Let $K$ be a perfect field. In what follows, an algebraic group $G/K$ is by definition a group scheme of finite type over $K$. 


The following seems to be well-known:


**Theorem:** Let $G/K$ be a connected smooth algebraic group. Then there is a connected smooth *affine* normal closed subgroup $N$ of $G$, an abelian variety $A/K$ and a homomorphism $G\to A$ with kernel $N$ such that the sequence 
$$0\to N\to G\to A\to 0$$
is exact for the fppf-topology (say). 


Can someone give me a proper reference or a hint why this is true? (Checking the literature I find on the one hand plenty of references treating affine algebraic groups, and on the other hand references containing the theory of abelian varieties, but I was surprised not to find a reference containing a proof of this Theorem about the "mixed case".)



Decomposition of an algebraic group in an affine and a proper part
This question is answered in the [Wikipedia page for algebraic groups](http://en.wikipedia.org/wiki/Algebraic_group). The article says that this is a difficult result of Chevalley, and it has a link to a [modern write-up by Brian Conrad](http://math.stanford.edu/~conrad/papers/chev.pdf) of Chevalley's result. So that is surely a proper reference. No particular hint leaps out at me for "why" it is true, but I can say something about what the proof is really saying. The subgroup $N$ appears as the common kernel of all algebraic homomorphisms from $G$ to all abelian varieties. So it is a functorial construction and $N$ is actually a characteristic subgroup, not just a normal subgroup. Relatively early on it is shown that there are no non-trivial algebraic homomorphisms from an affine group to an abelian variety, in fact not even any non-trivial algebraic morphisms that don't have to be homomorphisms. 


It is also relatively quick to show that $G/N$ is an abelian variety. The really hard part is to show that the kernel is affine. You might as well let $G = N$ and you might as well let $K$ be algebraically closed. The hard theorem is that if $G$ does not have any non-trivial homomorphisms to an abelian variety, then it is affine. It is important to remember that $G$ is affine if and only if it is linear, i.e., an algebraic subgroup of $\text{GL}(V)$ for some finite-dimensional vector space $V$. This vector space $V$ is the most difficult construction of the paper.


