implies that only finitely many points of $E(K)$ have bounded $h_x$.

Duplication descends to a rational map of the projective $x$-line of degree $4$. More generally, if a morphism $\varphi :{\mathbb{P}}^1\to {\mathbb{P}}^1$ has degree $d$, thenuniformly in $z$. Write $\varphi$ as two homogeneous degree-$d$ polynomials without common zeros. Bounding their coefficients gives the upper estimate at each place. A nonzero resultant, or homogeneous Bezout identities, prevents simultaneous cancellation and gives the lower estimate; only finitely many places contribute a nonzero constant. Applied to duplication, this gives the height growth under a morphism of the projective line estimate

Define the canonical height of an elliptic curve byThe factor $1/2$ is a convention making it the height for the degree-one symmetric divisor $O$. If $a_n=4^{-n}{h}_x(2^{n}P)$, then $|a_{n+1}-a_n|\le C_E{4}^{-(n+1)}$, so the limit exists by a geometric-series estimate. Summing that estimate also provesThus bounded canonical height still gives only finitely many $K$-points, and $\hat{h}(2P)=4\hat{h}(P)$ exactly.

The addition formula on the quotient by negation gives the uniform approximate parallelogram identityThis height parallelogram identity follows as follows. One way to obtain it is to express the unordered pair of their $x$-coordinates as a morphism ${\mathbb{P}}^1\times {\mathbb{P}}^1\to {\mathrm{Sym}}^2{\mathbb{P}}^1$ of bidegree $(2,2)$; the height of that unordered pair is the sum of the two heights, up to a bounded constant. It is also the height identity for the symmetric divisor $O$. Apply this identity to $2^{n}P,2^{n}Q$, divide by $2\cdot 4^n$, and take limits. The errors disappear, givingTogether with evenness and $\hat{h}(O)=0$, the recurrence implies $\hat{h}(mP)=m^2\hat{h}(P)$ for every integer $m$. The associated canonical height pairingis bilinear. Nonnegativity of the height implies its Cauchy-Schwarz inequality, by applying it to integer combinations and then rational approximations. In particular,The height vanishes precisely at torsion points: torsion gives a finite set of multiples, whereas if $\hat{h}(P)=0$, every multiple of $P$ has bounded naive height and therefore the set of multiples is finite by Northcott theorem.

Now apply the Weak Mordell-Weil theorem with $m=2$, proved cohomologically in the other essay. Choose a finite set $R$ of representatives for $E(K)/2E(K)$, and put $H={\max }_{R_i\in R}\hat{h}(R_i)$. Write an arbitrary point as $P=2Q+R_i$. ThenIf $\hat{h}(P)>H+1$, this decreases the height by more than $1/2$. Repeat until the current point has height at most $H+1$. There are only finitely many such points. Unwinding the equations $P=2Q+R_i$ shows that $E(K)$ is generated by this finite bounded-height set together with $R$. This is the height descent lemma, and proves the Mordell-Weil theorem.

Once finite generation is established, the structure theorem for finitely generated abelian groups gives the displayed decomposition. The canonical height pairing is positive definite on the real vector space of the free part: a null direction would, by approximation with integer combinations of generators, give infinitely many lattice points of bounded height, contradicting Northcott theorem. Its determinant is the regulator of an elliptic curve, and its geometry gives practical tools for searching for generators and comparing independence. Thus heights supply both the finiteness mechanism in the proof and the quadratic size function used in arithmetic computations.

For a nonzero isogeny of elliptic curves $\varphi :E_1\to E_2$, its degree of an isogeny is the degree of the induced extension of function fields. Since $\varphi$ commutes with negation, its first coordinate is a rational function of $x_1$ alone. Write it in lowest terms asThe coordinate maps $x_i:E_i\to {\mathbb{P}}^1$ are finite morphisms of degree two. Multiplying function field degrees in the commutative diagram $x_2\circ \varphi =R\circ x_1$ gives the degree of an isogeny from its x-coordinate map:This is the total degree, so the argument also covers an inseparable isogeny of elliptic curves. Counting distinct points in the kernel of an isogeny would give only its separable degree. The zero group homomorphism is assigned degree zero separately.

On a Short Weierstrass form $y^2=x^3+ax+b$ in characteristic of a field other than two, the elliptic-curve addition formula givesThere is no cancellation: at a root $r$ of $f(x)=x^3+ax+b$, the numerator equals $f^{\prime }(r{)}^2$, which is nonzero because the elliptic-curve discriminant is nonzero. The numerator has degree four and the denominator degree three. HenceFor completeness, the same answer holds in characteristic two, where the displayed short equation is not a nonsingular model. On a general Weierstrass equation of an elliptic curve usewhere $b_2=a_1^2+4a_2$, $b_4=a_1{a}_3+2a_4$, $b_6=a_3^2+4a_6$, and $b_8=a_1^2{a}_6+4a_2{a}_6-a_1{a}_3{a}_4+a_2{a}_3^2-a_4^2$. The resultant of this numerator and denominator is ${\Delta }^2$. Their resultant is therefore nonzero in every nonsingular case; even when the denominator drops degree, the numerator retains degree four.

Let $f(x)=a{\prod }_{i=1}^n(x-{\alpha }_i)$. Its polynomial discriminant isThe squared Vandermonde determinant is a symmetric polynomial in the roots. By the Fundamental theorem of symmetric polynomials, it is a polynomial in their elementary symmetric functions, which are the coefficients of $f$ by Vieta formulas; hence $\Delta (f)\in K$. Equivalently, the resultant formula puts it directly in $K$.

For $f=x^4+rx+s$, a resultant calculation with $f^{\prime }=4x^3+r$ gives

For $x=u/v\in \mathbb{Q}$ in lowest terms, the naive height on the projective line isHomogenize the coprime numerator and denominator of $\xi$ to degree $d$. The triangle inequality gives the upper estimate $H(\xi (x))\le c_{2}H(x{)}^d$. Since the two homogenized forms have no common projective zero, their resultant is nonzero, and the Bézout identities for the resultant express fixed multiples of $u^{2d-1}$ and $v^{2d-1}$ as combinations of their values with coefficients of degree $d-1$. After cancellation this gives $H(x{)}^d\le CH(\xi (x))$, which is the lower estimate with $c_1=C^{-1}$.

Now write $x=u/v$ in lowest terms and putSince $\gcd (N,v)=1$, the equation $y^2=N/v^3$ givesClearly $M\le \gamma H(x{)}^3$ for $\gamma =1+|a|+|b|$. Homogenizing the supplied polynomial identity givesIts coefficient sum is at most ${\gamma }^2$, so $|u{|}^5\le {\gamma }^{2}H(x{)}^{2}M$. The same lower bound is immediate from $|v{|}^3\le M$ when $|v|=H(x)$. Thus

For $P=[x:y]\in {\mathbb{P}}^1(\mathbb{Q})$ represented by coprime integers, its naive height on the projective line isWrite the degree-$d$ morphism as $\Phi =[F:G]$, where $F,G\in \mathbb{Z}[X,Y]$ are homogeneous of degree $d$ with no common projective zero. Bounding their coefficients gives

For the reverse inequality, the nonvanishing of the resultant of $F$ and $G$ gives homogeneous Bézout identities expressing fixed nonzero integer multiples of powers of $X$ and $Y$ as polynomial combinations of $F$ and $G$. Evaluating at $(x,y)$, removing the common divisor of $F(x,y)$ and $G(x,y)$, and taking the larger of $|x|,|y|$ givesThus $c_{1}H(P{)}^d\le H(\Phi (P))\le c_{2}H(P{)}^d$ for constants depending only on $\Phi$.

For projective plane curves without a common irreducible component, project from a point outside both curves. In coordinates where that point is $[1:0:0]$, their defining homogeneous polynomials have constant nonzero leading coefficients in $X$. The resultant in $X$ is nonzero by Gauss's lemma, and is homogeneous of degree $\mathrm{deg}C\mathrm{deg}D$ in the remaining two coordinates. Every intersection projects to one of its zeros. Each projection line has only finitely many intersections with the curves, so their intersection is finite. Choose the center again outside all joining lines of pairs of intersection points; the projection is now injective on that finite set, proving the bound.

For an algebraic integer $\alpha$ generating a number field, the polynomial discriminant of its minimal polynomial $f$ equals the discriminant of elements of a number field for its power basis. A squarefree reduction modulo $p$ has nonzero resultant with its derivative, so $p\nmid \mathrm{disc}(f)$. The discriminant-index formula for an integral lattice, $\mathrm{disc}(f)=[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]{]}^2{d}_K$, proves the assertion.

// Target: arithmetic.bigb

For polynomials $p,q$ of degrees $m,n$, the Sylvester matrix has $n$ shifted coefficient rows of $p$ and $m$ shifted rows of $q$, with coefficients in descending-power order. It represents, after transposition, the map $(A,B)\mapsto Ap+Bq$ for $\mathrm{deg}A<n$, $\mathrm{deg}B<m$. Its determinant is the resultant. Its kernel is nonzero exactly when a common polynomial factor exists: divide $p,q$ by that factor to obtain a nonzero cancelling pair; conversely, coprimality and $Ap=-Bq$ imply $p\mid B$, forcing the degree-bounded $B$ and then $A$ to vanish.