I am mostly writing this to correct my first comment. That comment addressed a completely different question: whether cohomology of a (split) generalized flag variety $G/B$ over a positive characteristic field $k$ equals cohomology of a (split) generalized flag variety of the same type over a characteristic $0$ field. For that question, **which is not the question asked by the OP**, the answer is positive by the Proper Base Change Theorem (you do not need to use affine pavings or "constancy of number").

Below I list some positive results about the question that the OP did ask in the classical types $A_n$, $B_n$,$C_n$, and $D_n$: in type $A_n$ and $C_n$, the cohomology is generated in degree $2$ in all characteristics, and for type $B_n$, $n\geq 3$, and for $D_n$, $n\geq 4$, the only "bad characteristc" is $2$. For the exceptional groups, there is an upper bound on the set of "bad primes" as in the answer by @NicholasKuhn.

**Edit.** The article of Demazure linked in the answer of @VictorPetrov proves much more: the ideal kernel is generated by Weyl invariants except in the case of bad primes. Also, the bad primes for type $G_2$ is $2$, the bad primes for type $E_6$, $E_7$, and $F_4$ are $2$ and $3$, and the bad primes for type $E_8$ are $2$, $3$, and $5$.

**Notation.** For every connected, simply connected topological space $M$, for every
coefficient ring $R$, the natural map $R\to H^0(M;R)$ is an isomorphism and $H^1(M;R)$ vanishes.
Denote by $q_{M;R}$ the natural homomorphism of graded $R$-algebras, $$q_{M;R}:\text{Sym}^*_R H^2(M;R) \to H^*(M;R).$$

**Definition.** The $R$-algebra $H^*(M;R)$ is $2$-**generated** if $q_{M;R}$ is surjective.

Let $G$ be a (split) semisimple complex Lie group. Denote by $T$ a maximal torus in $G$. Denote by $B$ a Borel subgroup of $G$ that contains $T$. The quotient space $B/T$ is isomorphic to affine space of complex dimension $m$ equal to the complex dimension of $G/B$. Thus the following surjection is an affine space bundle of relative dimension $m$, $$ \rho: G/T \to G/B. $$ Therefore, the pullback map $H^*(G/B;R)\to H^*(G/T;R)$ is an isomorphism. The conjugation action of the normalizer subgroup $N_G(T)$ on $G$ preserves left $T$-cosets. Thus, there is an induced action of $N_G(T)$ on $G/T$. By functoriality of cohomology, there is an induced action of $N_G(T)$ on $H^*(G/T;R)$. Of course the identity component $T$ of $N_G(T)$ acts trivially. Thus, there is an induced action on the graded $R$-algebra $H^*(G/B;R)$ by the Weyl group $W=N_G(T)/T$.

**Notation.** Denote by $\text{Sym}^*_R H^2(G/B;R)^W$ the graded $R$-subalgebra of Weyl-invariant elements of $\text{Sym}^*_R H^2(G/B;R)$. Denote by $\text{Sym}^*_R H^2(G/B;R)^W_+$ the graded ideal in this graded $R$-subalgebra generated by homogeneous Weyl-invariant elements of positive degree. Finally, denote by

$I_R$ the ideal in $\text{Sym}^*_R H^2(G/B;R)$ generated by $\text{Sym}^*_R H^2(G/B;R)^W_+$, $$I_R := \langle \left( \text{Sym}^*_R H^2(G/B;R)\right)^W_+ \rangle \subset \text{Sym}^*_R H^2(G/B;R).$$

**Question.** For which rings $R$ and for which type of simply connected, simple complex Lie group $G$ is the homomorphism $q_{G/B;R}$ surjective? Among those, for which rings does $\text{Ker}(q_{G/B;R})$ equal $I_R$?

**Borel's Theorem.** If $R$ contains $\mathbb{Q}$, then $q_{G/B;R}$ is surjective and the kernel ideal $\text{Ker}(q_{G/B;R})$ in the $r$-dimensional polynomial ring $\text{Sym}^*_{\mathbb{Q}} H^2(G/B;R)$ is a complete intersection ideal $I_R$ generated by the $r$ **fundamental invariants** $g_1,\dots,g_r$ of degrees $1+m_1,\dots,1+m_r$ with $1\leq m_1\leq \dots \leq m_r$. The fundamental invariants are primitive homogeneous elements in $\text{Sym}^*_{\mathbb{Z}}H^2(G/B;\mathbb{Z})^W_+$ such that the $R$-polynomial ring of Weyl-invariants equals the polynomial ring, $\left(\text{Sym}^*_R H^2(G/B;R)\right)^W=R[g_1,\dots,g_r].$

**Poincaré Polynomial Corollary.** The Poincaré polynomial $P_{\mathbb{Q}}(G/B,t) := \sum_{\ell=0}^{2n} b_{\ell}(G/B;\mathbb{Q})t^\ell$ equals $\prod_{i=1}^r (1+t^2+\dots + t^{2\ell}+\dots + t^{2m_i})$. In particular, the degrees of the fundamental invariants $m_i$ satisfy $m_1+\dots+m_r=m$ and $P_{\mathbb{Q}}(G/B,1) = (1+m_1)\cdots(1+m_r)$.

**Note.** Special cases of the corollary were known earlier to E. Cartan and to R. Brauer, but finally proved for all semisimple complex Lie groups by Borel and Chevalley. There is a wonderful history of this subject in the following article.

Pierre Cartier

A primer of Hopf algebras

http://preprints.ihes.fr/2006/M/M-06-40.pdf

**The Bruhat Decomposition and the Universal Coefficients Theorem.** The Bruhat decomposition of $G$ into double-$B$-cosets, $$G=\sqcup_{w\in W} BwB,$$ induces a partition of $G/B$ into locally closed subvarieties, $$G/B = \sqcup \mathbb{A}_w, \ \ \mathbb{A}_w:= (BwB)/B,$$ where each $\mathbb{A}_w$ is isomorphic to affine space $\mathbb{A}^{\ell(w)}$ for the **Bruhat length** $\ell(w)$ of $w$. In particular, the maximal length of an element equals $m$, the complex dimension of $G/B$.

**Bruhat Decomposition.** The homology of $G/B$ is a free $\mathbb{Z}$-module with one generator in degree $\ell(w)$ for each element $w\in W$. In particular, all nonzero homology groups have even degree, and $|W|$ equals $(1+m_1)\cdots (1+m_r)$. Moreover, the following cycle class map is an isomorphism of graded $\mathbb{Z}$-modules, $$\text{cl}:\text{CH}_*(G/B) \xrightarrow{\cong} H_*(G/B;\mathbb{Z}).$$

**Proof.** The first claim follows by the cellular decomposition of $G/B$ by the Bruhat cells. Thus $|W|$ equals $P(G/B,1)$, which equals $(1+m_1)\cdots (1+m_r)$ by the previous corollary. The final claim follows by Example 1.9.1, p. 23 of Fulton's *Intersection theory*. **QED**

**Corollary 0.** The integral cohomology of $G/B$ is a free Abelian group. Moreover, for every ring $R$, the following Change of Rings Homomorphism is an isomorphism, $$H^*(G/B;\mathbb{Z})\otimes_{\mathbb{Z}} R \to H^*(G/B;R).$$

**Proof.** This follows from the Universal Coefficients Theorem since the homology is a free Abelian group. **QED**

**Corollary 1.** The cokernel of $q_{G/B;\mathbb{Z}}$ is a graded Abelian group of finite order $N$. For every ring $R$, the homomorphism $q_{G/B;R}$ is surjective if and only if $N$ is invertible in $R$ (equivalently, each prime divisor of $N$ is invertible in $R$). In this case, the graded ideal $\text{Ker}(q_{G/B;R})$ has the same Poincaré polynomial as the ideal $\text{Ker}(q_{G/B;\mathbb{Q}})$ generated by $g_1,\dots,g_r$. If also $(g_1,\dots,g_r)$ form a regular sequence in the polynomial ring $\text{Sym}^*_R H^2(G/B;R)$, then $\text{Ker}(q_{G/B;R})$ equals the complete intersection ideal $\langle g_1,\dots,g_r\rangle$, which equals $I_R$.

**Corollary 2.** The kernel of $q_{G/B;\mathbb{Z}}$ contains $I_{\mathbb{Z}}$ as a subgroup of finite index $N'$. If both $N$ and $N'$ are invertible in $R$, then $q_{G/B;R}$ is surjective and $\text{Ker}(q_{G/B;R})$ equals $I_R$.

**Leray Spectral Sequences.** Consider a fiber bundle of topological spaces, $$\pi:E\to B,$$ with fibers $F_x$ for $x\in B$.
There is an induced Leray spectral sequence.

**Leray Lemma.** Assume that $B$ and a general fiber $F_x$ are connected and simply connected, that $H^*(B;R)$ is a free $R$-module of finite rank $b$, and that $H^*(F_x;R)$ is a free $R$-module of finite rank $f$. Then the free rank $e$ of $H^*(E;R)$ is at most $bf$, and $H^*(E;R)$ is a free $R$-module of rank $bf$ if and only if the Leray spectral sequence degenerates. In this case, if $H^*(E;R)$ is $2$-generated, then also $H^*(F_x;R)$ is $2$-generated. Conversely, if $H^*(B;R)$ and $H^*(F_x;R)$ are both $2$-generated and if $H^*(E;R)$ is a free $R$-module of rank $bf$, then also $H^*(E;R)$ is $2$-generated.

**Proof.** Since $B$ is simply connected, the Leray spectral sequence of $\pi$ is a stage-$2$ spectral sequence converging to the cohomology of $E$, $$E^{p,q}_2 = H^q(B;R)\otimes_R H^p(F_x;R) \Rightarrow H^{p+q}(E;R).$$ Thus $H^*(E;R)$ is a free $R$-module of rank $bf$, i.e., the free rank of $H^*(B;R)\otimes_R H^*(F_x;R)$, if and only if all of the differentials in the spectral sequence are zero.

When the spectral sequence degenerates, in particular all of the differentials out of $H^0(B;R)\otimes_R H^\ell(F_x;R)$ are zero. Thus, the natural pullback homomorphisms
$H^\ell(E;R)\to H^\ell(F_x;R)$ are surjective. Thus, we have a commutative square of homomorphisms of graded $R$-algebras, $$ \begin{array}{ccc} \text{Sym}^*_R H^2(E;R) & \xrightarrow{q_{E;R}} & H^*(E;R) \\ \downarrow & & \downarrow \\ \text{Sym}^*_R H^2(F_x;R) & \xrightarrow{q_{F_x;R}} & H^*(F_x;R) \end{array},$$ where the vertical arrows are surjective. If also $q_{E;R}$ is surjective, it follows that $q_{F_x;R}$ is surjective as well.

Conversely, suppose that the spectral sequence degenerates, and assume that both $H^*(B;R)$ and $H^*(F_x;R)$ are $2$-generated. The spectral sequence respects cup product with elements in $H^*(B;R)$. Thus, degeneration of the spectral sequence implies that $H^*(E;R)$ is a free graded $H^*(B;R)$-module, and any subset of $H^*(E;R)$ that maps to a generating set, resp. basis, of $H^*(F_x;R)$ as an $R$-module is a generating set, resp. basis, for $H^*(E;R)$ as an $H^*(B;R)$-module. By hypothesis, a generating set of $H^*(B;R)$ as an $R$-module is contained in the image of $q_{B;R}$, and thus its image in $H^*(E;R)$ is contained in the image of $q_{E;R}$.

By hypothesis, a generating set of $H^*(F_x;R)$ is contained in the image of $q_{F_x;R}$. Since the spectral sequence above degenerates, in particular, there is a short exact sequence, $$0 \to H^2(B;R)\to H^2(E;R)\to H^2(F_x;R)\to 0.$$ Thus, the generating set lifts to a subset of $H^*(E;R)$ that is in the image of $q_{E;R}$. Altogether, a generating set for $H^*(E;R)$ as an $R$-module is contained in the image of $q_{E;R}$. Therefore $q_{E;R}$ is surjective. **QED**

This suffices to settle the problem for many groups of classical type.

**$A_n$ Type Result.** For every $n$, in $A_n$-type the homomorphism $q_{G/B;R}$ is surjective and the kernel is generated by the fundamental invariants $g_1=s_2, \dots, g_n = s_{n+1}$, the elementary symmetric polynomials modulo $s_1$.

**Proof.** For every projective space $P$, the homomorphism $q_{P;R}$ is surjective. For $A_n$-type, since the flag variety is an iterated projective space bundle, by the Leray Lemma and by induction on the dimension, also $q_{G/B;\mathbb{Z}}$ is surjective. Moreover, since the First Fundamental Theorem of Invariant Theory holds on the integral level, the elementary symmetric polynomials $s_1,\dots,s_{n+1}$ are the generators of $I_{\mathbb{Z}}$. Therefore Borel's Theorem holds for every coefficient ring $R$ in $A_n$-type. **QED**

**$C_n$ Type Result.** For every $n\geq 1$, in $C_n$-type the homomorphism $q_{G/B;R}$ is surjective and the graded kernel ideal $\text{ker}(q_{G/B;R})$ has the same Poincaré polynomial as $\langle g_1,\dots,g_r\rangle \subset \text{Sym}^*_{\mathbb{Q}} H^2(G/B;\mathbb{Q})$.

**Proof.** This works in roughly the same way as $A_n$-type. By associating to every Borel the unique fixed point $x$ in $\mathbb{CP}^{2n-1}$, there is a fibration $\pi$ from $G/B$ to $\mathbb{CP}^{2n-1}$. When $n$ equals $1$, so that actually we are in type $C_1=A_1$, this complex projective space is the flag variety and the result holds by the argument for $A_n$-type.

Thus, by way of induction, assume that $n>1$, and assume that the result has been proved for $C_{n-1}$-type. In particular, the fibers $F_x$ of $\pi$ are flag varieties of type $C_{n-1}$. Thus, $H^*(F_x;R)$ is $2$-generated. Moreover, the sum of all Betti numbers for $G/B$ equals the order of the Weyl group (elements of the Weyl group index the Bruhat cells). Since the Weyl group in $C_n$-type is a semidirect product of the symmetric group $\mathfrak{S}_n$ and a free $\mathbb{Z}/2\mathbb{Z}$-module of rank $n$, it follows that the fraction of orders of the Weyl group for $C_n$-type and for $C_{n-1}$-type equals $2n$. This precisely equals the sum of all Betti numbers for $\mathbb{CP}^{2n-1}$ (the Euler characteristic of $\mathbb{CP}^{2n-1}$ equals $2n$). Thus, by the Leray Lemma, $q_{G/B;R}$ is surjective. By induction on $n$, for every flag variety of $C_n$-type, $q_{G/B;R}$ is surjective. **QED**

I have not checked whether the fundamental invariants generate the ideal $\text{Ker}(q_{G/B;\mathbb{Z}})$.

**$B_n$ Type Result.** For every $n\geq 2$, in $B_n$-type the homomorphism $q_{G/B;R}$ is surjective provided that $2$ is invertible in $R$.

**Proof.** This works in a similar way to type $C_n$. The fibration is now over the quadric hypersurface $Q$ of dimension $2n-1$ and degree $2$ in $\mathbb{CP}^{2n}$ parameterizing isotropic points for the symmetric, nondegenerate bilinear form on $\mathbb{C}^{2n+1}$. For $n=1$, we have $B_1=A_1$, and the result holds by the argument for $A_n$-type.

Thus, by way of induction, assume that $n>1$, and assume that the result has been proved for $B_{n-1}$-type. By the Lefschetz Hyperplane Theorem, again $Q$ has positive Betti numbers only in even degree, and the sum of the Betti numbers of $Q$ equals $2n$. However, the cohomology ring is not generated integrally by $c_1(\mathcal{O}(1))$. It is generated by $c_1(\mathcal{O}(1))$ over any ring $R$ in which the integer $2$ is invertible. Thus, assume that $2$ is invertible in $R$.

As in the $C_n$-type, the integer $2n$ equals the fraction between the orders of the Weyl group for $C_n$-type and for $C_{n-1}$-type. Thus, as in the $C_n$-type, the Leray Lemma implies surjectivity of $q_{G/B;R}$. Therefore, by induction on $n$, for every group of $B_n$-type and for every ring $R$ in which $2$ is invertible, the homomorphism $q_{G/B;R}$ is surjective. **QED**

Strangely, I seem to get a different conclusion than Allen Knutson. The simply connected group of $B_2$-type is isomorphic to the simply connected group of $C_2$-type. Thus, using the $C_2$-type argument above, I conclude that $q_{G/B;\mathbb{Z}}$ is surjective in $B_2$-type. It could well be that $q_{G/B;\mathbb{Z}}$ fails to be surjective in $B_n$-type beginning with $n\geq 3$, i.e., orthogonal groups for odd-dimensional inner product spaces of dimension $\geq 7$.

I have not checked whether the fundamental invariants generate the ideal $\text{Ker}(q_{G/B;\mathbb{Z}})$.

**$D_n$ Type.** The previous induction argument breaks down. For the corresponding quadric hypersurface of dimension $2n-2$ in $\mathbb{CP}^{2n-1}$, even the rational cohomology ring is not generated in degree $2$; there is an extra primitive cohomology class in the middle degree. However, there is a way to reduce the result for $D_n$-type to the result for $B_n$-type.

Let $(V,\beta:V\xrightarrow{\cong} V^\vee)$ be a symmetric bilinear pairing on a complex vector space $V$ of dimension $2n+1$, so that the corresponding group $G'=\text{Spin}(V,\beta)$ is a simply connected, simple complex Lie group of type $B_n$. The set of nonzero isotropic lines in $V$ form the points of a smooth quadric hypersurface $Q$ in $\mathbb{P}(V)\cong \mathbb{CP}^{2n}$. Inside the dual projective space $\mathbb{P}(V^\vee)$ parameterizing hyperplanes $H=\mathbb{P}(W_{2n})$ in $\mathbb{P}(V)$, there is a dual quadric $Q^\vee$ parameterizing tangent hyperplanes. This is a smooth quadric hypersurface. There is a degree $2$ branched cover $\overline{X}\to \mathbb{P}(V^\vee)$ branched over $Q^\vee$. The open complement $X$ of the branch locus is the universal cover of $\mathbb{P}(V^\vee)\setminus Q^\vee$, whose fundamental group is cyclic of order $2$. The cohomology of $X$ is one of the basic invariants that arises in the theory of vanishing cycles. Either by reviewing that theory (in SGA 7_2, Exposé XII, Table 7, p. 81) or by using the Gysin sequence for the pair $(\overline{X},Q^\vee)$, it follows that $H^*(X;R)$ is a free $R$-module of rank $2$ with one generator in degree $0$, $H^0(X;R)=R$, and with one generator in degree $2n$, $H^{2n}(X;R) \cong R$.

Denote by $F_\beta(1,\dots,n;V)$ the generalized flag variety parameterizing $\beta$-isotropic flags in $V$, say $(U_1\subset \dots \subset U_n\subset V)$. The natural action of $G'$ on $F_\beta(1,\dots,n;V)$ is transitive, and the stabilizers of points are Borel subgroups $B'$ of $G'$. Thus, $G'/B'$ is isomorphic to $F_\beta(1,\dots,n;V)$.

Now define $E\subset F_\beta(1,\dots,n;V)\times (\mathbb{P}(V)^\vee \setminus Q^\vee)$ to be the incidence correspondence parameterizing flags $(U_1\subset \dots \subset U_n \subset W_{2n} \subset V)$ such that $(U_1\subset \dots \subset U_n\subset V)$ is an isotropic flag and such that $\mathbb{P}(W_{2n})\subset \mathbb{P}(V)$ is a hyperplane whose intersection with $Q$ is smooth and contains $\mathbb{P}(U_n)$. There is a forgetful map, $$\rho:E\to F_\beta(1,\dots,n;V).$$ By extending a basis for $U_n$ first to a basis for the orthogonal complement $U_n^\perp$, and then to all of $V$, there exists a system of linear coordinates $(x_1,\dots,x_n,y,z_1,\dots,z_n)$ for $V$ such that $U_{n-\ell}$ equals $\text{Zero}(y,z_1,\dots,z_n,x_n,\dots,x_{n+1-\ell})$, such that $U_n^\perp$ equals $\text{Zero}(z_1,\dots,z_n)$, and such that the quadratic form of $\beta$ equals $x_1z_1+\dots + x_nz_n + y^2$. From this, it is straightforward to compute that $\rho$ is an affine space bundle of relative dimension $n$. In particular, the pullback map on cohomology is an isomorphism, $$H^*(\rho;R):H^*(G'/B';R)\xrightarrow H^*(E;R).$$

Now consider the other projection, to $\mathbb{P}(V)^\vee \setminus Q^\vee$. Since $E$ is simply connected, this morphism factors through the universal cover $X$, $$\pi:E\to X.$$ For each $[W]\in \mathbb{P}(V^\vee)\setminus Q^\vee$, denote by $\beta_W$ the restriction of $\beta$ to $W$. Since $[W]$ is in the complement of $Q^\vee$, this restriction is nondegenerate. The isotropic Grassmannian $F_{\beta_W}(n;W)$ has two connected components, and the fiber over $[W]$ in $X$ is naturally bijective to the set of connected components. Of course $U_n$ is one of these isotropic subspaces, thus it is contained in one of the connected components, and this induces the map $\pi$. For a fixed choice of $[W]$ and connected component, the corresponding fiber $F_W$ of $\pi$ is the generalized flag variety $G/B$ for the simply connected, complex Lie group $G=\text{Spin}(W,\beta_W)$ of type $D_n$.

**$D_n$ Type Result.** For every $n\geq 4$, in $D_n$-type the homomorphism $q_{G/B;R}$ is surjective provided that $2$ is invertible in $R$.

**Proof.** The $R$-module $H^*(G'/B';R)$ is a free $R$-module of rank equal to the size of the Weyl group $|W'|$. Thus, the same holds for $H^*(E;R)$. The base $X$ of $\pi$ is a connected, simply connected manifold with $H^*(X;R)$ a free $R$-module of rank $2$. Finally, the fibers $F_W$ of $\pi$ have cohomology $H^*(F_W;R) = H^*(G/B;R)$ a free $R$-module of rank equal to the size of the Weyl group $|W|$. Finally, $|W'|$ equals $2|W|$ for groups $G'$ and $G$ of respective types $B_n$ and $D_n$. Thus, by the Leray Lemma, the spectral sequence associated to $\pi$ degenerates. If $2$ is invertible in $R$, then $H^*(G'/B';R)$ is $2$-generated by the $B_n$-type result. Thus, by the Leray Lemma, also $H^*(F_W;R)$ is $2$-generated, i.e., $H^*(G/B;R)$ is $2$-generated for $G$ of type $D_n$. **QED**

It is interesting to examine this spectral sequence in relation to the degrees of the fundamental invariants of type $B_n$ and $D_n$. There is an invariant of degree $n$ in type $D_n$ that makes the cohomology in cohomological degree $2n$ of rank $1$ less than the corresponding cohomology in $B_n$-type. Of course, by the spectral sequence, the nonzero generator of $H^{2n}(X;R)$ explains the discrepancy.

Intersection theory). $\endgroup$ – Jason Starr Mar 7 at 15:34