# Composition of polynomial functions which gives the identity

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?

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No! e.g. $f_1(x,y)=x+y^{100}$, $f_2(x,y)=y$. You can reconstruct $x$ and $y$ as polynomials in $f_1$ and $f_2$. – Kevin Buzzard Dec 17 '11 at 22:01
The general theory of automorphisms of polynomial rings is quite subtle. See ams.org/journals/jams/2004-17-01/S0894-0347-03-00440-5/…. – Richard Stanley Dec 18 '11 at 1:31

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.