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

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.

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 $$\mathbb{Z/nZ}$$ !)

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