I think being just a little bit less explicit is better here: A morphism $\mathrm{Spec}S\to\mathrm{Spec}R$ has image in the complement of a closed subscheme $V(I)$ precisely when $SI=S$. In the case at hand $I=(X_1,\dots,X_{n+1})$ so that the map lies in the complement precisely when $(a_1,\dots,a_{n+1})=\mathbb Z$.

Unless I am mistaken these types of Weyl modules have the irreducible as unique minimal submodule rather than quotient. Consider the case of $S^pV$ which has $V^{(p)}$ as submodule not quotient module. For this reason they seem to be known as dual Weyl modules.

@Tom: I agree that $1$-dimensionality seems to make some difference. I still don't understand your argument however and have added a discussion of my problems above (as it was too long to put in a comment).

@Tom: You may be right but I don't see it myself. Don't you just conjugate the group action on the interval by a diffeomorphism? Why wouldn't then the same argument show that the action of $\mathrm{SL}_2(\mathbb R)$ on the open unit disc could be conjugated to extend to the sphere being the identity map outside of it? We know that this is not true.

(cont'd) The end game is then the same; the derived categories of $\mathbb Z$-modules and of $\mathbb Z$-modules are (I hope) equivalent.

I think Chris' view point is sufficiently different from the one I presented to merit mention. I however also agree that it is conceptually very close: In my case a map (of simplicial sets says) $X \to K(A,n)$ which by adjunction is the same as $\mathbb Z[X] \to K(A,n)$ and we have a map of abelian simplicial groups, i.e., of chain complexes. In Chris case we have a (stable) map $X \to K(A,n)$ which by adjunction is the same as $X\wedge H\mathbb Z \to K(A,n)$, a map of $H\mathbb Z$-module spectra. (cont'd)

@Tom: It isn't clear to me that your proposed action of $\mathrm{SL}_2(\mathbb R)$ on $\mathbb R^3$ will even be continuous, let alone smooth. For problems with continuity consider the action of $\mathrm{SL}_2(\mathbb R)$ on the open unit disc, its continuous extension to the boundary acts non-trivially on the boundary and hence can not bne continuously extended by the identity on the complement. In fact, $\mathrm{SL}_2(\mathbb R)$ has $S^1$ as a subgroup so we have a bound on the number of factors of $\mathrm{SL}_2(\mathbb R)$ that can act on $S^n$ (by the argument I gave).

@Victor: Great! I was hoping for a linear bound...