Is a quotient of a reductive group reductive?
Edit [Pete L. Clark]: As Minhyong Kim points out below, a more precise statement of the question is:
Is the quotient of a reductive linear group by a Zariskiclosed normal subgroup reductive?
Is a quotient of a reductive group reductive? Edit [Pete L. Clark]: As Minhyong Kim points out below, a more precise statement of the question is: Is the quotient of a reductive linear group by a Zariskiclosed normal subgroup reductive? 


For a linearly reductive (i.e. all representations are completely redudible) group G, any quotient G/H is also linearly reductive. Given a representation V of G/H, you get an induced representation of G (with the same underlying vector space). Then any decomposition of V as a representation of G is also a decomposition as a representation of G/H. Since Ginvariant subspaces of V are the same thing as G/Hinvariant subspace of V, the notions of irreducible subrepresentations agree. So you get a complete decomposition of V as a representation of G/H. In characteristic zero, reductive is equivalent to linearly reductive, so you have your answer. In general, a group is reductive if its unipotent radical is trivial. I feel like you should be able to argue that an element of the unipotent radical of G/H lifts to give you an element of the unipotent radical of G, but I'm not sure. 


Yes, I believe the quotient of a reductive group scheme is reductive. Perhaps it's possible to show directly as Anton was suggesting. I have an argument that shows that if G is geometrically reductive then G/H is geometrically reductive. The statement that G reductive implies G/H reductive follows from Haboush's theorem (ie. G is reductive if and only if G is geometrically reductive). 


I'm kind of late to the party, but anyway: being reductive means having no nilradical. Now nilradical cannot become larger as you take quotient, ergo your question has a positive answer. 


Presumably the argument ilya was driving at is this: If one had a surjective map of group schemes $G \to H$, then consider the preimage of the nilradical of $H$. This is an algebraic group with a surjective map to a unipotent group. Since there are no group homomorphisms from reductive groups to unipotent ones, the nilradical of this preimage (which is contained in the nilradical of $G$) must surject onto the nilradical of $H$. So if the former is trivial, so is the latter. 


Allow me to insert a trivial word of caution: If analytic quotients are allowed, then the assertion is false even in characteristic zero, as the expression $$\mathbb{G}_m/q^{\mathbb{Z}}$$ for an elliptic curve will show. 


It is important in answering this question that one can extend scalars to a perfect (e.g., algebraically closed) ground field, as was implicit in many of the other answers even if not said explicitly. Indeed, if $k$ is an imperfect field then it always happens that there exist many examples of pairs $(G,H)$ with $G$ a smooth connected affine $k$group containing no nontrivial smooth connected unipotent normal $k$subgroup and $H$ a smooth connected normal $k$subgroup in $G$ such that $G/H$ contains a nontrivial smooth connected unipotent normal $k$subgroup. This can even happen when $G$ and $H$ are perfect (i.e., equal to their own derived groups), which is really disorienting if one is accustomed to working with reductive groups. For a simple commutative example, let $k'/k$ be a purely inseparable extension of degree $p = {\rm{char}}(k)$ and let $G$ be the Weil restriction ${\rm{Res}} _{k'/k}(\mathbf{G} _m)$, which is just "${k'}^{\times}$ viewed as a $k$group". This contains a natural copy of $\mathbf{G}_m$, and since $k'/k$ is purely inseparable this $k$subgroup $H$ is the unique maximal $k$torus and the quotient $G/H$ is unipotent of dimension $p1$ (over $\overline{k}$ it is a power of $\mathbf{G} _a$ via truncation of $\log(1+x)$ in degrees $< p$, as one sees using the structure of the $\overline{k}$algebra $\overline{k} \otimes_k k'$). The main point then is that $G$ itself contains no nontrivial smooth connected unipotent $k$subgroups, which is true because we are in characteristic $p > 0$ and $G$ is commutative with $G(k_s)[p] = {k'_s}^{\times}[p] = 1$! Note: the unipotent quotient $G/H$ is an example of a smooth connected unipotent $k$group (even commutative and $p$torsion) which contains no $\mathbf{G}_a$ as a $k$subgroup (proof: commutative extensions of $\mathbf{G}_a$ by $\mathbf{G}_m$ split over any field, due to the structure of ${\rm{Pic}}(\mathbf{G}_a)$ and a small calculation); that is, $G/H$ is a "twisted form" of a (nonzero) vector group, which never happens over perfect fields. Making examples with perfect $G$ and $H$ is less straightforward; see Example 1.6.4 in the book "Pseudoreductive groups". As for the suggestion to use Haboush's theorem (whose proof I have never read), I wonder if that is circular; it is hard to imagine getting very far into the theory of reductive groups (certainly to the point of proving Haboush's theorem) without needing to already know that reductivity is preserved under quotients (a fact that is far more elementary than Haboush's theorem, so at the very least it seems like killing a fly with a sledgehammer even if it is not circular). Finally, since nobody else has mentioned it, look in any textbook on linear algebraic groups (Borel, Springer, etc.) for a proof of the affirmative answer to the original question. For example, 14.11 in Borel's book. Equally important in the theory is that for arbitrary smooth connected affine groups, formation of images also commutes with formation of maximal tori and especially (schemetheoretic) torus centralizers; see corollaries to 11.14 in Borel's book. 

