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How does the Frobenius elements play a crucial role in Galois representation?

As far as I have understood that Frobenius endomorphisms specially Frobenius automorphism is a generator of a Galois group. But how ?

Let $F_q=F_{p^{\large e}}$ be a finite field of $q=p^e$ elements and let $F_{p^f}$ be a finite of extension of $F_q$. Now Frobenius mapping is $x \to x^p$. The Frobenius iterates are $x \to x^p \to (x^p)^p=x^{p^2} \to (x^p)^{p^2}=x^{p^3} \to \cdots $

Now I have information that the Galois group of a finite extension is generated by the above iterate of the Frobenius automorphism.

But how?

Please someone nicely relate the Frobenius elements with Galois representation.

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There are two things to note to see why the Frobenius generates the Galois group :

1) Let $\varphi : L\to L$ be the Frobenius on the finite field $L$; if $\varphi^n = id_L$, then $m\mid n$, where $|L| = p^m$

Indeed, $\varphi^n = id_L$ means $x^{p^n}=x$ for all $x\in L$, so this means every element of $L$ is a root of $X^{p^n}-X$, so $L$ is a subfield of $\mathbb{F}_{p^n}$, so $m\mid n$.

Conversely, it is clear that $\varphi^m = id_L$, so that $\langle \varphi \rangle \simeq \mathbb{Z/mZ}$.

2) $Gal(L/\mathbb{F}_p)$ has cardinality at most $m$.

Indeed, remember that $L^\times $ is cyclic, so that there is an element $\theta\in L$ such that $\mathbb{F}_p(\theta)= L$. But if $\pi_\theta$ is the minimal polynomial of $\theta$ over $\mathbb{F}_p$, $\mathbb{F}_p(\theta) \simeq \mathbb{F}_p[X]/(\pi_\theta)$, so $\deg \pi_\theta = m$.

Therefore, $\pi_\theta$ has at most $m$ roots in $L$ (actually, it's exactly $m$ roots and it's not hard to show, but we don't need that), and since an automorphism of $L$ is determined by where it sends $\theta$, and it has to send $\theta$ to a root of $\pi_\theta$, it follows that there are at most $m$ automorphisms of $L$.

Now since $\langle \varphi\rangle \subset Gal(L/\mathbb{F}_p)$ and $|\langle \varphi\rangle| = m, |Gal(L/\mathbb{F}_p)|\leq m$, it follows that $\langle \varphi\rangle = Gal(L/\mathbb{F}_p)$ : the Frobenius generates the Galois group ! In particular $Gal(L/\mathbb{F}_p) \simeq \mathbb{Z/mZ}$, so the representations of this group are very easy to understand.

Moreover, it's easy to go from that computation to the fact that $Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p) \simeq \widehat{\mathbb{Z}}$, the profinite integers, so you have some information about the representations of this group if you know $\widehat{\mathbb{Z}}$ (especially when you're looking at actions on finite groups, which factor through some $\mathbb{Z/nZ}$ !)

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  • $\begingroup$ Thank you for nice explanation $\endgroup$ – M. A. SARKAR Feb 12 at 8:28

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