Functor of points of the cone over the Grassmanian

Question

$\newcommand{\Gr}{\operatorname{Gr}}$Let $\Gr(d,n)$ be the Grassmanian (I am assuming the field is $\mathbb{C}$). Let us denote by $\widehat{\Gr(d,n)}$ the cone of the Grassmanian under the plucker embedding. As a variety I know the $\mathbb{C}$ valued points correspond to decomposable tensors in $\bigwedge^d \mathbb{C}^n$. I am trying to understand the $R$-valued points of $\widehat{\Gr(d,n)}$ where $R$ is an algebra over $\mathbb{C}$. The decomposable elements in $\bigwedge^d R^n$ definitely lies in $\widehat{\Gr(d,n)}(R)$, but it is not clear to me if every element in $\widehat{\Gr(d,n)}(R)$ is a decomposable tensor in $\bigwedge^d R^n$. The case I am primarily interested in are the algebras $R_m := \frac{\mathbb{C}[t]}{t^{m+1}}$. In these cases we can represent any element $\alpha \in \widehat{G(d,n)}(R_m)$ in the form $\alpha_0 + t \alpha_1 + \cdots + t^m \alpha_m$ where $\alpha_0 \in \widehat{\Gr(d,n)}(\mathbb{C})$. If $\alpha_0 \neq 0$, then it seems likely that $\alpha$ is a decomposable tensor. I do not know what happens when $\alpha_0 = 0$. Any help would be much appreciated.

Answer 1

The tangent space to $\widehat{\mathrm{Gr}(d,n)}$ at the vertex is the whole space ${\bigwedge}^d(\mathbb{C}^n)$, hence any tangent vector at the origin gives an $R_1$-point of $\widehat{\mathrm{Gr}(d,n)}$ with $\alpha_0 = 0$. And for a general tangent vector the coresponding $\alpha$ is not decomposable.