About n-tuple unimodular

Question

Let ($\mathcal{O}$, $\mathcal{M}$, k) be an DVR and $F_{1},...,F_{n} \in \mathcal{O}[X_{1},...,X_{n}]$ such that detJF = 1 where JF is the matrix $(\frac{\partial F_{i}}{\partial X_{j}})$. Suppose there exists $d_{1},...,d_{n} \in \mathcal{O}$ such that

$$u_{1}F_{1}(d_{1},...,d_{n})+\cdots+u_{n}F_{n}(d_{1},...,d_{n}) = 1$$

for some $u_{1},...,u_{n} \in \mathcal{O}$. In this case, we say that $F = (F_{1},...,F_{n})$ is an $n$-tuple unimodular. Let $G_{1},...,G_{n} \in \mathcal{O}[X_{1},...,X_{n}]$ defined by $G_{j} = F_{j}(F_{1},...,F_{n})$ for all $j$.

Is it $G = (G_{1},...,G_{n})$ be an $n$-tuple unimodular?

Answer 1

An $n$-tuple $F=(F_1,\dots,F_n)$ is "unimodular" iff $\exists x \in k^n, F(x) \neq 0$ in $k$.

Thus an equivalent question is: if $F =(F_1,\dots,F_n) \in k[X_1,\dots,X_n]^n$ satisfies $J(F) = 1$ and $F_{| k^n}$ is not identically $0$, then is $F \circ F_{|k^n}$ necessarily not identically $0$ ?

If $k$ is infinite then this is true since $F \circ F_{|k^n} = 0$ implies $F \circ F = 0$ and thus $JF \times JF\circ F = 0$, which contradicts the invertibility of $JF$.

If $k$ has cardinality $q < \infty$ then this is false. Let $h : k \rightarrow k$ be any non-zero map such that $h \circ h =0$, and let $H \in k[X]$ be any polynomial such that $H_{|k} = h$. Then the polynomial $F(X) = X - X^q + H(X^q)$ satisfies $JF = 1$ and $F_{|k} \neq 0$ while $F \circ F_{|k} =0$. This yields a counter example to your question by taking $\mathcal{O} = k[[T]]$ and $n=1$.