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?