# Simple abelian varieties over non algebraically closed fields.

I was wondering what people would normally mean by a simple abelian variety $A$ where $A$ is defined over a field $k$ that is not algebraically closed.

The definition I found in for example on the bottom of page 29 in Langs book Abelian Varieties states:

An abelian variety $A$ is simple if $0$ and $A$ are the only abelian subvarieties of A.

But this definition is not entirely unambiguous for me. And other sources that I read only define being simple over algebraically closed fields.

My main question is: Would $A$ still be called simple if $A$ contains a nontrivial subvariety $B$ that is defined over $\bar k$ but such that $0$ and $A$ are the only subvarieties that are defined over $k$?

The reason I am asking is because I want to apply the results of http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.172.6978 to the weil restriction of an elliptic curve $A=Res_{K/\mathbb Q}E$. One of the hypothesis of that article is that $A$ must be simple. But depending on the definition of simple it might be so that $Res_{K/\mathbb Q}E$ is never simple as variety over $\mathbb Q$

• In many contexts, one usually one distinguishes between being "absoutely simple" (i.e., simple under extension to an algebraic closure) and "simple" tout court. Abelian varieties might be an exception, though! – Mariano Suárez-Álvarez May 27 '13 at 4:49
• The same arises for finite group representations over a field that isn't algebraically closed ($\mathbf{Q}$, $\mathbf{R}$, $\mathbf{F}_p$...): "irreducible" versus "absolutely irreducible". And for connected semisimple algebraic groups over fields. And anywhere else rationality issues are relevant. So Lang is right (and the answer to your main question is "yes"): the word "simple" is a condition within the category under consideration (isogeny category of abelian varieties over $k$, etc.). But it's nice to add clarity by saying "$k$-simple", as Pete suggests. – user29283 May 27 '13 at 6:07
• Indeed $A=\operatorname{Res}_{K/\mathbf{Q}} E$ is never absolutely simple, since $A_{\overline{K}}$ is isomorophic over $\overline{K}$ to the product of the conjugates of $E$. – François Brunault May 27 '13 at 7:27
• It depends whether the author is using terminology based on Weil's Foundations or the terminology of modern (i.e., post 1960) algebraic geometry. See my comment on Pete Clark's answer. In each school, the terminology is unambiguous. The problem is determining whether the author is using the terminology of pre-1960 algebraic geometry or post-1960 algebraic geometry. – anon May 27 '13 at 12:22

You're right: "simple abelian variety $A/K$" is ambiguous when $K$ is not algebraically closed. One should say geometrically (or absolutely) simple or K-simple to emphasize which is meant.
If you do encounter "simple abelian variety $A/K$" in the literature, then in my experience it usually means "geometrically simple". Somehow the typical perspective is that the geometric properties of the abelian variety are considered and described first, and the arithmetic properties only later. A big example of this is that $\operatorname{End} A$ almost always means $\operatorname{End} A_{/\overline{K}}$, and then you can endow it with the structure of an $\operatorname{Aut}(\overline{K}/K)$-module if you like. In my mind the notable exception is Honda-Tate theory, where the isogenies, the Weil polynomials, the endomorphism algebras and so forth really do all take place over a fixed finite field $\mathbb{F}_q$, and extending the ground field can change everything.
Having said all that I did glance through Waldschmidt's paper, and my impression is that he means $\mathbb{Q}$-simple abelian variety, so yes, including certain Weil restrictions. I think that if you read the paper more carefully, this will become even more apparent: the simplicity assumption must come up in the proof, after all! Further, he mentions several times that he is restricting to this special case for simplicity (!!), and makes references to a more general result holding for any commutative group $G_{/\mathbb{Q}}$. So at some point you should try to track that down: maybe any abelian variety will do...