1. Brief review on Weierstrass $\wp$-function. We first begin with a lattice $L:=\mathbb{Z}\omega_1+\mathbb{z}\omega_2$, where $\omega_1,\omega_2\in \mathbb{C}$ are linearly independent over $\mathbb{R}$. The fundamental domain $\mathcal{F}$ of $L$ is the set $\mathcal{F}:=\{a\omega_1+b\omega_2|a,b\in [0,1)\} = \mathbb{C}/L$. One can consider $\mathcal{F}$ as a group by the following addition law: $z_1+z_2 = z_1+z_2\mod L$. The fundamental domain $\mathcal{F}$ is called a torus associated to $L$. The Weierstrass $\wp$-function associated to $L$ is defined as follow:
$$\wp(z) := \frac{1}{z^2}+\sum_{\omega\in L\setminus \{0\}}[\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}]$$
It can be seen that $\wp(z)$ is a non-constant meromorphic even function on $\mathbb{C}$. More clearly, poles of $\wp(z)$ is exactly the lattice points of $L$, and $\wp(z)=\wp(-z)$. If we take the derivative of $\wp(z)$, we get
$$\wp'(z) = -2\sum_{\omega\in L}\frac{1}{(z-\omega)^3}$$
We can see that $\wp'(z)$ is an odd function, and both $\wp(z)$ and $\wp'(z)$ are doubly periodic, i.e. for all $\omega\in L$, we have $\wp(z+\omega)=\wp(z)$, and $\wp'(z+\omega) = \wp'(z)$. We next define the $k$-th Eisentein's series associated to $L$ as follows.
$$G_k=\sum_{\omega\in L\setminus \{0\}}\frac{1}{\omega^k}$$
Around a point $\omega$ in $L$, we can can see
$$\wp(z) = \frac{1}{z^2}+\sum_{\omega\in L\setminus \{0\}}\omega^{-2}[\frac{1}{(z/\omega-1)^2}-1]$$
Using the geometric series, we get
$$\wp(z) = \frac{1}{z^2} + \sum_{\omega\in L\setminus \{0\}}\omega^{-2}(\sum_{n\ge 1}(z/\omega)^n)^2=\frac{1}{z^2}+\sum_{\omega\in L\setminus \{0\}}\sum_{n\ge 1}(n+1)\frac{z^n}{\omega^{n+2}}=$$
$$=\frac{1}{z^2}+\sum_{n\ge 1}(n+1)z^n\sum_{\omega\in L\setminus \{0\}}\frac{1}{\omega^{n+2}}=\frac{1}{z^2}+\sum_{n\ge 1}G_{n+2}(n+1)z^n$$
Because $\wp(z)$ is an even function, we get
$$\wp(z) = \frac{1}{z^2}+\sum_{n\ge 1}G_{2n+2}(2n+1)z^{2n}$$
Using this identity, we can obtain the power series expansion for $\wp(z)$ as follows
$$\wp'(z) = \frac{-2}{z^3}+\sum_{n\ge 1}G_{2n+2}(2n+1)(2n)z^{2n-1}$$
and by direct computation, we have
$$f(z):=\wp'(z)^2 - 4\wp(z)^3 + 60G_4\wp(z) + 140G_6 = c_1z+c_2z^2 + ...$$
That means $f(z)$ is a holomorphic function, this yields $f(z)$ is bounded in the fundamental domain $\mathcal{F}$, and it is doubly periodic (since $\wp(z)$ and $\wp'(z)$ are). Now, it follows that $f(z)$ is an entire funcion and bounded. By Louville's theorem, $f(z)=f(0)$ is a constant function. Because $f(0)=0$, we have $f(z)\equiv 0$. This yields a following important identity:
$$\wp'(z)^2 = 4\wp(z)^3 - g_2\wp(z) - g_3$$
where $g_2 = 60G_4, g_3 = 140 g_6$. That means, $(\wp(z),\wp'(z))$ is a point on a curve $ (E): y^2= x^3 - g_2x - g_3$. This curve is non-singular, and for any $(x,y)$ on the curve, there exists $z$ such that $(x,y)=(\wp(z),\wp'(z))$. Moreover, one obtain the following important bijection $\mathbb{C}/L \ to (E)$ by sending $z\ne 0 \mapsto (\wp(z):\wp'(z):1)$ and $0\mapsto (0:1:0)$. This gives a natural group law on $(E)$ induced from the group law on $\mathbb{C}/L$. For any $z\ne 0 \in \mathbb{C}/L$, because $z+0=0+z=z$, we can define
$$(\wp(z):\wp'(z):1) + (0:1:0) = (\wp(z):\wp'(z):1)$$
If $z=0$, we have $(0:1:0)+ (0:1:0) = (0:1:0)$. If $z_1,z_2\in \mathbb{C}/L\setminus\{0\}$, and $z_1\ne z_2$, then $z_1+z_2$ correspond to the point $(\wp(z_1+z_2):\wp'(z_1+z_2):1)$ on the curve. So, we can define
$$(\wp(z_1):\wp'(z_1):1)+(\wp(z_2):\wp'(z_2):1):=(\wp(z_1+z_2):\wp'(z_1+z_2):1)$$
Similarly, when $z_1 = -z_2 in \mathbb{C}/L$, and both are non-zero, because $z_1+z_2=0$, we can define
$$(\wp(z_1), \wp'(z_1):1)+(\wp(z_2):\wp'(z_2):1)=(0:1:0)$$
This gives $(E)$ the natural group structure, and $(E)\cong \mathbb{C}/L$ as abelian groups.
NOTE. We are familiar with the fact that the group law on an elliptic curve comes from the intersection of curves with lines. Yes, it is true also in this context. Let us consider the line goes through two (affine) points $(\wp(z_1), \wp'(z_1))$ and $(\wp(z_2),\wp'(z_2))$ with equation: $y-ax-b=0$, it will cut the curve at the third point: $(\wp(-z_1-z_2), \wp'(-z_1-z_2))$.
To prove this fact, let $f(z) = \wp'(z) - a\wp(z) - b$ (correspond to our line: $y-ax-b$), then we know that $f(z)=0$ at two points $z_1,z_2 in \mathbb{C}/L$, and it has unique pole in this domain (at 0) with multiplicity 3 (since $\wp'(z)$ has). We now use a theorem from complex analysis, if $f$ is a meromorphic map from $\mathbb{C}/L$ to $\mathbb{C}$, and it has zero and poles at $z_i$ with multiplicity $n_i$, then $\sum_{i}n_i=0$, and $\sum_{i}n_iz_i=0$. From this, we can see $f$ has another zero $z$ such that $z_1+z_2+z+0=0$, i.e. $z=-z_1-z_2$.
This will yields the third intersection point between the line and our curve is $(\wp(-z_1-z_2), \wp'(-z_1-z_2))$. So, taking the inverse point, we get $(\wp(-z_1-z_2), -\wp'(-z_1-z_2))$. Because $\wp(z)$ is an even function, and $\wp'(z)$ is an odd function, we get $((\wp(-z_1-z_2), -\wp'(-z_1-z_2))=(\wp(z_1+z_2),\wp'(z_1+z_2))$. And this is identical with our familiar definition on group law.
2. Complex theory of isogenies. Let $L_1, L_2$ be lattices in $\mathbb{C}$ with fundamental domain $E_1:=\mathbb{C}/L_2,E_2:=\mathbb{C}/L_2$, which are considered as elliptic curves defined over $\mathbb{C}$. Let $\alpha\in \mathbb{C}$ such that $\alpha L_1\subset L_2$, we define the isogeny $[\alpha]: E_1\to E_2$ by sending $z\mod L_1$ to $\alpha z\mod L_2$. This is a well-defined map, and is a homomorphism of groups.
If $\alpha\ne 0$, it can be seen that $\alpha L_1$ is a sublattice of $L_2$, i.e. a subgroup of rank 2 of $L_2$. This will yields the index $[L_2:\alpha L_1]$ is finite. In the case $\alpha\ne 0$, the degree of $[\alpha]$ is defined by the index $[L_2:\alpha L_1]$. If $\alpha=0$, we define the degree of $[\alpha]$ as 0.
Assume that $\alpha\ne 0$ and $n$ is the degree of the map $\alpha$, we have $n L_2\subset \alpha L_1$. This implies $\hat{\alpha} L_2\subset L_1$, where $\hat{\alpha} := (n/\alpha)$ and it will yields the isogeny $[\hat{\alpha}]$ from $E_2$ to $E_1$, which is called the dual of $[\alpha]$. We will prove that $\deg [\alpha] = \deg [\hat{\alpha}]$.
In fact, let $\{\omega_1,\omega_2\}$ is the basis for $L_1$, and $\{\omega_3,\omega_4\}$ is the basis for $L_2$. Because $\alpha L_1\subset L_2$, one can represent $\alpha \omega_1 = a\omega_3 + b\omega_4$, $\alpha \omega_2=c\omega_3 + d\omega_4$, with $a,b,c,d\in \mathbb{Z}$. Let us denote $A$ the $2\times 2$ matrix with the first row is $a,b$ and second row is $c,d$. Then the degree of $[\alpha]$ is $|\det A| = n$ Also, since $n/\alpha L_2\subset L_1$, we can also represent $n \omega_3 = e (\alpha\omega_1) + f (\alpha\omega_2) = (ea+fc)\omega_3 + (eb+fd)\omega_4$, and $n \omega_3 = g(\alpha \omega_1) + g(\alpha\omega_2) = (ga+hc)\omega_3 + (gb+hd)\omega_4$. If we denote $B$ the matrix with the first row $e,f$ and the second row $g, h$, then $\deg([\hat{\alpha}])=|\det B|$. Furthermore, it can be seen that $BA$ is the matrix send $(\omega_3, \omega_4)$ to $(n\omega_3, n\omega_4)$, i.e. $|\det B||\det A|=n^2$. This yields $\deg [\hat{\alpha}]=\deg \alpha$.
One can also easily deduce that $[\hat{\hat{\alpha}}]=\alpha$, and $[\alpha]\circ [\hat{\alpha}] = [\deg \alpha]$ is an isogeny from $E_1$ to $E_1$. Also, $[\hat{\alpha}]\circ [\alpha]=[\deg \alpha]=[\deg \hat{\alpha}]$ is an isogeny from $E_2$ to $E_2$. The kernel of $[\alpha]$, it is exactly $z\in E_1$ such that $\alpha z\in L_2$.
Furthermore, we can deduce the complex version of Velu's formula.
Proposition 2.1. Let $G\subset E_1$ is a finite subgroup. Then there exists a lattice $L_2$ and an isogeny from $E_1\to E_2$ such that its kernel is $G$.
Proof. Due to the correspondence theorem of groups, there exists $L_2\subset \mathbb{C}$ such that $G=L_2/L_1$. If we denote $\#G= n$, then it can be seen that $L_1\subset L_2\subset (1/n)L_1$, i.e. $L_2$ is also a lattice. The map $E_1\to E_2$ sending $z\mod L_1$ to $z\mod L_2$ has the kernel $G$. (Q.E.D)
We can also obtain the complex version of the following statement
Proposition 2.2. Let $f$ be a holomorphic map between $E_1$ and $E_2$, then $f(z)$ is of the form $\alpha z+\beta$ for some $\alpha,\beta \in \mathbb{C}$, i.e. $f$ is a composition of an isogeny and a translation map.
Proof. We will use a bit theory of Riemann surfaces, because $\mathbb{C}$ is a universal covering of $E_1, E_2$, with projection maps $\pi_1, \pi_2$. The holomorphic map between $E_1$ and $E_2$ will induce the holomorphic map from $\tilde{f}: \mathbb{C}\to \mathbb{C}$ such that $\tilde{f}(z\mod L_1)=\tilde{f}(z)\mod L_2$, i.e. it makes the diagram commutes (the reader should draw it, I cannot draw it here).
From this, for any $\omega\in L_1$, we have
$$\tilde{f}(z+\omega) \equiv \tilde{f}(z+\omega)=\tilde{f}(z)\equiv \tilde{f}(z)\mod L_2$$
If we let $g(z) = \tilde{f}(z+\omega) - \tilde{f}(z)$, then $g(z)$ is a holomorphic function, and $g(z)$ takes values only on $L_2$, which is a discrete subset of $\mathbb{C}$. Hence, $g(z)$ is a constant function, then $0=g'(z) = \tilde{f}'(z +\omega)-\tilde{f}'(z)$, i.e. $\tilde{f}'(z) = \tilde{f}'(z+\omega)$, i.e. $\tilde{f}'$ is a bounded entire function. By Louville's theorem again, $\tilde{f}'(z)$ is a constant function. This yields, $\tilde{f}(z) = \alpha z + \beta$, for some $\alpha,\beta \in \mathbb{C}$. (Q.E.D)
3. Brief review on divisors. We will recall something about intersection numbers and divisors. Let $E$ be any projective curve in $\mathbb{P}^2$, the formal sum $D:=\sum_{P\in E}n_P [P]$, where $n_P\in \mathbb{Z}$ and $n_P\ne 0$ at finitely many points $P\in E$, is called a divisor on $E$. The set of all divisors on $E$ is denoted $Div(E)$. It can be seen that $Div(E)$ has the abelian group structure induced from $\mathbb{Z}$. The degree of $D$, denoted by $\deg D$ is defined as $\sum_{P\in E}n_P$.
If $E, F$ are projective curves in $\mathbb{P}^2$, without common irreducible components. it can be seen that $E\cap F$ is a finite set (since $E,F$ is of dimension 1, their intersection is of dimension 0, and hence, discrete and finite). At a point $P\in E$, we can define the intersection multiplicity (with $F$) at $P$, which is denoted $I_P(E,F)$, it is a non-negative integer, and is positive if $P\in E\cap F$. Furthermore, it has the following properties: $I_P(E,F)=I_P(F,E)$ and $I_P(E,FG) = I_P(E,F)+I_P(E,G)$, where we identify $FG$ with the zeros of $fg$, with $F, G$ are given by $f=0, g = 0$ resp. From this, one obtains the Bezout's theorem:
$$\deg E \deg F = \sum_{P\in E\cap F} I_P(E,F)$$
If $E$ is a curve, and $f:=f_1/f_2$ is a rational function, where $f_1,f_2\in k[x_0,x_1,x_2]\setminus \{0\}$ are homogeneous polynomial of the same degree, such that $F_i$ (the curve defined by $f_i=0$) has no common component with $E$. We define
$$div(f) := \sum_{P\in E\cap F_1}I_P(E,F_1)[P] - \sum_{Q\in E\cap F_2}I_P(E,F_2)[Q]$$
If $E, F$ are projective curves in $\mathbb{P}^2$, without common irreducible components. it can be seen that $E\cap F$ is a finite set (since $E,F$ is of dimension 1, their intersection is of dimension 0, and hence, discrete and finite). At a point $P\in E$, we can define the intersection multiplicity (with $F$) at $P$, which is denoted $I_P(E,F)$, it is a non-negative integer, and is positive if $P\in E\cap F$. Furthermore, it has the following properties: $I_P(E,F)=I_P(F,E)$ and $I_P(E,FG) = I_P(E,F)+I_P(E,G)$, where we identify $FG$ with the zeros of $fg$, with $F, G$ are given by $f=0, g = 0$ resp. From this, one obtains the Bezout's theorem:
$$\deg E \deg F = \sum_{P\in E\cap F} I_P(E,F)$$
If $E$ is a curve, and $f:=f_1/f_2$ is a rational function, where $f_1,f_2\in k[x_0,x_1,x_2]\setminus \{0\}$ are homogeneous polynomial of the same degree, such that $F_i$ (the curve defined by $f_i=0$) has no common component with $E$. We define
$$div(f) := \sum_{P\in E\cap F_1}I_P(E,F_1)[P] - \sum_{Q\in E\cap F_2}I_P(E,F_2)[Q]$$
Because $\deg F_1=\deg F_2$, due to Bezout's theorem, we get $\deg (div(f))=0$. If $D\in Div(E)$, and there exists $f=f_1/f_2$, where $f_1,f_2\in k[x_0,x_1,x_2]\setminus \{0\}$ are homogeneous polynomial of the same degree, such that $F_i$ (the curve defined by $f_i=0$) has no common component with $E$, such that $D = div(f)$, then $D$ is called a principal divisor. Due to the properties of the intersection numbers, we have $div(fg)=div(f)+div(g)$, i.e. the set of all principal divisors form a subgroup of $Div(E)$. We denote this group $Prin(E)$. If $D_1, D_2\in Div(E)$, and $D_1-D_2\in Prin(E)$, we denote $D_1\sim D_2$. It is obvious to see that $\sim$ is an equivalent relation.
It follows from above that any principal divisor has degree 0. Hence, $Prin(E)\subset Div^0(E)$, which is the subgroup of $Div(E)$ containing all degree zero divisors.The quotient group $Div^0(E)/Prin(E)$ is denoted $Pic^0(E)$.
If $E$ is an elliptic curve (we now view $E$ a projective curve), and $L$ is a line. There is 3 possibilities:
1. $L$ cuts $E$ at three distinct points
2. $L$ cuts $E$ at two distinct points, and the intersection multiplicity at a point is $2$, i.e. $L$ is a tangent line of $E$ at this point.
3. $L$ cuts $E$ only at 1 point, and the intersection multiplicity at this point is $3$. In this case, $L$ is also a tangent line of the curve at this point. For example, if we consider the Weierstrass form of $(E): y^2z=x^3+Axz^2+Bz^3$, then the line at infinity $(L): z=0$ cuts the curve only at $(0:1:0)$. Hence, $I_{(0:1:0)}((E), L)=3$. Another example is $(E): y^2z+yz^2=x^3$. The line $(L): y=0$ cuts the curve only at $(0:0:1)$, and $I_{(0:0:1)}(E, L)=3$.
Let $P,Q$ are two distinct points on $E$, the line $L_1$ given by equation $f_1=0$ connecting $P,Q$ will cut the curve at the third point $R$. If this line is the tangent line at $P$ (or $Q$), then $R$ can be considered as $P$ (or $Q$, resp.). If we take the line $L_2$ connecting $R$ and $-R$, given by equation $f_2=0$ then it will cut the curve at the third point $\infty$. So, in this case, by definition,
$$div(f_1/f_2)=[P]+[Q]+[R]-([R]+[-R]+[\infty])= [P]+[Q] - [-R] - [\infty]$$
It follows from above that any principal divisor has degree 0. Hence, $Prin(E)\subset Div^0(E)$, which is the subgroup of $Div(E)$ containing all degree zero divisors.The quotient group $Div^0(E)/Prin(E)$ is denoted $Pic^0(E)$.
If $E$ is an elliptic curve (we now view $E$ a projective curve), and $L$ is a line. There is 3 possibilities:
1. $L$ cuts $E$ at three distinct points
2. $L$ cuts $E$ at two distinct points, and the intersection multiplicity at a point is $2$, i.e. $L$ is a tangent line of $E$ at this point.
3. $L$ cuts $E$ only at 1 point, and the intersection multiplicity at this point is $3$. In this case, $L$ is also a tangent line of the curve at this point. For example, if we consider the Weierstrass form of $(E): y^2z=x^3+Axz^2+Bz^3$, then the line at infinity $(L): z=0$ cuts the curve only at $(0:1:0)$. Hence, $I_{(0:1:0)}((E), L)=3$. Another example is $(E): y^2z+yz^2=x^3$. The line $(L): y=0$ cuts the curve only at $(0:0:1)$, and $I_{(0:0:1)}(E, L)=3$.
Let $P,Q$ are two distinct points on $E$, the line $L_1$ given by equation $f_1=0$ connecting $P,Q$ will cut the curve at the third point $R$. If this line is the tangent line at $P$ (or $Q$), then $R$ can be considered as $P$ (or $Q$, resp.). If we take the line $L_2$ connecting $R$ and $-R$, given by equation $f_2=0$ then it will cut the curve at the third point $\infty$. So, in this case, by definition,
$$div(f_1/f_2)=[P]+[Q]+[R]-([R]+[-R]+[\infty])= [P]+[Q] - [-R] - [\infty]$$
This implies $[P]+[Q]\sim [-R]+[\infty]$. If $P\equiv Q$, we take the tangent line $L_1$ of $E$ at $P$, this will cut the curve at $R$ ($R\equiv P$ in the case the tangent line cuts the curve only at 1 point), we again take the line connecting $R$ and $-R$, this will cut the curve at $\infty$. We again get $div(f_1/f_2)=2[P] - [-R] - [\infty]$, i.e. $2[P]\sim [-R]+[\infty]$.
What we can see from this is that we can define the group law on $E$ as $P + Q = -R$. It is identical with our familiar definition for the group law on an elliptic curve, and it is more accurate, because in the case of tangent line of order 3, we are tricked by our geometric intuition. And we do not have to prove the associativity law, because it is natural induced from the group law on $Pic^0(E)$. Furthermore, one obtains the bijection (and hence, a group isomorphism) between $E$ and $Pic^0(E)$ given by $P\mapsto [P]-[\infty]$.
4. A quick look on algebraic theory of isogenies. Let $E_1,E_2$ are two elliptic curves over an algebraically closed field $k$. An isogeny is both a non-constant morphism and a group homomorphism from $E_1$ to $E_2$, i.e. if we look at affine parts of $E_1$ and $E_2$, an isogeny a rational map between $E_1$ and $E_2$ that sends $\infty$ to $\infty$. Let $\alpha: E_1\to E_2$ be an isogeny, it will induce the injective field homomorphism $\alpha^*: k(E_2)\to k(E_1)$. The field extension $k(E_1)$ of $\alpha^*(k(E_2))$ is finite, and the degree of $\alpha$ is defined as the degree of the field extension. If the field extension is separable, $\alpha$ is called separable isogeny. Otherwise, $\alpha$ is called inseparable.
It follows from Chapter II of Washington's book that:
1. An isogeny is always surjective (We can use a little bit of algebraic geometry to deduce this fact: $E_1$ is projective variety, and hence, is complete, and $\alpha(E_1)$ is closed and irreducible in $E_2$ (since $E_1$ is irreducible), i.e. $\alpha(E_1)=E_2$, since $E_2$ is also irreducible and of dimension 1, and $\alpha$ is non-constant).
2. The kernel of an isogeny $\alpha$ is always finite, and $\# \ker \alpha =\deg \alpha$ if $\alpha$ is separable, i.e. $\deg \alpha$ is actually the number of points in a fiber $\alpha^{-1}(\infty)$. It can be proved that in the case $\alpha$ is an separable isogeny, the fiber of $\alpha$ at every point is equal, and it is $\deg \alpha$.
3. If $\alpha$ is non-separable, then $\deg \alpha >\#ker \alpha$. An example is the Frobenius endomorphism, its degree is $q$ (in the case $E$ are defined over $\mathbb{F}_q$), and it is bijection from $E(\overline{\mathbb{F}_q})$ to $E(\overline{\mathbb{F}_q})$, hence, the its kernel is just $\infty$.
So from 2 and 3, we can see that the kernel of an isogeny is always finite.
4. If $\alpha: E_1\to E_2$ is a morphism, then it induces the push-forward map $\alpha_*$ from $Div(E_1)\to Div(E_2)$ defined by $\alpha_*(\sum_{P\in E} n_P [P])=\sum_{P\in E} n_P [\alpha(P)]$. This can be seen that $\alpha_*$ is a group homomorphism. If furthermore, $\alpha(\infty)=\infty$, then it can be proved $\alpha_*$ maps principal divisors to principal divisors, i.e. the induced map $\alpha_*: Pic^0(E)\to Pic^0(E)$ is well-defined.
Using the isomorphism between $E$ and $Pic^0(E)$, we can prove a beautiful
Proposition 4.1. Let $E_1,E_2$ as above, and $\alpha:E_1\to E_2$ is a non-constant morphism, and $\alpha(\infty)=\infty$, then $\alpha$ is an isogeny.
Proof. It is sufficient to prove that $\alpha$ is a group homomorphism. From the isomorphism $\phi_i$ between $E_i$ and $Pic^0(E_i)$, which sends $P\to [P]-[\infty]$, and Remark 4 above, we have
$$\phi_2^{-1}\circ \alpha_*\circ \phi_1(P) = \phi_2^{-1}\circ \alpha_* ([P]-[\infty])=\phi_2^{-1}([\alpha(P)]-[\infty])=\alpha(P)$$
Because $\phi_i$ and $\alpha_*$ are group homomorphism, we have $\alpha$ is also a group homomorphism. (Q.E.D)
In section 3, we give the proof of the Velu's theorem for complex elliptic curves. It is the original version of Velu's theorem.
Proposition 4.2. Let $E_1$ be an elliptic curve defined over $k$, and $G\subset E_1$ a finite subgroup. Then there exists the curve $E_2$ and an isogeny $\alpha: E_1\to E_2$ such that $\ker(\alpha)=G$ and $E_2, \alpha$ can be computed explicitly via $G$ and the equation of $E_1$.
The existence of dual isogenies (for the separable case) can be deduced by the proposition above and the following suprising
Lemma 4.3. Let $E_1, E_2, E_3$ be three elliptic curves, with $\alpha_2: E_1\to E_2$ and $\alpha_3: E_1\to E_3$ are separable isogenies, and $\ker \alpha_2=\ker \alpha_3$. Then $E_2\cong E_3$, via the isomorphism $\beta$ such that $\alpha_3=\beta\circ\alpha_2$.
We now deduce the existence of dual isogeny for separable case (note that this also holds for the non-separable case).
Proposition 4.4. Let $\alpha: E_1\to E_2$ be a separable isogeny, then there exists an isogeny $\hat{\alpha}:E_2\to E_1$ such that $\hat{\alpha}\circ \alpha= [\deg \alpha]$.
Proof. Let $n:=\deg \alpha$, we will prove this proposition in the case $char(k)$ does not divide $n$. Since $\alpha$ is separable, we have $\deg\alpha=\#\ker\alpha$. It follows $n=\#\ker\alpha$, and hence, $\ker\alpha\subset E_1[n]$, where $E_1[n]$ is the $n$-torsion subgroup of $E_1$. It can be seen then $\alpha(E_1[n])\cong E_1[n]/\ker\alpha$, i.e. $\#\alpha(E_1[n])=n$, since $E_1[n]\cong \mathbb{Z}_n\oplus \mathbb{Z}_n$ in the case $char(k)\nmid n$.
Due to Velu's formula, there exists a curve $E_3$ with an isogeny $\alpha_1$ from $E_2$ to $E_3$ such that $\ker\alpha_1 = \alpha(E_1[n])$. Hence, the composition map $\alpha_1\circ \alpha$ form $E_1$ to $E_3$ has the kernel $E_1[n]$. This is also the kernel of the map $[n]: E_1\to E_1$ sending $P$ to $nP$. Due to Lemma 4.3, we have an isomorphism $\beta$ between $E_3$ and $E_1$, such that $\beta\circ \alpha_1\circ \alpha=[n]$. Now, we just let $\hat{\alpha}:=\beta\circ \alpha_1$ (Q.E.D)
What we can see from this is that we can define the group law on $E$ as $P + Q = -R$. It is identical with our familiar definition for the group law on an elliptic curve, and it is more accurate, because in the case of tangent line of order 3, we are tricked by our geometric intuition. And we do not have to prove the associativity law, because it is natural induced from the group law on $Pic^0(E)$. Furthermore, one obtains the bijection (and hence, a group isomorphism) between $E$ and $Pic^0(E)$ given by $P\mapsto [P]-[\infty]$.
4. A quick look on algebraic theory of isogenies. Let $E_1,E_2$ are two elliptic curves over an algebraically closed field $k$. An isogeny is both a non-constant morphism and a group homomorphism from $E_1$ to $E_2$, i.e. if we look at affine parts of $E_1$ and $E_2$, an isogeny a rational map between $E_1$ and $E_2$ that sends $\infty$ to $\infty$. Let $\alpha: E_1\to E_2$ be an isogeny, it will induce the injective field homomorphism $\alpha^*: k(E_2)\to k(E_1)$. The field extension $k(E_1)$ of $\alpha^*(k(E_2))$ is finite, and the degree of $\alpha$ is defined as the degree of the field extension. If the field extension is separable, $\alpha$ is called separable isogeny. Otherwise, $\alpha$ is called inseparable.
It follows from Chapter II of Washington's book that:
1. An isogeny is always surjective (We can use a little bit of algebraic geometry to deduce this fact: $E_1$ is projective variety, and hence, is complete, and $\alpha(E_1)$ is closed and irreducible in $E_2$ (since $E_1$ is irreducible), i.e. $\alpha(E_1)=E_2$, since $E_2$ is also irreducible and of dimension 1, and $\alpha$ is non-constant).
2. The kernel of an isogeny $\alpha$ is always finite, and $\# \ker \alpha =\deg \alpha$ if $\alpha$ is separable, i.e. $\deg \alpha$ is actually the number of points in a fiber $\alpha^{-1}(\infty)$. It can be proved that in the case $\alpha$ is an separable isogeny, the fiber of $\alpha$ at every point is equal, and it is $\deg \alpha$.
3. If $\alpha$ is non-separable, then $\deg \alpha >\#ker \alpha$. An example is the Frobenius endomorphism, its degree is $q$ (in the case $E$ are defined over $\mathbb{F}_q$), and it is bijection from $E(\overline{\mathbb{F}_q})$ to $E(\overline{\mathbb{F}_q})$, hence, the its kernel is just $\infty$.
So from 2 and 3, we can see that the kernel of an isogeny is always finite.
4. If $\alpha: E_1\to E_2$ is a morphism, then it induces the push-forward map $\alpha_*$ from $Div(E_1)\to Div(E_2)$ defined by $\alpha_*(\sum_{P\in E} n_P [P])=\sum_{P\in E} n_P [\alpha(P)]$. This can be seen that $\alpha_*$ is a group homomorphism. If furthermore, $\alpha(\infty)=\infty$, then it can be proved $\alpha_*$ maps principal divisors to principal divisors, i.e. the induced map $\alpha_*: Pic^0(E)\to Pic^0(E)$ is well-defined.
Using the isomorphism between $E$ and $Pic^0(E)$, we can prove a beautiful
Proposition 4.1. Let $E_1,E_2$ as above, and $\alpha:E_1\to E_2$ is a non-constant morphism, and $\alpha(\infty)=\infty$, then $\alpha$ is an isogeny.
Proof. It is sufficient to prove that $\alpha$ is a group homomorphism. From the isomorphism $\phi_i$ between $E_i$ and $Pic^0(E_i)$, which sends $P\to [P]-[\infty]$, and Remark 4 above, we have
$$\phi_2^{-1}\circ \alpha_*\circ \phi_1(P) = \phi_2^{-1}\circ \alpha_* ([P]-[\infty])=\phi_2^{-1}([\alpha(P)]-[\infty])=\alpha(P)$$
Because $\phi_i$ and $\alpha_*$ are group homomorphism, we have $\alpha$ is also a group homomorphism. (Q.E.D)
In section 3, we give the proof of the Velu's theorem for complex elliptic curves. It is the original version of Velu's theorem.
Proposition 4.2. Let $E_1$ be an elliptic curve defined over $k$, and $G\subset E_1$ a finite subgroup. Then there exists the curve $E_2$ and an isogeny $\alpha: E_1\to E_2$ such that $\ker(\alpha)=G$ and $E_2, \alpha$ can be computed explicitly via $G$ and the equation of $E_1$.
The existence of dual isogenies (for the separable case) can be deduced by the proposition above and the following suprising
Lemma 4.3. Let $E_1, E_2, E_3$ be three elliptic curves, with $\alpha_2: E_1\to E_2$ and $\alpha_3: E_1\to E_3$ are separable isogenies, and $\ker \alpha_2=\ker \alpha_3$. Then $E_2\cong E_3$, via the isomorphism $\beta$ such that $\alpha_3=\beta\circ\alpha_2$.
We now deduce the existence of dual isogeny for separable case (note that this also holds for the non-separable case).
Proposition 4.4. Let $\alpha: E_1\to E_2$ be a separable isogeny, then there exists an isogeny $\hat{\alpha}:E_2\to E_1$ such that $\hat{\alpha}\circ \alpha= [\deg \alpha]$.
Proof. Let $n:=\deg \alpha$, we will prove this proposition in the case $char(k)$ does not divide $n$. Since $\alpha$ is separable, we have $\deg\alpha=\#\ker\alpha$. It follows $n=\#\ker\alpha$, and hence, $\ker\alpha\subset E_1[n]$, where $E_1[n]$ is the $n$-torsion subgroup of $E_1$. It can be seen then $\alpha(E_1[n])\cong E_1[n]/\ker\alpha$, i.e. $\#\alpha(E_1[n])=n$, since $E_1[n]\cong \mathbb{Z}_n\oplus \mathbb{Z}_n$ in the case $char(k)\nmid n$.
Due to Velu's formula, there exists a curve $E_3$ with an isogeny $\alpha_1$ from $E_2$ to $E_3$ such that $\ker\alpha_1 = \alpha(E_1[n])$. Hence, the composition map $\alpha_1\circ \alpha$ form $E_1$ to $E_3$ has the kernel $E_1[n]$. This is also the kernel of the map $[n]: E_1\to E_1$ sending $P$ to $nP$. Due to Lemma 4.3, we have an isomorphism $\beta$ between $E_3$ and $E_1$, such that $\beta\circ \alpha_1\circ \alpha=[n]$. Now, we just let $\hat{\alpha}:=\beta\circ \alpha_1$ (Q.E.D)
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