The following question was part of my abstract algebra assignment and I was clueless on how to solve the problem.
So, I am posting it here.
Let $F= \mathbb F_3 [X] / (x^3 + 2x -1)$, where $\mathbb F_3$ is the field with 3 elements. Then find the automorphism group of $F$.
I had no idea on how this question should be approached
So, I posted it here and it was linked with another question:
This question has a answer here: Automorphism group of $F=\mathbb F_3[x]/(x^3+2x-1)$.
But I have a question I answer of Shivani Goel and she has been unactive for quite few days and I urgently need to figure out my question in her answer.
My question is I know why $x \to x $, $x \to x^3 $ and $x\to x^9$ are automorphisms but why $x \to x^{27} $ and $x\to x^{81}$ and so higher powers of 3 are not automorphisms ?
Kindly give reason for that.
Best Answer
The general theory goes as follows. Let $p$ be a prime number. Then for all powers $q=p^r$ there is only one field $\mathbb{F}_q$ with $q$ elements up to isomorphism. The automorphism group $$ {\rm Aut}(\mathbb{F}_q)={\rm Gal}(\mathbb{F}_q/\mathbb{F}_p) $$ is cyclic of order $r$, generated by the Frobenius automorphism $$ {\rm Frob}_p:\mathbb{F}_q\longrightarrow\mathbb{F}_q,\qquad {\rm Frob}_p(x)=x^p. $$ The fact that ${\rm Frob}_p^r={\rm id}$ is very consistent with the fact that $\mathbb{F}_q^\times$ is a group, in fact cyclic, of order $q-1$.
In the question's case $p=3$ and $q=3^3=27$ since the given polynomial is irreducible. Thus $x\to x^{27}$ is simply the identity map.