I try to solve the following problem:
For a polynomial algebra $\mathbb{K}^{[n]}=\mathbb{K}[x_1,\ldots , x_n]$ and its locally nilpotent derivation $D$, there exists an extension of $D$ to a triangular derivation $T$ on some larger polynomial algebra $\mathbb{K}^{[N]}$, together with an integral ideal $I\subset\mathbb{K}^{[N]}$, such that $B/I=\mathbb{K}^{[n]}$ and $T/I=D$.
Is this problem unsolved? I think there might be a trivial proof of it. My ideas are the following:
Since $D$ is locally nilpotent, we have $k_i=\nu_D(x_i)<\infty$, thus it is possible to construct an algebra $\mathbb{K}^{[N]}=\mathbb{K}[\{ X_{i1},X_{i2},\ldots, X_{ik_i} \}_{1\leq i\leq n}]$ and a triangluar derivation $T$ with $T(X_{ij}) = X_{i(j+1)}$, $T(X_{ik_i}) = 0$. Consider the homomorphism $\pi : \mathbb{K}^{[N]}\to\mathbb{K}^{[n]}$, where $\pi(X_{ij}) = D^{j-1}(x_i)$. So $\ker\pi=I$ is an integral ideal of $\mathbb{K}^{[N]}$ with $\mathbb{K}^{[N]}/I = \mathbb{K}^{[n]}$ and $T/I = D$.
Is this proof correct?