The following is a question I posted about a week ago on Maths stackexchange [there](https://math.stackexchange.com/questions/3694924/about-the-type-of-the-polarization-of-an-abelian-variety), but it didn't bring any discussion nor comment. For this reason I am posting it here also.


Let $X$ be an abelian variety of dimension $g$ over an algebraically closed field $k$ of characteristic different from $2$, and consider $\lambda:X\rightarrow \hat X$ a polarization of degree $d$. Assume that $d$ is prime to the characteristic of $k$. Then it is known that the kernel $\mathrm{Ker}(\lambda)$ is an étale, constant group scheme over $k$. Moreover because $\lambda$ is symmetric, its kernel also has the structure of a symplectic module. We deduce the existence of a unique sequence of integers $d\_1|\ldots |d\_n$ such that $d\_1\geq 2$ and
$$\mathrm{Ker}(\lambda)\simeq \left( \mathbb Z/d\_1\mathbb Z \times \ldots \times \mathbb Z/d\_n\mathbb Z\right)^2$$
as group schemes over $k$. (In particular, $d$ is the square of the product of all the $d\_i$'s). 


On many occasions in the litterature, I see that the integer $n$ is taken to be equal to the dimension $g$ of $X$, up to adding some $1$ at the beginning of the sequence $(d\_1,\ldots ,d\_n)$. We then call $D = (d\_1,\ldots ,d\_g)$ the type of the polarization. I am all fine with that when $n\leq g$, but wouldn't it be possible for $n$ to actually be bigger than $g$ in the first place ? Am I missing something obvious ?


The only way I can think of relating the kernel of $\lambda$ with the dimension of $X$ would be by mean of the Tate module, attached to any prime $l$ different from the characteristic of $k$. Indeed, this module $\mathrm T\_l(X)$ has rank $2g$ over $\mathbb Z\_l$, and it is equipped with the Weil symplectic pairing which involves the polarization $\lambda$. Considering the restricted product of these modules, we obtain a symplectic space over the ring $\mathbb A\_f^p$ of finite adèles away from $p$. In PEL moduli problems, we impose the condition that this symplectic pairing should also have type $D$, ie. it should be represented by the matrix $\left(
\begin{matrix}
0 & \mathrm{Diag}(D) \\
-\mathrm{Diag}(D) & 0
\end{matrix}
\right)$ in some appropriate basis. This also suggests that $n$ shouldn't be bigger than $g$, but I have been failing to write down convincing arguments to show it.




---


Some references where $n$ is taken to be the dimension $g$ of $X$ without any specific explanation: 


* Genestier and Ngo's lecture on Shimura varieties, available [here](https://www.math.uchicago.edu/~ngo/Shimura.pdf). See the definition of the moduli problem in 2.3 page 13. The condition (3) implicitly implies that $n=g$. This moduli problem corresponds to that studied in Mumford's GIT, where no such condition was imposed to my understanding.
* Olsson's workshop notes on abelian varieties, available [here](https://math.berkeley.edu/~molsson/Hangzhou-workshop-notes.pdf). See remark 6.13.
* Hulek and Sankaran's paper on the geometry of Siegel modular varieties, available [here](https://projecteuclid.org/download/pdf_1/euclid.aspm/1546230378). See p.93 (ie. p.5 of the pdf). In the case of abelian varieties over $\mathbb C$ described as projective tori, the definition of a polarization seems to be slightly adapted, and there it is clear that the number of integers in the sequence $(d\_1,\ldots,d\_n)$ is precisely the dimension of $X$.



About the type of a polarization of an abelian variety
Let $\lambda: A\rightarrow A^{\vee}$ be any polarization of degree prime to the characteristic, not necessarily self-dual.
There exists an $\lambda^{\vee} : A^{\vee}\rightarrow A$ such that $\lambda^{\vee}\circ \lambda = [n]$ for some integer $n$ where $n$ is invertible in $k$. 
So $\ker(\lambda) \subset A[n] \simeq \left(\mathbb{Z}/n\mathbb{Z} \right)^{2g}$.
By linear algebra over $\mathbb{Z}$, any subgroup of a finite group generated by $2g$ elements is generated by at most $2g$ elements, and we can choose the generators to be compatible with a given symplectic form. This proves the claim.


