Frobenius elements

Question

Can someone please provide me a clear and basic definition of the concept of Frobenius elements,denoted by Frob_p?

Answer 1

The following is standard in any algebraic number theory textbook.

Frobenius elements are first defined for finite fields. Let $k$ be a finite field with $q$ elements (necessarily $q$ is a power of a prime number), and let $l$ be a finite extension of $k$. Then $l/k$ is automatically Galois with cyclic Galois group. There is a unique element $\sigma$ of $\operatorname{Gal}(l/k)$, called the Frobenius element, with the property that

$$\sigma.x = x^q$$

for all $x \in l$. Of course, $x^q = x$ for all $x \in k$ by Lagrange's theorem.

Now let $L/K$ be a finite Galois extension of number fields. Let $A$ and $B$ be the ring of integers of $K$ and $L$ respectively. Let $\mathfrak P$ be a maximal ideal of $B$. Then $\mathfrak p = \mathfrak P \cap A$ is also a maximal ideal of $A$. The inclusion $A \subset B$ induces an inclusion of fields

$$k := A/\mathfrak p \subset l := B/\mathfrak P.$$

This is a finite extension. The decomposition group $\operatorname{Gal}(L/K)_{\mathfrak P}$ is the subgroup of $\sigma \in \operatorname{Gal}(L/K)$ such that $\sigma(\mathfrak P) = \mathfrak P$. Each $\sigma \in \operatorname{Gal}(L/K)_{\mathfrak P}$ induces an element $\overline{\sigma} \in \operatorname{Gal}(l/k)$ by the formula

$$\overline{\sigma}(x+\mathfrak P) = \sigma(x) + \mathfrak P$$.

The map $\sigma \mapsto \overline{\sigma}$ is a surjective homomorphism $\operatorname{Gal}(L/K)_{\mathfrak P} \rightarrow \operatorname{Gal}(l/k)$. The kernel $K_{\mathfrak P}$ of this homomorphism is called the inertia group.

A Frobenius element for $\mathfrak P$ is defined to be any $\sigma \in \operatorname{Gal}(L/K)_{\mathfrak P}$ whose image $\overline{\sigma}$ in $\operatorname{Gal}(l/k)$ is the Frobenius element in the above sense.

Remarks:

• A Frobenius element $\sigma$ in $\operatorname{Gal}(L/K)_{\mathfrak P}$ is not generally unique, as every other Frobenius element is equal to $\sigma \tau$ for some $\tau \in K_{\mathfrak P}$. However, most of the time, it is unique. For all but finitely maximal ideals $\mathfrak P$ of $B$, the inertia group $K_{\mathfrak P}$ is the trivial group.

• If $\mathfrak p$ is fixed, and $\mathfrak P'$ is another maximal ideal of $B$ such that $\mathfrak P' \cap A = \mathfrak p$, a Frobenius element of $\operatorname{Gal}(L/K)_{\mathfrak P'}$ is generally not the same as a Frobeinus element of $\operatorname{Gal}(L/K)_{\mathfrak P}$, and in fact these subgroups of $\operatorname{Gal}(L/K)$ are probably not equal. However, Frobenius elements are always conjugate, which is to say there exists a $\phi \in \operatorname{Gal}(L/K)$ such that $\phi(\mathfrak P) = \mathfrak P'$ (and consequently $\phi \operatorname{Gal}(L/K)_{\mathfrak P} \phi^{-1} = \operatorname{Gal}(L/K)_{\mathfrak P'}$), such that $\sigma \in \operatorname{Gal}(L/K)_{\mathfrak P}$ is a Frobenius element for $\mathfrak P$ if and only if $\phi \sigma \phi^{-1}$ is one for $\operatorname{Gal}(L/K)_{\mathfrak P'}$.

• If $\operatorname{Gal}(L/K)$ is abelian, then conjugation does nothing, the decomposition groups for $\mathfrak P$ and $\mathfrak P'$ are equal, and it does make sense to talk about "the Frobenius element of $\mathfrak p$," rather than of $\mathfrak P$ or $\mathfrak P'$. But generally, "the Frobenius element of $\mathfrak p$" is only a conjugacy class in $\operatorname{Gal}(L/K)$.

• If $\pi: \operatorname{Gal}(L/K) \rightarrow \operatorname{GL}_n(\mathbb C)$ is a representation, and $\mathfrak p$ is a maximal ideal of $A$ which is unramified in $B$ (that is, the inertia group $K_{\mathfrak P}$ is trivial for some, or equivalently every, maximal ideal $\mathfrak P$ of $B$ such that $\mathfrak P \cap A = \mathfrak p$), then $$\operatorname{det}(\pi(\operatorname{Frob}_{\mathfrak p}))$$ is a well defined complex number, where $\operatorname{Frob}_\mathfrak p$ is the Frobenius element of any maximal ideal $\mathfrak P$ of $B$ such that $\mathfrak P \cap A = \mathfrak p$, even though $\operatorname{Frob}_{\mathfrak p}$ is only defined up to conjugacy. This is because of the property that $\operatorname{det}(PAP^{-1}) = \operatorname{det}(A)$ for any invertible matrices $P$ and $A$. This property is important essential to how Artin L-functions are defined.