It is an elementary exercise to show that for irrational $\alpha \in \mathbb{R}$, the sequence $\{ \alpha n \mod 1 \}_{n \in \mathbb{N}}$ is dense in $T := \mathbb{R}/\mathbb{Z}$. With more work, it can be shown that the sequence is in fact equidistributed.
This result has two natural generalizations.
The first of them concerns the higher dimensions: suppose that $1,\alpha_1,\alpha_2,\dots,\alpha_d$ are linearly independent over $\mathbb{Q}$ (in other words, no nontrivial combination of $\alpha_i$ with rational coefficients is a rational number). Then the sequence $\{ (\alpha_i n)_{i=1}^d \mod 1 \}_{n \in \mathbb{N}}$ is equidistributed in $T^d$.
The second concerns more general functions; this result is due to Weyl, I think. Suppose that $p(n) = \sum_{k=1}^r a_k n^k$ is a polynomial with $0$ constant term and at least one irrational coefficient. Then, the sequence $\{ p(n) \mod 1 \}_{n \in \mathbb{N}}$ is equidistributed in $T$.
The following common generalisation comes to mind: Suppose that $p_1(n), p_2(n), \dots, p_d(n)$ are polynomials with $p_i(0) = 0$ such that any linear combination of $p_i$ with rational coefficients is a polynomial with at least one irrational coefficients not counting the constant term. (Equivalently, the polynomials $p_1(n), p_2(n), \dots, p_d(n), n,n^2,\dots$ are linearly independent over $\mathbb{Q}$). Then, the sequence $\{ (p_i(n))_{i=1}^d \mod 1 \}_{n \in \mathbb{N}}$ is equidistributed in $T^d$.
Question: Is this more general result true? Also, is it known?