Tensor products of Weyl modules in positive characteristic Let $G$ be a simple algebraic group over a field $k$, and let $U$ be the unipotent radical of a Borel subgroup $B$. Because $B$ normalises $U$, the group $H = B/U$ acts on the coordinate ring $\mathcal{O} = k[X]$ of the basic affine space $X = G/U$ via $(h.f)(x) = f(xh)$. We get a decomposition of $\mathcal{O}$ into a direct sum 
$\mathcal{O} = \oplus_{\lambda \in \Lambda^+} \mathcal{O}^\lambda$
where the Weyl module $\mathcal{O}^\lambda$ is the set of all $f \in \mathcal{O}$ such that $h.f = \lambda(h)f$ for all $h \in H$. Because the action of $H$ commutes with the action of $G$ on $k[X]$ given by $g.f(x) = f(g^{-1}x)$, each $\mathcal{O}^\lambda$ is a $G$-submodule of $\mathcal{O}$. We can also identify $\mathcal{O}^\lambda$ with the space of global sections $H^0(G/B, \lambda)$.
Next, multiplication in $\mathcal{O}$ induces a $G$-module map $\mathcal{O}^\lambda \otimes \mathcal{O}^\mu \to \mathcal{O}^{\lambda + \mu}$ for any $\lambda, \mu \in \Lambda^+$. Since $\mathcal{O}$ has no zero-divisors, this map is non-zero. 
Now if the base field $k$ has characteristic zero, it is well-known that the $G$-modules $\mathcal{O}^\lambda$ for $\lambda \in \Lambda^+$ are irreducible, so the multiplication map above must be surjective. Does this remain true when the characteristic of $k$ is positive, when the Weyl modules $\mathcal{O}^\lambda$ are no longer irreducible in general?
 A: The question has an affirmative answer and a fairly long history as well, but the
proof uses some nontrivial ideas.   The notation used here is nonstandard relative to that found in Jantzen's book Representations of Algebraic Groups (second edition, AMS, 2003).   Also, a "Weyl module" (in the usual sense)  of a given highest weight is the dual of the module of global sections for a related line bundle on the flag variety, using Kempf's vanishing theorem (1976).   The term "Weyl module" was coined by Carter and Lusztig in their paper on special linear groups, partly because the formal character is given by Weyl's formula.
A Weyl module has a unique simple quotient, while the corresponding module has this module as its unique simple submodule.   
There was a series of papers by Lakshmibai-Musili-Seshadri on the geometry of flag varieties in prime characteristic, in which they stated along the way that the tensor product of
these dual Weyl modules maps onto the one specified by the sum of highest weights.  (Their proof may not be rigorous.  In any case, Kempf's theorem comes into play here.)  A focused reference is the paper in J. Algebra 27 (1982) by Jian-pan Wang, "Sheaf cohomology on $G/B$ and tensor product of Weyl modules".   That paper followed up a suggestion of mine that such a tensor product should have a filtration with appropriate Weyl modules as subquotients.   The paper by Olivier  Matthieu in Duke Math. J. 59 (1989) used Frobenius splitting techniques to prove this in full generality after the partial results by Wang and then by Steve Donkin in Springer Lecture Notes 1140 (1985).  Eventually all of this gets folded into the general theory of "tilting modules" for reductive algebraic groups (Chapter G in Jantzen).
[ADDED] As Ekedahl just pointed out, a treatment is given in the more recent and more extensive book by Brion and Kumar along with history.   
A: Yes, it is true in general.
I found it as Thm 3.1.2 of Brion, Michel(F-GREN-IF); Kumar, Shrawan(1-NC)
Frobenius splitting methods in geometry and representation theory.
Progress in Mathematics, 231. Birkhäuser Boston, Inc., Boston, MA, 2005. x+250 pp. ISBN: 0-8176-4191-2. The result itself is earlier (see historical remarks at the end of Chapter 3).
A: Yes.  
A map to a Weyl module is surjective if and only if the composition with the projection to its irreducible quotient is surjective (since any submodule of the Weyl module has trivial image there), so you're back to working with irreducibles, where the result is obvious.
