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( with $\mathbb{Z}_n$ I mean $\frac{ \mathbb{Z} }{ n\mathbb{Z} }$ )

Theorem :

Let $\psi,\phi \in \mathbb{K}[x]$ be such that

$$\psi(\phi(x)) = x = \phi(\psi(x)) \; \forall x \in \mathbb{K}$$

Then there exists $m,q \in \mathbb{K}$

$$\psi(x) = mx + q \; \forall x \in \mathbb{K}$$

A way to prove this would be this one :

Let $\psi(x) = a_0 + a_1 + \dots + a_nx^n$ and $\phi(x) = b_0 + b_1x + \dots + b_mx^m$ with $a_n \neq 0 \neq b_m$. Suppose by contradiction that $n \geq 2$, then $n^m \geq 2$

$$x = \phi(\psi(x)) = (\text{part of degree } \leq n^m - 1 ) + b_m(a_n)^m x^{n^m}$$

But this is a contradiction because if two polynomials are equal all the coefficients need to be equal so $b_m(a_n)^m = 0$ but I know that $b_m(a_n)^m \neq 0$.

The problem with this proof is that It works only on infinite fields $\mathbb{K}$

Take for example $\mathbb{Z}_3$, then the polynomials

$$\psi(x) = x^3$$ $$\phi(x) = x$$

Are inverses of each other, but $\psi$ is of degree $3$, In fact I have that

$$x = x^3 \; \forall x \in \mathbb{Z}_3$$

but clearly the coefficients aren't equal.

My question is : how can I prove the theorem?

Paul
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    It does seem to be true in $\mathbb Z_2$ and $\mathbb Z_3$. But in $\mathbb Z_5$ we have $x^3$, which is its own inverse, as a counter-example. For more, see permutation polynomials. – mr_e_man Nov 08 '22 at 23:26
  • The first proof works for all fields $\Bbb{K}$. The correct test (=the definition) says that polynomials are equal if and only if all their coefficients are equal. It is NOT about whether the two polynomials have the same value for all inputs from $\Bbb{K}$. – Jyrki Lahtonen Nov 09 '22 at 04:57
  • Look up this old thread for a longer explanation. – Jyrki Lahtonen Nov 09 '22 at 05:00
  • @JyrkiLahtonen the prof doesn't work for finite fields, when I say $\psi(\psi^{-1}(x)) = x$ I mean that they are equal as functions! Not as polynomials! So what what you're saying answers a different question than the one I asked – Paul Nov 09 '22 at 15:14
  • Ok. So you work with polynomial functions instead of polynomials. – Jyrki Lahtonen Nov 09 '22 at 15:37
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    But then the claim is obviously false. If $\Bbb{K}$ is a finite field with $q$ elements then every function from $\Bbb{K}$ to itself is a polynomial (by Lagrange interpolation formula). In particular, there are $q!$ bijective polynomial functions, all with an inverse that is also a polynomial function. But there are only $q(q-1)$ linear polynomials, so the rest of the invertible functions come from higher degree polynomials. Observe that $q(q-1)<q!$ when $q>3$. – Jyrki Lahtonen Nov 09 '22 at 15:41

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