The Vandermonde determinant for real moment-curve vectors

By induction on $n$, every iterated ring $\mathbb{R}[t_1,\dots ,t_n]$ is an integral domain: the base $\mathbb{R}$ is a field, hence a domain, and the successor step is exactly [F1]. In each iteration the new indeterminate has coefficient $1$ in degree $1$, whereas every preceding indeterminate is embedded as a constant polynomial; since $1\mathrm{\ne }0$ in $\mathbb{R}$, all the $t_j$ are pairwise distinct.

For $d=1$, $V_d=[1]$ and the empty product is $1$, so the formula holds. For the inductive step let $d\ge 2$ and suppose the formula holds in size $d-1$. Put $C=\mathbb{R}[t_1,\dots ,t_{d-1}]$ and $T=t_d$, so the recursive construction identifies $\mathbb{R}[t_1,\dots ,t_d]=C[T]$. Regard $D_d:=\det V_d$ as a polynomial $f(T)\in C[T]$.

The Leibniz expansion in [F2] shows that only powers $T^0,\dots ,T^{d-1}$ occur in $f$, and that the coefficient of $T^{d-1}$ is $D_{d-1}:=\det V_{d-1}$: precisely the terms with exponent $d-1$ in the last column are those permutations fixing the last row, and their remaining signed sum is the Leibniz determinant of $V_{d-1}$. For each $j<d$, substituting $T=t_j$ makes the last column equal column $j$, so alternation in [F2] gives $f(t_j)=0$.

The elements $t_1,\dots ,t_{d-1}$ are distinct in the domain $C$ by step 1.1. Starting with $f$, apply [F3] successively at $t_1,\dots ,t_{d-1}$. At stage $k$, if $f=({\prod }_{j<k}(T-t_j))g_k$, evaluation at $t_k$ gives $0=f(t_k)=({\prod }_{j<k}(t_k-t_j))g_k(t_k)\mathrm{.}$ Every factor $t_k-t_j$ is nonzero, so their product is nonzero in the domain $C$; hence $g_k(t_k)=0$, and [F3] gives $T-t_k\mid g_k$. Induction over these finitely many stages yields $f(T)=g(T){\prod }_{j<d}(T-t_j)$ for some $g(T)\in C[T]$.

By the induction hypothesis, $D_{d-1}={\prod }_{1\le p<q<d}(t_q-t_p)$, which is nonzero because each difference is nonzero in the domain $C$. Thus the coefficient of $T^{d-1}$ in $f$ is nonzero. The product ${\prod }_{j<d}(T-t_j)$ is monic of degree $d-1$. If $g$ had any nonzero coefficient at a positive power, its highest such power times this monic product would give a nonzero coefficient of $f$ above degree $d-1$, contradicting step 1.3. Therefore $g$ is constant, and comparison of the $T^{d-1}$ coefficients gives $g=D_{d-1}$. This proves $D_d=D_{d-1}{\prod }_{j<d}(t_d-t_j)={\prod }_{1\le p<q\le d}(t_q-t_p)\mathrm{.}$ By induction the determinant formula holds for every $d\ge 1$.

For distinct real nodes $s_1,\dots ,s_d$, the evaluation homomorphism in [F3] substitutes them into the formula and gives $\det V_d(s_1,\dots ,s_d)={\prod }_{p<q}(s_q-s_p)\mathrm{\ne }0$. By [F4], the square matrix with these moment-curve vectors as columns is invertible, hence has trivial coefficient kernel and linearly independent columns by [F5].