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A prolongation of the question 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 SOME $f_i$ or $g_j$ has degree 1?

edited Dec 18 2011 at 15:14

Adrien
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asked Dec 18 2011 at 15:10

Boris Novikov
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	@ Mahdi Majidi-Zolbanin - Of course, $n>1$. – Boris Novikov Dec 18 2011 at 15:39
	@Boris: Sorry I deleted my comment, because even though the answer to your question is yes for $n=1$, your question still makes sense without saying $n>1$. – Mahdi Majidi-Zolbanin Dec 18 2011 at 15:42

Angelo · Accepted Answer · 2011-12-18 15:38:02Z UTC

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Of course not. For example, consider the automorphism of $\mathbb Q[x, y]$ given by $(x, y) \mapsto (x+y^2, y + (x+y^2)^2)$.

answered Dec 18 2011 at 15:38

Angelo
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Is it really necessary to start an answer with an "of course"? – Ramiro de la Vega Dec 18 2011 at 17:40

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Of course not, is it? – Ramiro de la Vega Dec 18 2011 at 17:41

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Of course it is not necessary; but when you start playing around with automorphisms of polynomial rings you discover these examples very quickly. – Angelo Dec 18 2011 at 17:49

@ Angelo - in your example $g_2=y$? – Boris Novikov Dec 19 2011 at 0:11

@Boris: no. The inverse map is given by $(x-(y-x^2)^2,y-x^2)$. – Vladimir Dotsenko Dec 19 2011 at 11:37

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Jacques Carette · Answer 2 · 2011-12-19 01:12:33Z UTC

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Here is another way to say what Angelo said. Pick any set of polynomials $\left(g_1,\ldots,g_n\right)$ such that it has Groebner basis $\left[x_1,\ldots,x_n\right]$ under lexicographical order $\left[x_n,\ldots,x_1\right]$. Then you can get the $f_j$'s from the cofactors (pdf).

[This is merely a more constructive way of phrasing the answer, so that you may construct more examples for yourself easily, which is not as easy to do given Angelo's answer.]

edited Dec 19 2011 at 1:12

answered Dec 18 2011 at 16:01

Jacques Carette
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Is it true then, that every $f_i$ and every $g_i$ must have a degree $1$ term? – Mahdi Majidi-Zolbanin Dec 18 2011 at 16:12

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@Mahdi: yes on both counts. Easy proof by contradiction and combinatorics -- just look at the (total) degrees of each term in the result. – Jacques Carette Dec 18 2011 at 16:29

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To Mahdi: or, more geometrically, notice that if some $f_i$ had no term of degree, the differential of the polynomial map given by the $f_i$ would would vanish at some point. – Angelo Dec 18 2011 at 16:39

@ Jacques Carette - Your answer is too laconic for me – Boris Novikov Dec 19 2011 at 0:22

@Boris: added links for all the important parts. If you don't already know about Groebner bases, you're better off reading the Wikipedia entry than having me try to expand my answer into a tutorial on the topic. – Jacques Carette Dec 19 2011 at 1:14

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