TITLE: How to convert a permutation to permutation polynomial?
QUESTION [5 upvotes]: Let Fq be the finite field with q elements, where q is a prime power.
A permutation on Fq is a bijection from Fq to itself.
Let Fq[x] be the ring of polynomials in a single indeterminate x over Fq.
A polynomial f \member Fq[x] is called a permutation polynomial of Fq if it induces a one-to-one map from Fq to itself.
It is well known that every permutation on Fq can be expressed as a
permutation polynomial over Fq.
Problem 1: How to convert a permutation to a permutation polynomial? Please note that Lagrange polynomial as polynomial interpolation is possible. Any other method?
Problem 2: as problem (1) but with Fq and q composite (not a prime).

REPLY [3 votes]: Well, you can just use the fact that $1-(x-a)^{q-1}$ is the indicator function of the point $a$, so the formula $\sum \sigma(a)\cdot(1-(x-a)^{q-1})$ will do.
Generally, constructing permutation polynomials is a complicated problem. There is a chapter on the subject in Lidl, Niederreiter finite fields. There are a lot of papers on the subject, e.g. http://arxiv.org/abs/1303.2229 . The answer, of course,  depends on the general form of resulting formula that you want.
I don't see why any permutation is a polynomial over $\mathbb{Z}/n\mathbb{Z}$. Lets take any permutation that moves $2$ to $0$ and $0$ to $1$ in $\mathbb{Z}/4\mathbb{Z}$. The corresponding polynomial must have its last coefficient both odd and even.