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?
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asked Dec 18, 2011 at 15:10
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$\begingroup$ @ Mahdi Majidi-Zolbanin - Of course, $n>1$. $\endgroup$– Boris NovikovCommented Dec 18, 2011 at 15:39
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$\begingroup$ @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 $n>1$. $\endgroup$– Mahdi Majidi-ZolbaninCommented Dec 18, 2011 at 15:42
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2 Answers
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Of course not. For example, consider the automorphism of $\mathbb Q[x, y]$ given by $(x, y) \mapsto (x+y^2, y + (x+y^2)^2)$.
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3$\begingroup$ Is it really necessary to start an answer with an "of course"? $\endgroup$ Commented Dec 18, 2011 at 17:40
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2$\begingroup$ Of course it is not necessary; but when you start playing around with automorphisms of polynomial rings you discover these examples very quickly. $\endgroup$– AngeloCommented Dec 18, 2011 at 17:49
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$\begingroup$ @ Angelo - in your example $g_2=y$? $\endgroup$ Commented Dec 19, 2011 at 0:11
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$\begingroup$ @Boris: no. The inverse map is given by $(x-(y-x^2)^2,y-x^2)$. $\endgroup$ Commented Dec 19, 2011 at 11:37
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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.]
answered Dec 18, 2011 at 16:01
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$\begingroup$ Is it true then, that every $f_i$ and every $g_i$ must have a degree $1$ term? $\endgroup$ Commented Dec 18, 2011 at 16:12
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1$\begingroup$ @Mahdi: yes on both counts. Easy proof by contradiction and combinatorics -- just look at the (total) degrees of each term in the result. $\endgroup$ Commented Dec 18, 2011 at 16:29
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1$\begingroup$ 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. $\endgroup$– AngeloCommented Dec 18, 2011 at 16:39
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$\begingroup$ @ Jacques Carette - Your answer is too laconic for me $\endgroup$ Commented Dec 19, 2011 at 0:22
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$\begingroup$ @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. $\endgroup$ Commented Dec 19, 2011 at 1:14