We work over $k=\bar{k}$ and $\operatorname{char}k=0$ for simplicity. It is known that the Grassmannian scheme represents the functor $\operatorname{Sch}/k\rightarrow \operatorname{Set}$ by sending a scheme $S$ to isomorphism classes of surjections $\mathscr{O}_S^n\rightarrow \mathscr{V}$, where $\mathscr{V}$ is a vector bundle of rank $n-d$. In particular, if we $S=\operatorname{Spec}k$, the closed points of $Gr(d,n)$ correspond to $d$-dimensional subspaces of $k^n$. It is also known that this scheme is projective, so there is some some sufficiently nice graded $k$ algebra $A$ such that $Gr(d,n)=\operatorname{Proj}A$, is it known what this ring is?
Taking inspiration from the fact that $\mathbb P(V)=\operatorname{Proj}(\operatorname{Sym}V^*)$, I think that perhaps a suitable candidate could be $A=\operatorname{Sym}(\Lambda^dV^*)$. My reasoning is as follows: if $W\subset V$ with $V\cong k^n$, and $\dim W=d$, we can send it do the following homogenous prime ideal: $$\mathfrak p_W=\left\langle\omega\in \Lambda^dV^*:\omega(W)=0\right\rangle$$ I claim that this is closed. Otherwise there exists an ideal $I$ which does not contain the irrelevant ideal but does contain $\mathfrak p_W\subset I$. In particular, we have that $\mathfrak p_W\cap \Lambda^dV^*\subset I\cap\Lambda^d V^*$, but $\mathfrak p_W\cap \Lambda^d V^*$ is $\dim \Lambda^dV^*-1$ dimensional, so either $I\cap \Lambda^dV^*= \Lambda V^*$ or $I\cap \Lambda^dV^*=\mathfrak p_W\cap \Lambda^dV^*$. Since $\mathfrak p_W$ is generated in degree $1$ we have the claim. This shows there is a map from subspaces of $V$ to closed points of $\operatorname{Proj}A$.
If $\mathfrak p\in \operatorname{Proj}A$ is a closed point, then since $k=\bar k$, Hilbert's Nullstellensatz implies that $\mathfrak p\cap \Lambda^dV^*\neq 0$. In particular, since $\mathfrak p$ is maximal amongst homogenous prime ideals not containing $\Lambda^dV^*$ we now immediately have that $$\dim \mathfrak p\cap \Lambda^dV^*=\dim \Lambda^dV^*-1$$ and so $\mathfrak p\cap \Lambda^dV^*$ cuts out a one dimensional subspace of $\Lambda^dV$. I think we can take this one dimensional subspace to be spanned by $v_1\wedge \cdots \wedge v_d$ (if this is true, how can I show this?), and then $\langle v_1,\dots, v_d\rangle\subset V$ is a $d$-dimensional linear subspace of $W$.
These feel like inverses of one another, but I can't quite show it. Is this the right idea? Or should I try a different route.
Edit: I know that this is related to the Plucker equations, but I am having trouble interpreting what they are doing geometrically in constructing the Grassmannian as a projective scheme. Moreover, with this formalism we are working the dual space so I am also having trouble navigating that part.
