1. Dual Isogenies. Let $f: A\to B$ be a homomorphism between abelian variety, then one has $(f,id): A\times \hat{B}\to B\times \hat{B}$ is a morphism. Let $P_B$ be the Poincare's bundle on $B\times \hat{B}$, and $M:=(f,id)^*P_B$ the pullback of $P_B$. One has
$$M|_{A\times \{ \hat{b}\}}=(f,id)^*P_B|_{A\times \{ \hat{b}\}}\cong f^*P_{ \hat{b}}\in Pic^0(A)$$
due to the following
Lemma 1.1. Let $f: A\to B$ be a homomorphism between abelian varieties, then $f^*: Pic^0(B)\to Pic^0(A)$ is well-defined, and it is a group homomorphism.
Proof. Let $L\in Pic^0(B)$, and for all $a\in A$, we have $\phi_{f^*L}(a)=\tau_a^*f^*L\otimes f^*L^{-1}$. But then, $\tau_a^*f^*=(f\circ \tau_a)^*=(\tau_{f(a)}\circ f)^*=f^*\tau_{f(a)}^*$. This yields $\phi_{f^*L}(a)=f^*(\tau_{f(a)}^*L\otimes L^{-1})$, and it is trivial, since $L\in Pic^0(A)$. Hence, $f^*L\in Pic^0(A)$. (Q.E.D)
From the definition of $M$, one can also see $M|_{0\times \hat{B}}$ is trivial. And due to the universal property of Poincare's bundle, there exists a unique map $ \hat{f}: \hat{B}\to \hat{A}$ such that $M=(id, \hat{f})^*P_A$, where $P_A$ is the Poincare's bundle on $A\times \hat{A}$. In particular, $\hat{f}$ is the unique map from $ \hat{B}$ to $ \hat{A}$ such that $(f,id)^*P_B\cong (id, \hat{f})^*P_A$. The map $ \hat{f}$ is called the dual of $f$. At the level of line bundles on $Pic^0$, one can see $f^*$ is actually the map sends $L\in Pic^0(B)$ to $f^*L\in Pic^0(A)$.
Now, let $L,M$ be ample line bundles on $A,B$, respectively, and $\lambda_L:A\to \hat{A}$, and $\lambda_M: B\to \hat{B}$ the corresponding maps. Then one can hope $\lambda_L= \hat{f}\circ \lambda_M\circ f$, but this does not hold in general. In fact, if we change $L$ by $f^*L$ then the last identity holds.
Lemma 1.2. Let $M$ be any line bundle on $B$, then $\lambda_{f^*M}= \hat{f}\circ \lambda_M\circ f$.
Proof. The proof is very similar to parts of the proof of Lemma 1.1. (Q.E.D)
Now, in case $f$ is an isogeny, i.e. $f$ is surjective and $\dim A=\dim B$, $\hat{f}$ is also an isogeny from $\hat{B}$ to $\hat{A}$, and even more interestingly, we have $\deg f=\deg \hat{f}$. And if $M$ is an ample line bundle on $B$, then $f^*M$ is an ample line bundle on $A$ (We will prove this fact later). This yields $K(\lambda_{f^*M})$ is finite, and it will be killed by some $n\in \mathbb{Z}$, i.e. $K(\lambda_{f^*M})\subseteq A[n]$. From this, one can define the map $\pi: \hat{A}\to A$ sending $\phi_{f^*M}(a)$ to $na$. It is an isogeny, and $\pi\circ \lambda_{f^*M}=[n_A]$. By the previous lemma, we have $[n_A]=\pi\circ \hat{f}\circ \lambda_M\circ f$, i.e. there exists an isogeny $g:B\to A$ such that $g\circ f=[n_A]$. For another version of this observation, we want $n=\deg f$.
Theorem 1.3. Let $f: A\to B$ be an isogeny between abelian varieties, with $n=\deg f$. Then there exists $g: B\to A$ such that $g\circ f=[n_A], f\circ g=[n_B]$.
Proof. First, it can be seen that any element in $\ker f$ is killed by $n$. This can be deduce easily if $f$ is an separable isogeny. Otherwise, one can factor $f$ as $h\circ g$ for some $g: A\to C$ is purely inseparable, and $h: C\to B$ is separable, and from this, $\ker f$ is one-to-one correspondent with $\ker h$. And hence, any element in $\ker f$ is killed by $\deg h$. This also implies any element in $\ker f$ is killed by $\deg f$, since $\deg f=\deg g\deg h$.
One can construct from this the map $g: B\to A$ that sends $f(a)$ to $na$. The composition of $g\circ f$ is exactly $n_A$, and $n_B\circ f=f\circ n_A=f\circ g\circ f$. Now, $f$ is an epimorphism of schemes implies that $n_B=f\circ g$. (Q.E.D)
2. Polarization. Let $A$ be an abelian variety, and $f: A\to \hat{A}$ is a homomorphism. We call $f$ a polarization if $f$ has the form $\phi_L$ for some ample line bundle $L$ on $A$, and $f$ is called principal polarization if $f$ is a polarization, and $\deg f=1$, i.e. $f$ is an isomorphism.
When $A$ is an elliptic curve, one has $P\mapsto [P]-[\infty]$ is the principal polarization. But this does not hold in general that any abelian variety has a principal polarization (See ???). But it is true that any abelian variety is isogeneous with a principal polarized abelian variety. This can be proved with the help of Weil's pairing. The rest of this note is devoted for main ideas of the proof.
Let $[m_A]:A\to A$ the multiplication by $m$ map, where $(m,char(k))=1$. It then induces the dual map $[m_{\hat{A}}]:\hat{A}\to \hat{A}$, and actually, $\hat{A}[m]\cong Hom(A[m], k^\times)=Hom(A[m], \mu_m)$, where $\mu_m$ is the group of $m$-th root of unity in $k^\times$. This yields the pairing
$$e_m: A[m]\times \hat{A}[m]\to \mu_m$$
Similar to the case of elliptic curves, the general Weil's pairing also has a nice interpretation. Let $L\in Pic^0(A)[m]$ (we now identify $\hat{A}$ and $Pic^0(A)$). Then because $L$ is anti-symmetric, we have $[m]^*L\cong L^{\otimes m}\cong \mathscr{O}_A$, i.e. there exists two rational functions $f_L,g_L$ on $A$ such that $div(f_L)=L^{\otimes m}, div(g_L)=[m]^*L$. From this,
$$div(f_L\circ [m])=[m]^*L^{\otimes m}=([m]^*L)^{\otimes m}=div(g_L^m)$$
This implies there exists a constant $c\in k^\times$ such that $g_L^m(x)=c f_L(mx)$. And then, for all $a\in A[m]$, we have
$$g_L(x+a)^m=c f_L(mx+ma)=c f_L(mx)=g_L(x)^m$$
This implies $\frac{g_L}{g_L\circ\tau_a}$ is a $m$-th root of unity. We now define $e_m(a,L)=\frac{g_L}{g_L\circ\tau_a}$. By using this, we first prove
Lemma 2.1. Let $a\in A[mn], L\in Pic^0(mn)$, then $e_{mn}(a,L)^n=e_m(na,L^{\otimes n})$
Proof. Assume that $(mn)^*L=div(g)$, and $m^*(L^{\otimes n})=div(g')$, then we can see
$$div(g'\circ n)=n^*m^*(L^{\otimes n})=((mn)^*L)^{\otimes n}=div(g^n)$$
This yields there exists $c\in k^\times$ such that $g'(nx)=cg^n(x)$. And from this,
$$e_{mn}(a,L)^n=\frac{g^n(x)}{g^n(x+a)}=\frac{g'(nx)}{g'(nx+na)}=e_m(na,nL)$$
(Q.E.D)
By very similar proofs in Silverman, the general Weil pairing we have just defined is also bilinear, non-degenerate and compatible with the Galois action. Note that because an elliptic curve $E$ has principal polarization, that mean we can identify $Pic^0(E)$ and $E$, and the Weil pairing now can be defined as $e_m: E[m]\times E[m]\to \mu_m$, and it is skew-symmetric, i.e. $e_m(a,b)=e_m(a,b)^{-1}$.
Now, let $\lambda: A\to \hat{A}$ be any homomorphism, then we can define the pairing
$$e_m^\lambda:A[m]\times A[m]\to \mu_m$$
by sending $(a,a')$ to $e_m(a,\lambda(a'))$. Recall that for any $L\in Pic(A)$, one can construct $\lambda_L: A\to Pic^0(A)$ by sending $a$ to $\tau_a^*L\otimes L^{-1}$, and if $\lambda_L\equiv 0$ iff $L\in Pic^0(A)$. This yields if $L,L'\in Pic(A)$, and $L\equiv L'\mod Pic^0(A)$, then $\lambda_L,\lambda_{L'}$ are just the same. This observation gives right to define the Neron-Severi's group of $A$, $NS(A):=Pic(A)/Pic^0(A)$. Now, an important fact mentioned in Milne's note is that if $\lambda$ comes from $\lambda_L$, for some $L\in NS(A)$, then $e_m^\lambda$ is skew-symmetric.
We now reduce to the case $\lambda$ is a polarization, i.e. $\lambda$ comes from $\lambda_L$ for some ample line bundle $L\in Pic(A)$. We can make change a little bit on the definition of the Weil's pairing, by defining it on the kernel of $\lambda$. Assume that $\ker\lambda$ is killed by some integer $m$, then for all $a,a'\in \ker\lambda$, and $b\in A$ such that $mb=a'$, then we define
$$e^\lambda(a,a'):=e_m(a,\lambda(b))$$
We will prove that this definition makes sense, by proving that it is independent on $m,b$. First, $m\lambda(b)=\lambda(mb)=\lambda(a)=0$, so $\lambda(b)\in Pic^0(A)[m]$, so $e_m(a,\lambda(b))$ is well-defined. Second, if we replace $m$ by $mn$, and replace $b$ by some $b'$ such that $mnb'=a$. Let $a_1\in A$ such that $na_1=a$, then it follows by Lemma 2.1 that
$$e_{mn}(a,\lambda(b'))=e_{mn}(na_1,\lambda(b'))=e_{mn}(a_1,\lambda(b'))^n=e_{m}(a,\lambda(nb'))$$
And then by the skew-symmetric property, we have,
$$e_m(a,\lambda(nb'))/e_m(a,\lambda(b))=e_m(a,\lambda(nb'-b))=e_m^\lambda(a,nb'-b)$$
$$=e_m^\lambda(nb'-b,a)^{-1}=e_m(nb'-b,\lambda(a))=1$$
An important application of $e^\lambda$ is that
Theorem 2.2. Let $\lambda: A\to \hat{A}$ be a
polarization, and $f: A\to B$ an isogeny, then $\lambda=\lambda_{f^*L}$,
for some ample line bundle $L$ on $B$ if and only if $\ker f\subseteq \ker
\lambda$, and $e^\lambda$ is trivial on $\ker f\times \ker f$.
Now, with specific $a,a'\in \ker\lambda$, we can even choose some $m$ such that $m$ kills both $a,a'$, then the definition above still works, and we also obtain $e^\lambda(a,a')$ in this case. In the case $a\equiv a_m,a'\equiv a'_m$ is killed by $l^m$, for some prime $l\ne char(k)$, we have
$$e^\lambda(a_m,a'_m)=e_{l^m}(a_m,l^m\lambda(a'_{2m}))=e_{l^m}(l^ma_{2m}, l^m\lambda(a'_{2m}))=e_{l^{2m}}^\lambda(a_{2m},a'_{2m})$$
And this then yields $e^\lambda$ is skew-symmetric on $A[l^m]$, for all prime $l\ne char(k)$. Using this, we obtain our main result of this note
Theorem 2.3. Any abelian variety is isogeneous with a polarized abelian variety.
Proof. Let $\lambda: A\to \hat{A}$ be a polarization with $\deg \lambda$ is co-prime to $char(k)$. If $\lambda$ is principal, then there is nothing to prove. Otherwise, there exists a prime $l|\deg\lambda$, and let $K$ be the subgroup of $\ker \lambda$, whose element is $l$. Then because $e^\lambda$ is skew-symmtric on $N$, it is trivial on $N\times N$. Let $B:=A/N$ be another abelian variety, and $f: A\to B$ is the projection map. Then $N=\ker f\subseteq \ker\lambda$, and $e^\lambda$ is trivial on $N\times N$. This follows from Theorem 2.1 that $\lambda=\lambda_{f^*L}$ for some ample line bundle $L$ on $B$. This follows $\lambda_{f^*L}=\hat{f}\circ \lambda_L\circ f$, i.e. $\deg \lambda=\deg \lambda_L.(\deg f)^2$, and now, this implies $B$ has a polarization of degree $\deg \lambda/l^2<\deg \lambda$. Continue this process, we obtain an abelian $C$, which has a principal polarization. (Q.E.D)
Some questions up to this:
1.For an abelian variety $A$, and a finite subgroup $N$ of $A$, how can we produce an abelian variety $B\cong A/N$.
2. Can we produce an algorithm for Theorem 2.3?
3. What about abelian varieties with complex multiplciations? They are all principally polarized?
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