Let $ k$ be a field of characteristic zero. Given integers $ 2 \leq s \leq r < n$ , define the variety $ X_n$ in $ P_K^n$ with coordinates $ y_0, \cdots, y_n$ and $ K=k(u_0, \cdots, u_n)$ ( where $ u_i$ ‘s are independent transcendental vairables) by the folloing $ n-r$ equatuions:

$ $ f_{i-r}(y_0, y_1, \cdots, y_n)= \begin{vmatrix} 1 & 1 & \cdots & 1 & 1 \ u_0& u_1& \cdots & u_r & u_i\ \vdots & \vdots & \cdots & \vdots & \vdots \ u_0^r & u_1^r & \cdots & u_r^r& u_i^r\ y_0^s & y_1^s & \cdots & y_r^s & y_i^s \end{vmatrix} =0, \ (r+1 \leq i \leq n). $ $

It is easy to see that for fixed $ r$ , $ s$ , and $ n > (sr + 1)/(s−1)$ , $ X_n$ is a smooth complete intersection variety of general type with $ \dim(X_n)=r$ . For a set of pairwise distinct elements $ b=\{b_0, b_1, \cdots, b_n\}$ in $ k$ , letting $ u_i=b_i$ , one can get $ X_{b, n}$ a smooth variety defined over $ k$ .

Question: Is it possible to show that $ X_{b, n}$ is a Geometric Mordellic variety, i.e., $ X_{b, n}$ does not contain subvarieties which are not of general type over $ \bar{k}$ the algebraic closure of $ k$ ?