I am trying to solve a problem about bijections on finite fields, but i am stuck in this problem and don't see how can i solve it. So, here it is:

Given is a polynomial $g(y):=y^{3}+3y^{2}+3y+3\in \mathbb{Z}[y]$. Show that it defines a bijection from $\mathbb{F}_{11^{31415}}$ to itself.

To show the bijection $\mathbb{F}_{11^{31415}}\rightarrow \mathbb{F}_{11^{31415}}$, i've been thinking that i have to find an inverse map, but i don't know how to do it. I think this should be better than proving that the map is injective and surjective. I also noticed that this polynomial is irreducible over $\mathbb{Z}$ and that the number $31415=5\times 61\times 103$, but i don't know what it brings.

Can anybody, please, help me with this problem? I will be glad to read your ideas and remarks. Thanks in advance!


Observe that $g(y)=(y+1)^3+2$, so its inverse is something like $g^{-1}(x)=\sqrt[3]{x-2}\,-1$.

The only problematic part is the cubic root, so all you shoud show is that $x\mapsto x^3$ is a bijection in the given field. Since that is finite, its multiplicative group is cyclic, of order $11^{31415}-1$.

Since we have $11\equiv -1 \pmod3$, $\ 11^{31415}\equiv -1\pmod{3}$, so that the group order is coprime to $3$, and therefore $x\mapsto x^3$ is indeed a bijection:

By the Bezout identity, there are integers $a,b$ such that $a\cdot N+b\cdot 3=1$ where $N$ denotes the order of the group now. So that $x\mapsto x^b$ will be an inverse for $x\mapsto x^3$.

  • $\begingroup$ Thank you for your answer! Only the last part of your proof is not clear for me...computation is okay, but how does the bijection follow from the group order coprime to 3? $\endgroup$ – Lullaby Oct 17 '13 at 21:48
  • $\begingroup$ Before the edit, i've beet thinking that you're using this theorem: groupprops.subwiki.org/wiki/… $\endgroup$ – Lullaby Oct 17 '13 at 21:54
  • $\begingroup$ Yes, it is just that one. $\endgroup$ – Berci Oct 17 '13 at 22:09

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