Consider the surface in $\mathbb{C}^{d+1}$ parametrized by the map
$$\theta:\mathbb{C}^2\rightarrow\mathbb{C}^{d+1}$$ defined by $(s,t)\rightarrow(s^d,s^{d-1}t,\dots,st^{d-1},t^d)$
And $J_d=\langle x_ix_{j+1}-x_{i+1}x_{j}:0\le i<j\le d-1\rangle\subseteq \mathbb{C}[x_0,\dots,x_d]$
Also we know $\widehat{C}_d=\theta(\mathbb{C}^2)$. Then I want to show that $\theta(\mathbb{C}^2)=V(J_d)$.
One direction is clear. Let take $P_i\in \theta(\mathbb{C}^2)$ and suppose that $P_i=s^{d-i}t^i$ for all $i\in\{0,d\} \mbox{ and } s,t\in\mathbb{C}$. So it is clear that $P_i\in V(J_d).$
But i think the other direction is not easy.
We know that $V(J_d)=\left\{P\in\mathbb{C}^{d+1}:f(P)=0 \mbox{ for all } f \in J_d\right\}$. Let $P\in V(J_d)$. Suppose that $P=(P_0,\dots,P_d)\in\mathbb{C}^{d+1}$. So we get $f(P)=P_iP_{j+1}-P_{i+1}P_j=0$ for all $f\in J_d$ and $0\le i< j\le d-1$.
Then it is clear that $P_iP_{j+1}=P_{i+1}P_j$.
Now, how can I show that $P\in \theta(\mathbb{C}^2)$? Because I cant see the transition between $\theta(\mathbb{C}^2)$ and point P.
I need to some idea for this direction.
Because $P_0,P_1$ are complex numbers, there is $s,t\in \mathbb{C}$ such that $P_0=s^d$ and $P_1=s^{d-1}t$. Now use the recurence relations $P_iP_{j+1}=P_{i+1}P_j$ to get $P_k=t^ks^{d-k}$.