TITLE: When are generalized Severi-Brauer varieties Grassmannians?
QUESTION [7 upvotes]: Let $F$ be a field and $A$ be an $F$-central simple algebra of degree $n$. Let $0< k< n$ and let $SB_k(A)$ denote the generalized Severi-Brauer variety: if $E/F$ is a field extension, $SB_k(A)(E)$ consists of the right ideals of dimension $kn$ of $A_E=A\otimes_F E$.
If $A$ is split, i.e. $A\simeq M_n(F)$, then $SB_k(A)=Gr(k,n)$, the Grassmannian.

Is the converse true? If not, can you provide a counterexample?

The result is true if $k=1$, since $SB_k(A)$ has a rational point over $F$ iff the index of $A$ divides $k$, and Grassmannians have rational points over $F$.
This is cross-posted from Math.SE, where I posted it some days ago, with a few upvotes but no answers or comments. I hope it is ok for MO.
https://math.stackexchange.com/questions/618616/when-are-generalized-severi-brauer-varieties-trivial

REPLY [4 votes]: Daniel Litt's answer is completely correct, of course.  You can also see the result using projective geometry (perhaps this is how Chow proves the result).  For every $1$-dimensional subspace $L$ of $F^{\oplus n}$, there is an induced embedding, $$\phi_L:\text{Grass}(k-1,F^{\oplus n}/L) \hookrightarrow \text{Grass}(k,F^{\oplus n}),$$ obtained by associating to every subspace $\overline{V}$ of $F^{\oplus n}/L$ the inverse image $V$ in $F^{\oplus n}$.  Denote by $P(t)$ the Hilbert polynomial of $\text{Image}(\phi_L)$ with respect to the ample invertible sheaf $\omega^\vee$, where $\omega$ is the dualizing sheaf of $\text{Grass}(k,F^{\oplus n})$.  Then there is a well-defined morphism, $$\Phi:\text{Grass}(1,F^{\oplus n}) \to \text{Hilb}^{P(t)}_{\text{Grass}(k,F^{\oplus n})}, \ \ [L]\mapsto [\text{Image}(\phi_L)].$$
I claim that $\Phi$ gives an isomorphism of the domain with one of the connected components of the target (when $n\neq 2k$, there is only one connected component, so that $\Phi$ is an $F$-isomorphism).  Moreover, it is equivariant for the natural actions of $\textbf{PGL}_n$.  Therefore, for every central simple algebra $A$ of "degree" $n$, there is an induced isomorphism of $F$-schemes,
$$ SB_1(A) \to \text{Hilb}^{P(t)}_{SB_k(A)},$$
where again the Hilbert polynomial is with respect to the very ample invertible sheaf $\omega^\vee$ on $SB_k(A)$.  
In particular, if $SB_k(A)$ is isomorphic to $\text{Grass}(k,F^{\oplus n})$, then every connected component of $\text{Hilb}^{P(t)}_{SB_k(A)}$ has an $F$-rational point.  Thus, also $SB_1(A)$ has an $F$-rational point.  That implies that $A$ is split by the result quoted by the OP.  
Of course one still needs to verify that $\Phi$ is an isomorphism.  However, since this is a purely geometric result, I imagine that most of us are happier with this claim than the original problem.  Also it is fairly easy to prove using some "homogeneity" and cohomology vanishing.