Let $k$ be a field with $\operatorname{char}(k)=0$. Let $G$ be a connected quasi-split reductive algebraic group definied over $k$, with maximally split maximal torus $T$, Borel $B \supset T$ and the Weyl group $W=N_G(T)/T$ with $B$-simple reflections $S$. For $I \subset S$, we define $W_I \subset W$ to be the subgroup of $W$ generated by the elements of $I$. Then $P:=BW_IB$ is a parabolic subgroup of $G$ containing $B$. Furthermore let $W^I:=\{w \in W \mid l(ws)>l(w) \space \forall s\in I\}$ be the set of minimal coset representatives for $W/W_I$, which has an element of maximal length $w_0^P$. Denote by $X(w_0^P)=Bw_0^PP/P$ the big cell in $G/P$. Then $Z:=G/P-X(w_0)$ is of codimension 1 and the union of all Schubert divisors in $G/P$.

If $k$ is algebraically closed then Billey and Laakshmibai state in "Singular Loci of Schubert Varieties", p.44 (MSN), that $Z=G/P \cap H$, where $H \subset \operatorname{Proj}(H^0(G/P,L(\lambda)))$ is a hyperplane defined by an extremal weight vector of $H^0(G/P,L(\lambda))$. Here $L(\lambda)$ is a very ample line bundle on $G/P$ associated to a dominant weight $\lambda$ s.t $P=P_\lambda$.

Does this also hold for $k$ not algebraically closed? I was wondering if the union of all Schubert divisors in $G/P$ coming from a linear homogeneous polynomial.

not$\Gamma$-stable, then $P=P_I$ is not defined over $k$, and so I would expect that your set $Z=Z_I$ is not defined over $k$, and therefore, it is not the set of zeros of a polynomial defined over $k$, so the answer is No. $\endgroup$ – Mikhail Borovoi 2 days agois$\Gamma$-stable, then maybe Yes. Please edit your question. The last sentence is grammatically incorrect, and I don't understand what you are asking. $\endgroup$ – Mikhail Borovoi 2 days ago