2
$\begingroup$

Let $f_1,\ldots,f_n, g_1,\ldots, g_n$ be polynomials in $\mathbb{Q}[X_1,\ldots,X_n]$ such that if $g=(g_1,\ldots,g_n)$ then $f_i(g(x_1,\ldots,x_n))=x_i$ for all $i=1,\ldots,n$. Does it follow that $f$ and $g$ have degree 1?

$\endgroup$
2
$\begingroup$

For instance, triangular polynomial automorphisms, that is, polynomial maps of the form $$f_i(x_1,\dots x_n)=a_i x_i + \varphi_i(x_{i+1},\dots,x_n)\, ,$$ with $a_i\neq 0$, $1\le i \le n\, ,$ are a group under composition. Note that these maps have the last component $f_n$ linear, but this property is easily lost after composing with a linear (non triangular) map.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.